Electrochemistry

When an electrode potential (which is a potential difference that develops between the electrode and electrolyte), has values of the concentrations of all the species involved in a half-cell equal to unity, it is known as standard electrode potential.

The standard hydrogen electrode which is assigned a zero potential for all temperatures to measure the potential of the given half-cell. The electrode is represented as Pt(s)|H₂(g)|H⁺(aq) and the equation is given as H⁺(aq) + e^-→ 1/2H₂(g).

This leads to the conclusion that the pressure of hydrogen gas is 1 bar and the concentration of hydrogen ions in solution is 1 M. We place the SHE on the left and copper electrode on the right, which will help us find the standard reduction potential of copper electrode.

If the concentration of copper’s reduced and oxidized forms are at unity, then the reduction potential is equal to standard potential. So, for the half-cell equation,

Cu²⁺ (aq, 1M) + 2e^–→ Cu(s), the cell that will measure the electrode potential is

Pt(s)| H₂ (g, 1 bar) | H⁺ (aq., 1 M) || Cu²⁺ (aq., 1 M) | Cu.

Thus, the correct option is (iii).

When an electrode potential (which is a potential difference that develops between the electrode and electrolyte), has values of the concentrations of all the species involved in a half-cell equal to unity, it is known as standard electrode potential.

The standard hydrogen electrode which is assigned a zero potential for all temperatures to measure the potential of the given half-cell. The electrode is represented as Pt(s)|H₂(g)|H⁺(aq) and the equation is given as H⁺(aq) + e^-→ 1/2H₂(g).

This leads to the conclusion that the pressure of hydrogen gas is 1 bar and the concentration of hydrogen ions in solution is 1 M. We place the SHE on the left and copper electrode on the right, which will help us find the standard reduction potential of copper electrode.

If the concentration of copper’s reduced and oxidized forms are at unity, then the reduction potential is equal to standard potential. So, for the half-cell equation,

Cu²⁺ (aq, 1M) + 2e^–→ Cu(s), the cell that will measure the electrode potential is

Pt(s)| H₂ (g, 1 bar) | H⁺ (aq., 1 M) || Cu²⁺ (aq., 1 M) | Cu.

Thus, the correct option is (iii).

The given equation is the Nernst equation. Here is the cell potential at a given concentration and log [Mg²⁺] is the log of the concentration of Mg²⁺. They are directly proportional to each other as increase in concentration leads to an increase in cell potential.

The given equation is equivalent to equation of straight line y = mx + c. Hence, the graph will be a straight line with as the y-intercept and a positive slope 0.059/2.

The equation is arranged in the y = mx+c form as = . The correct option is (ii).

The given equation is the Nernst equation. Here is the cell potential at a given concentration and log [Mg²⁺] is the log of the concentration of Mg²⁺. They are directly proportional to each other as increase in concentration leads to an increase in cell potential.

The given equation is equivalent to equation of straight line y = mx + c. Hence, the graph will be a straight line with as the y-intercept and a positive slope 0.059/2.

The equation is arranged in the y = mx+c form as = . The correct option is (ii).

The relation between Δ_rG and E_cell is given by ∆_rG = –nFE_cell. ∆_rG is the Gibbs energy of the reaction and E_cell is the cell potential. nF is the amount of charge passed in the cell.

E_cell is an intensive property, it does not depend on the quantity of the substance, while Gibbs energy is extensive and thermodynamic property, which depends on the number of charges. Option (iii) is correct.

The relation between Δ_rG and E_cell is given by ∆_rG = –nFE_cell. ∆_rG is the Gibbs energy of the reaction and E_cell is the cell potential. nF is the amount of charge passed in the cell.

E_cell is an intensive property, it does not depend on the quantity of the substance, while Gibbs energy is extensive and thermodynamic property, which depends on the number of charges. Option (iii) is correct.

Cell potential is defined as the potential difference between the two electrodes of a galvanic cell and it is measured in volts. It is the difference between the electrode potentials (reduction potentials) between the cathode and anode. The cell potential is known as electromotive force or emf when no current is drawn through the cell. The correct answer is (ii).

Cell potential is defined as the potential difference between the two electrodes of a galvanic cell and it is measured in volts. It is the difference between the electrode potentials (reduction potentials) between the cathode and anode. The cell potential is known as electromotive force or emf when no current is drawn through the cell. The correct answer is (ii).

Inert electrodes are made of less reactive metals such as gold and platinum, which do not participate in the electrochemical reaction but provide a surface for the reactions to occur. Their surface is provided for either oxidation or reduction reactions or for conduction of electrons. Redox reactions, a combination of oxidation and reduction cannot occur on the surface. The correct option is (iv).

Inert electrodes are made of less reactive metals such as gold and platinum, which do not participate in the electrochemical reaction but provide a surface for the reactions to occur. Their surface is provided for either oxidation or reduction reactions or for conduction of electrons. Redox reactions, a combination of oxidation and reduction cannot occur on the surface. The correct option is (iv).

A galvanic cell is a cell which converts chemical energy liberated in a redox reaction to electrical energy. If an externally opposite potential is applied in a slowly increasing manner to the cell, the reaction continues to occur till the potential difference is equal to the value of the cell’s potential. Then, the reaction stops and no current flows through the cell. Further increase in the potential leads to the start of the reaction again but in the opposite direction, making it function as an electrolytic cell which is a cell for using electrical energy to drive non-spontaneous chemical reactions. The correct option is (iii).

A galvanic cell is a cell which converts chemical energy liberated in a redox reaction to electrical energy. If an externally opposite potential is applied in a slowly increasing manner to the cell, the reaction continues to occur till the potential difference is equal to the value of the cell’s potential. Then, the reaction stops and no current flows through the cell. Further increase in the potential leads to the start of the reaction again but in the opposite direction, making it function as an electrolytic cell which is a cell for using electrical energy to drive non-spontaneous chemical reactions. The correct option is (iii).

When ionic substances/electrolytes are dissolved in water, they dissolve and increase the conductivity of water. The conductivity of electrolytes is dependent on various factors just like conductivity of metal solids. The various factors include nature of the electrolyte added, size of their ions produced and solvation, which is the interaction of the solute molecules with solvent, the nature of the solvent and its viscosity, concentration of the electrolyte, and temperature of the solution. Conductivity does depend on the solvation of ions due to the interaction of solute-solvent molecules. The correct answer is (iii).

When ionic substances/electrolytes are dissolved in water, they dissolve and increase the conductivity of water. The conductivity of electrolytes is dependent on various factors just like conductivity of metal solids. The various factors include nature of the electrolyte added, size of their ions produced and solvation, which is the interaction of the solute molecules with solvent, the nature of the solvent and its viscosity, concentration of the electrolyte, and temperature of the solution. Conductivity does depend on the solvation of ions due to the interaction of solute-solvent molecules. The correct answer is (iii).

If the standard potential of an ion is greater than zero, then the reduced form is more stable compared to hydrogen gas. On the other hand, if the standard potential is lesser than zero, then hydrogen gas is more stable than reduced form. The positive value of the standard electrode potential means that the ions of the element can oxidise hydrogen, or hydrogen ions can reduce the atoms. The negative value of the standard electrode potential indicates that hydrogen gas can oxidise the ion, or the atom can reduce hydrogen ions. So, below the value of zero, the more negative the value of the standard electrode potential, the stronger the hydrogen gas can oxidise the ion, and the stronger the ion can be as a reducing agent. Standard electrode potential of Cr is -0.74V, which is a negative value and lower than the other mentioned values, hence it is the strongest reducing agent. The correct option is (ii).

If the standard potential of an ion is greater than zero, then the reduced form is more stable compared to hydrogen gas. On the other hand, if the standard potential is lesser than zero, then hydrogen gas is more stable than reduced form. The positive value of the standard electrode potential means that the ions of the element can oxidise hydrogen, or hydrogen ions can reduce the atoms. The negative value of the standard electrode potential indicates that hydrogen gas can oxidise the ion, or the atom can reduce hydrogen ions. So, below the value of zero, the more negative the value of the standard electrode potential, the stronger the hydrogen gas can oxidise the ion, and the stronger the ion can be as a reducing agent. Standard electrode potential of Cr is -0.74V, which is a negative value and lower than the other mentioned values, hence it is the strongest reducing agent. The correct option is (ii).

As mentioned in the explanation of the previous question, if the standard electrode potential is higher than zero, then reduced form is more stable than the H₂ gas. The positive value of the standard electrode potential means that the ions of the element can oxidise hydrogen, or hydrogen ions can reduce the atoms. The higher the positive value, the more the ability of oxidation. The equation of formation of reduced form of MnO^4- is given by the equation MnO₄^- + 8H⁺ + 5e^-→ Mn²⁺ + 2e^-. The standard electrode potential is 1.51V which is highest and most positive. Option (iii) is the correct answer.

As mentioned in the explanation of the previous question, if the standard electrode potential is higher than zero, then reduced form is more stable than the H₂ gas. The positive value of the standard electrode potential means that the ions of the element can oxidise hydrogen, or hydrogen ions can reduce the atoms. The higher the positive value, the more the ability of oxidation. The equation of formation of reduced form of MnO^4- is given by the equation MnO₄^- + 8H⁺ + 5e^-→ Mn²⁺ + 2e^-. The standard electrode potential is 1.51V which is highest and most positive. Option (iii) is the correct answer.

Again referring to data and explanation of Q. 8, if the standard potential of an element is negative value, then H₂ gas is more stable than the reduced form of the element, which means these elements can reduce hydrogen ions H⁺ ions. The more negative the value, the more its reducing power. So the order of reducing values in decreasing order is Cr > Cr³⁺ > Cl^- > Mn²⁺. The correct option is (ii).

Again referring to data and explanation of Q. 8, if the standard potential of an element is negative value, then H₂ gas is more stable than the reduced form of the element, which means these elements can reduce hydrogen ions H⁺ ions. The more negative the value, the more its reducing power. So the order of reducing values in decreasing order is Cr > Cr³⁺ > Cl^- > Mn²⁺. The correct option is (ii).

The value of standard reduction potential of Mn²⁺ ion is more positive than the other ions and hydrogen, hence it is most stable in its reduced form and is a strong oxidizing agent. The correct option is (iv).

The value of standard reduction potential of Mn²⁺ ion is more positive than the other ions and hydrogen, hence it is most stable in its reduced form and is a strong oxidizing agent. The correct option is (iv).

The value of standard reduction potential of Cr³⁺ ion is more negative than the other ions and hydrogen, hence it is most stable in its oxidized form and is a strong reducing agent. The correct option is (i).

The value of standard reduction potential of Cr³⁺ ion is more negative than the other ions and hydrogen, hence it is most stable in its oxidized form and is a strong reducing agent. The correct option is (i).

The quantity of electricity passed in a circuit is given by the formula

Q = It

Where,

Q is the charge developed in the circuit,

I is the current in the circuit in amperes,

t is the time period for which current is allowed to pass through in the circuit.

Electrolysis of Al₂O₃ gives Al³⁺, which can be reduced to Al(s) by 3 moles of electrons. 1 Faraday or 1F is the charge on one mole of electrons.

The charge on one electron is 1.6 × 10^-19C. Charge on a mole of electrons is 1.6 × 10^-19 C × 6.023 × 10²³ mol^-1 = 96487 C mol^-1.

Thus, for three moles of aluminum, 3F of charge is required. The correct option is (iii).

The quantity of electricity passed in a circuit is given by the formula

Q = It

Where,

Q is the charge developed in the circuit,

I is the current in the circuit in amperes,

t is the time period for which current is allowed to pass through in the circuit.

Electrolysis of Al₂O₃ gives Al³⁺, which can be reduced to Al(s) by 3 moles of electrons. 1 Faraday or 1F is the charge on one mole of electrons.

The charge on one electron is 1.6 × 10^-19C. Charge on a mole of electrons is 1.6 × 10^-19 C × 6.023 × 10²³ mol^-1 = 96487 C mol^-1.

Thus, for three moles of aluminum, 3F of charge is required. The correct option is (iii).

The resistivity and conductivity of an ionic solution can be measured using a conductivity cell which uses alternating current. The resistance of the solution is given by the formula R =

Where

l is the distance between the electrodes,

A is the area of cross section of cell.

The term is known as the cell constant.

It depends on the length and area of cross section of cell, hence it remains constant for a cell, regardless of nature of electrolyte.

The correct option is (iv).

The resistivity and conductivity of an ionic solution can be measured using a conductivity cell which uses alternating current. The resistance of the solution is given by the formula R =

Where

l is the distance between the electrodes,

A is the area of cross section of cell.

The term is known as the cell constant.

It depends on the length and area of cross section of cell, hence it remains constant for a cell, regardless of nature of electrolyte.

The correct option is (iv).

Lead storage battery are secondary storage cells which can be recharged after use by passing current through it in the opposite direction and can be used again. The cell reactions when the battery is in use are as follows:

Anode: Pb(s) + SO₄^2–(aq) → PbSO₄(s) + 2e^–

Cathode: PbO₂(s) + SO₄^2–(aq) + 4H⁺(aq) + 2e^–→ PbSO₄(s) + 2H₂O(l)

When the battery is charged, the reactions are reversed and PbSO₄(s) on the anode and cathode is converted to Pb and PbO₂. From the anodic reaction, it is seen that the reverse is reduction. The correct option is (i).

Lead storage battery are secondary storage cells which can be recharged after use by passing current through it in the opposite direction and can be used again. The cell reactions when the battery is in use are as follows:

Anode: Pb(s) + SO₄^2–(aq) → PbSO₄(s) + 2e^–

Cathode: PbO₂(s) + SO₄^2–(aq) + 4H⁺(aq) + 2e^–→ PbSO₄(s) + 2H₂O(l)

When the battery is charged, the reactions are reversed and PbSO₄(s) on the anode and cathode is converted to Pb and PbO₂. From the anodic reaction, it is seen that the reverse is reduction. The correct option is (i).

HCl + NH₄OH → NH₄Cl + H₂O is an acid-base neutralization reaction. We can find using Kohlrausch law of independent migration of ions. This law states that limiting molar conductivity of an electrolyte is equal to the sum of the individual limiting molar conductivities of the anion and cation of the electrolyte.

NH₄OH is a weak base. We can calculate its limiting molar conductivity using limiting molar conductivities of strong salt NH₄Cl and NaCl. The cation NH₄⁺ is from strong electrolyte NH₄Cl and anion is from strong base NaOH.

The correct option is (ii).

HCl + NH₄OH → NH₄Cl + H₂O is an acid-base neutralization reaction. We can find using Kohlrausch law of independent migration of ions. This law states that limiting molar conductivity of an electrolyte is equal to the sum of the individual limiting molar conductivities of the anion and cation of the electrolyte.

NH₄OH is a weak base. We can calculate its limiting molar conductivity using limiting molar conductivities of strong salt NH₄Cl and NaCl. The cation NH₄⁺ is from strong electrolyte NH₄Cl and anion is from strong base NaOH.

The correct option is (ii).

The products of molten sodium chloride and aqueous solution of sodium chloride are not the same. In molten NaCl electrolysis, the products are only Na and Cl₂. In the electrolysis of aqueous NaCl, apart from Na⁺ and Cl^- ions, H⁺ and OH^- ions also exist because of presence of H₂O. At the cathode, there is competition between two reduction reactions –

(i) Na⁺(aq) + e^–→ Na(s) where standard electrode potential E_cell = –2.71 V

(ii) H⁺(aq) + e^– → 1/2 H₂(g) E_cell = 0.00 V

The reaction with higher potential value is preferred, hence the second equation takes place. But H⁺ is produced as a product of dissociation of H₂O, which is given by H₂O(l) → H⁺ + OH^-. The net reaction at the cathode is written as H₂O (l) + e^–→ 1/2H₂ (g) + OH^–.

At anode, two oxidation reactions are possible.

(i) Cl^– (aq) → 1/2 Cl₂ (g) + e^– E_cell = 1.36 V

(ii) 2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e^– E_cell = 1.23 V

Here, the issue of overpotential arises. Some electrochemical processes which are possible at an electrode, are so slow kinetically that at lower voltages these cannot take place and an extra potential (called overpotential) needs to be applied. The reaction at anode with lower value of E _cell is preferred and accordingly, water should get oxidised in preference to Cl^–(aq). However, because of overpotential of oxygen, second reaction is preferred. Hence, the correct option is (ii).

The products of molten sodium chloride and aqueous solution of sodium chloride are not the same. In molten NaCl electrolysis, the products are only Na and Cl₂. In the electrolysis of aqueous NaCl, apart from Na⁺ and Cl^- ions, H⁺ and OH^- ions also exist because of presence of H₂O. At the cathode, there is competition between two reduction reactions –

(i) Na⁺(aq) + e^–→ Na(s) where standard electrode potential E_cell = –2.71 V

(ii) H⁺(aq) + e^– → 1/2 H₂(g) E_cell = 0.00 V

The reaction with higher potential value is preferred, hence the second equation takes place. But H⁺ is produced as a product of dissociation of H₂O, which is given by H₂O(l) → H⁺ + OH^-. The net reaction at the cathode is written as H₂O (l) + e^–→ 1/2H₂ (g) + OH^–.

At anode, two oxidation reactions are possible.

(i) Cl^– (aq) → 1/2 Cl₂ (g) + e^– E_cell = 1.36 V

(ii) 2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e^– E_cell = 1.23 V

Here, the issue of overpotential arises. Some electrochemical processes which are possible at an electrode, are so slow kinetically that at lower voltages these cannot take place and an extra potential (called overpotential) needs to be applied. The reaction at anode with lower value of E �_cell is preferred and accordingly, water should get oxidised in preference to Cl^–(aq). However, because of overpotential of oxygen, second reaction is preferred. Hence, the correct option is (ii).

The standard reduction potential of Cu²⁺/Cu is 0.34 and is a positive value. The value lies higher than the potential of hydrogen electrode. The reduction potential value being higher than hydrogen is a stronger oxidizing agent than H⁺/H₂. It also means that it is more stable when reduced i.e. more stable on gaining electrons. They get more easily reduced than hydrogen ions. This also means hydrogen ions cannot oxidize Cu, alternatively hydrogen gas can reduce copper ion. The reverse cannot occur, hence Cu cannot reduce H⁺. Options (ii) and (iv) are correct.

In electrolysis of dilute sulphuric acid, the half reactions at cathode and anode are as follows,

Anode: 2H₂O (l) → 4H⁺ (aq) + O₂ (g) + 4e^–

4OH^– (aq) → 2H₂O (l) + O₂ (g) + 4e^–

The standard reduction potential of H₂ is higher than that of SO₄^2- ion, hence oxidation of H₂O takes place instead of sulphate.

Cathode: 2H⁺ (aq) + 2e^–→ H₂ (g)

Here the H⁺ ions are attracted to the cathode and are reduced to form H₂ gas.

For concentrated sulphuric acid, there is very low amount of water present, so at the anode, the sulphate ions get attracted and forms di-anion.

2SO₄⁻⟶S₂O₈²⁻+2e⁻ is the equation. Thus, at anode sulphuric acid gets oxidized, not water.

The correct options are (i) and (iii).

The electrode potential of any cell compared to standard hydrogen electrode is given by the Nernst equation, . Where E⁰ is the standard cell potential, R is the universal gas constant, T is the temperature in Kelvin, F is the Faraday constant, [Mⁿ⁺] is the concentration of the electrolyte.

For a Daniell cell, the cell potential is given in the following equation after converting the natural logarithm to the base 10 and substituting the values of R, F and T = 298 K –

E_cell = - .

At equilibrium,

E_cell = 0. So, E^Ө_cell =

K_c = [Zn²⁺]/[Cu²⁺]

So the equation can also be written as

E^Ө_cell =

From these equations, it can be calculated that (ii) and (iii) are correct.

The conductivity of a substance is the inverse of resistivity and is the conductance of a substance when it is one metre long and area of cross section is one m². Conductivity of electrolytic solutions also can be measured in a cell and even pure water has a low level of conductivity. The conductivity of electrolytic solutions depends on the nature of the electrolyte/solute added, the size of the ions produced, their solvation, the nature of the solvent and its viscosity, the concentration of the solute and the temperature. The conductivity of a substance is constant at a constant temperature and pressure. The correct options are (i) and (ii).

We can find the solution using Kohlrausch law of independent migration of ions. This law states that limiting molar conductivity of an electrolyte is equal to the sum of the individual limiting molar conductivities of the anion and cation of the electrolyte. There are three equations mentioned in the options which are acid-base neutralization reactions. They are –

HNO₃ + NaOH → NaNO₃ + H₂O

HCl + NaOH → NaCl + H₂O

HCl + NH₄OH → NH₄Cl + H₂O

Λ^o_m(H2O) = Λ^o_m(HNO3) + Λ^o_m(NaOH) – Λ^o_m(NaNO3)

Λ^o_m(H2O) =Λ^o_m(HCl) + Λ^o_m(NaOH) – Λ^o_m(NaCl)

Λ^o_m(H2O) =Λ^o_m(HCl) + Λ^o_m(NH4OH) – Λ^o_m(NH4Cl)

The correct options are (i), (iii) and (iv).

Dissolving CuSO₄ in water will lead to dissociation of CuSO₄ to Cu²⁺ and SO₄^2-. Water will dissociate as H⁺ and OH^-. On electrolysis using inert platinum electrodes, the cathode and anode reactions along with their standard electrode potential are as follows:

At anode, two reactions are possible –

(i) 2H₂O (l) → 4H⁺ (aq) + O₂ (g) + 4e^– OR 4OH^– (aq) → 2H₂O (l) + O₂ (g) + 4e^–, E^Ө_cell = 1.23V

(ii) 2SO₄^2- (l) → S₂O₈^2- + 2e^-, E^Ө_cell = 1.96 V

The reaction with lower electrode potential will be favoured at the anode, so O₂ is released at anode.

At cathode, two reactions are possible –

(i) Cu²⁺ + 2e^- → Cu(s), E^Ө_cell = 0.34V

(ii) H⁺ + e^-→ H₂(g), E^o_cell = 0.00 V

The reaction with higher electrode potential is favoured at the cathode, so Cu metal is deposited at the anode.

The correct options are (i) and (iii).

Dissolving CuSO₄ in water will lead to dissociation of CuSO₄ to Cu²⁺ and SO₄^2-. Water will dissociate as H⁺ and OH^-. In this case, copper electrodes are used instead of platinum inert electrodes. So the reactions will be different. Electrolysis of CuSO₄ using Cu electrodes will deposit copper metal at the cathode and the copper at the anode dissolves to form cations in solution.

At the anode, the negatively charged ions SO₄^2- and OH^- are attracted but they are stable to exist on their own and do not get oxidized, so the copper on the anode gets oxidized to form Cu²⁺ ions. The reaction is Cu(s) → Cu²⁺ + 2e^-.

At the cathode, the positively charged Cu²⁺ ions get attracted and get reduced to form Cu metal, which gets deposited on the copper cathode. It is favoured more than H⁺ to get reduced because of higher electrode potential. The reaction is given as Cu²⁺ + 2e^- → Cu(s).

The correct options are (i) and (ii).

Conductivity is the inverse of resistivity ρ, and is represented by the formula

κ =

=

= l/RA =

The quantity l/A is called cell constant denoted by the symbol, G*.

So,

κ = G*/R.

is the molar conductivity and is given by = κ/c.

Options (i) and (ii) are correct.

Molar conductivity of an ionic solution is defined as the conductivity of an electrolytic solution divided by the concentration of the electrolytic solution. It is represented by. Where is the conductivity and c is the concentration of the solution.

It is proportional to the resistance of the solution, hence dependent on the temperature. It is also dependent on the concentration of the solution. The correct options are (i) and (iii).

According to convention, an electrochemical reaction can be represented as Reaction at anode || Reaction at cathode. The || represents the salt bridge. The reaction in question shows the anodic reaction of Mg which gets oxidized to Mg²⁺ and the cathodic reaction is Cu²⁺ which gets reduced to Cu and is deposited on the cathode. The combined half-cell reactions can be represented as

Mg → Mg²⁺ + 2e^- at anode and Cu²⁺ + 2e^-→ Cu(s) at cathode. Combined, it forms Mg + Cu²⁺→ Mg²⁺ + Cu.

Hence, the correct options are (ii) and (iii).

The standard reduction potential of Cu²⁺/Cu is 0.34 and is a positive value. The value lies higher than the potential of hydrogen electrode. The reduction potential value being higher than hydrogen is a stronger oxidizing agent than H⁺/H₂. It also means that it is more stable when reduced i.e. more stable on gaining electrons. They get more easily reduced than hydrogen ions. This also means hydrogen ions cannot oxidize Cu, alternatively hydrogen gas can reduce copper ion. The reverse cannot occur, hence Cu cannot reduce H⁺. Options (ii) and (iv) are correct.

In electrolysis of dilute sulphuric acid, the half reactions at cathode and anode are as follows,

Anode: 2H₂O (l) → 4H⁺ (aq) + O₂ (g) + 4e^–

4OH^– (aq) → 2H₂O (l) + O₂ (g) + 4e^–

The standard reduction potential of H₂ is higher than that of SO₄^2- ion, hence oxidation of H₂O takes place instead of sulphate.

Cathode: 2H⁺ (aq) + 2e^–→ H₂ (g)

Here the H⁺ ions are attracted to the cathode and are reduced to form H₂ gas.

For concentrated sulphuric acid, there is very low amount of water present, so at the anode, the sulphate ions get attracted and forms di-anion.

2SO₄⁻⟶S₂O₈²⁻+2e⁻ is the equation. Thus, at anode sulphuric acid gets oxidized, not water.

The correct options are (i) and (iii).

The electrode potential of any cell compared to standard hydrogen electrode is given by the Nernst equation, . Where E⁰ is the standard cell potential, R is the universal gas constant, T is the temperature in Kelvin, F is the Faraday constant, [Mⁿ⁺] is the concentration of the electrolyte.

For a Daniell cell, the cell potential is given in the following equation after converting the natural logarithm to the base 10 and substituting the values of R, F and T = 298 K –

E_cell = - .

At equilibrium,

E_cell = 0. So, E^Ө_cell =

K_c = [Zn²⁺]/[Cu²⁺]

So the equation can also be written as

E^Ө_cell =

From these equations, it can be calculated that (ii) and (iii) are correct.

The conductivity of a substance is the inverse of resistivity and is the conductance of a substance when it is one metre long and area of cross section is one m². Conductivity of electrolytic solutions also can be measured in a cell and even pure water has a low level of conductivity. The conductivity of electrolytic solutions depends on the nature of the electrolyte/solute added, the size of the ions produced, their solvation, the nature of the solvent and its viscosity, the concentration of the solute and the temperature. The conductivity of a substance is constant at a constant temperature and pressure. The correct options are (i) and (ii).

We can find the solution using Kohlrausch law of independent migration of ions. This law states that limiting molar conductivity of an electrolyte is equal to the sum of the individual limiting molar conductivities of the anion and cation of the electrolyte. There are three equations mentioned in the options which are acid-base neutralization reactions. They are –

HNO₃ + NaOH → NaNO₃ + H₂O

HCl + NaOH → NaCl + H₂O

HCl + NH₄OH → NH₄Cl + H₂O

Λ^o_m(H2O) = Λ^o_m(HNO3) + Λ^o_m(NaOH) – Λ^o_m(NaNO3)

Λ^o_m(H2O) =Λ^o_m(HCl) + Λ^o_m(NaOH) – Λ^o_m(NaCl)

Λ^o_m(H2O) =Λ^o_m(HCl) + Λ^o_m(NH4OH) – Λ^o_m(NH4Cl)

The correct options are (i), (iii) and (iv).

Dissolving CuSO₄ in water will lead to dissociation of CuSO₄ to Cu²⁺ and SO₄^2-. Water will dissociate as H⁺ and OH^-. On electrolysis using inert platinum electrodes, the cathode and anode reactions along with their standard electrode potential are as follows:

At anode, two reactions are possible –

(i) 2H₂O (l) → 4H⁺ (aq) + O₂ (g) + 4e^– OR 4OH^– (aq) → 2H₂O (l) + O₂ (g) + 4e^–, E^Ө_cell = 1.23V

(ii) 2SO₄^2- (l) → S₂O₈^2- + 2e^-, E^Ө_cell = 1.96 V

The reaction with lower electrode potential will be favoured at the anode, so O₂ is released at anode.

At cathode, two reactions are possible –

(i) Cu²⁺ + 2e^- → Cu(s), E^Ө_cell = 0.34V

(ii) H⁺ + e^-→ H₂(g), E^o_cell = 0.00 V

The reaction with higher electrode potential is favoured at the cathode, so Cu metal is deposited at the anode.

The correct options are (i) and (iii).

Dissolving CuSO₄ in water will lead to dissociation of CuSO₄ to Cu²⁺ and SO₄^2-. Water will dissociate as H⁺ and OH^-. In this case, copper electrodes are used instead of platinum inert electrodes. So the reactions will be different. Electrolysis of CuSO₄ using Cu electrodes will deposit copper metal at the cathode and the copper at the anode dissolves to form cations in solution.

At the anode, the negatively charged ions SO₄^2- and OH^- are attracted but they are stable to exist on their own and do not get oxidized, so the copper on the anode gets oxidized to form Cu²⁺ ions. The reaction is Cu(s) → Cu²⁺ + 2e^-.

At the cathode, the positively charged Cu²⁺ ions get attracted and get reduced to form Cu metal, which gets deposited on the copper cathode. It is favoured more than H⁺ to get reduced because of higher electrode potential. The reaction is given as Cu²⁺ + 2e^- → Cu(s).

The correct options are (i) and (ii).

Conductivity is the inverse of resistivity ρ, and is represented by the formula

κ =

=

= l/RA =

The quantity l/A is called cell constant denoted by the symbol, G*.

So,

κ = G*/R.

is the molar conductivity and is given by = κ/c.

Options (i) and (ii) are correct.

Molar conductivity of an ionic solution is defined as the conductivity of an electrolytic solution divided by the concentration of the electrolytic solution. It is represented by. Where is the conductivity and c is the concentration of the solution.

It is proportional to the resistance of the solution, hence dependent on the temperature. It is also dependent on the concentration of the solution. The correct options are (i) and (iii).

According to convention, an electrochemical reaction can be represented as Reaction at anode || Reaction at cathode. The || represents the salt bridge. The reaction in question shows the anodic reaction of Mg which gets oxidized to Mg²⁺ and the cathodic reaction is Cu²⁺ which gets reduced to Cu and is deposited on the cathode. The combined half-cell reactions can be represented as

Mg → Mg²⁺ + 2e^- at anode and Cu²⁺ + 2e^-→ Cu(s) at cathode. Combined, it forms Mg + Cu²⁺→ Mg²⁺ + Cu.

Hence, the correct options are (ii) and (iii).

No, the absolute electrode potential of an electrode cannot be measured independently. A half-cell reaction never occurs all by itself. Two half-cells together make an electrolytic reaction possible. We can only measure the difference in electrode potential between the two half-cells. We can also measure electrode potential difference with respect to a standard electrode.

No, the absolute electrode potential of an electrode cannot be measured independently. A half-cell reaction never occurs all by itself. Two half-cells together make an electrolytic reaction possible. We can only measure the difference in electrode potential between the two half-cells. We can also measure electrode potential difference with respect to a standard electrode.

_{No, E}^°_Cell or ∆_r G^° can never be equal to zero. The only standard electrode potential which is arbitrarily assigned the value zero is the standard hydrogen electrode (SHE). Since everything else is measured with respect to SHE; the E^°_cell can never be zero.

_{We know that,}

_∆r G^°= -n F E^°_cell,

_Where∆r G^° is the standard Gibbs energy of the reaction

_{n is the number of electrons transferred}

_{F is Faraday’s constant and}

_E^°_cell is the standard electrode potential of a cell

_{so ∆r} G^° can also never be zero.

_{Note: There is one possibility for E}^°_cell=0_, that is when both cathode and anode are made of the same metal.

_{No, E}^°_Cell or ∆_r G^° can never be equal to zero. The only standard electrode potential which is arbitrarily assigned the value zero is the standard hydrogen electrode (SHE). Since everything else is measured with respect to SHE; the E^°_cell can never be zero.

_{We know that,}

_∆r G^°= -n F E^°_cell,

_Where∆r G^° is the standard Gibbs energy of the reaction

_{n is the number of electrons transferred}

_{F is Faraday’s constant and}

_E^°_cell is the standard electrode potential of a cell

_{so ∆r} G^° can also never be zero.

_{Note: There is one possibility for E}^°_cell=0_, that is when both cathode and anode are made of the same metal.

Electrolysis happens when a redox reaction occurs. Like all reactions, redox reaction to moves towards equilibrium. At equilibrium condition, the cell has discharged completely, and cell potential drops to zero.

_∆rG= -nF E _cell =0

_Where∆rG^° is the standard Gibbs energy of the reaction

_{n is the number of electrons transferred}

_{F is Faraday’s constant and}

_E^°_cell is the standard electrode potential of a cell

Electrolysis happens when a redox reaction occurs. Like all reactions, redox reaction to moves towards equilibrium. At equilibrium condition, the cell has discharged completely, and cell potential drops to zero.

_∆rG= -nF E_cell =0

_Where∆rG^° is the standard Gibbs energy of the reaction

_{n is the number of electrons transferred}

_{F is Faraday’s constant and}

_E^°_cell is the standard electrode potential of a cell

By convention, all standard electrode potentials are reduction potentials. A positive value suggests that the reduced form of the species is more stable than hydrogen gas and a negative value suggests that the hydrogen gas is more stable than the reduced form of species. Here, the reduced form (Zn) is not stable. It is difficult to reduce Zn²⁺ to Zn. So, the reverse is more likely to happen. Zn would rather get oxidized to Zn²⁺ and H⁺ will get reduced.

Thus, Zn is more reactive than Hydrogen.

By convention, all standard electrode potentials are reduction potentials. A positive value suggests that the reduced form of the species is more stable than hydrogen gas and a negative value suggests that the hydrogen gas is more stable than the reduced form of species. Here, the reduced form (Zn) is not stable. It is difficult to reduce Zn²⁺ to Zn. So, the reverse is more likely to happen. Zn would rather get oxidized to Zn²⁺ and H⁺ will get reduced.

Thus, Zn is more reactive than Hydrogen.

Faraday’s second law states that the mass of substances deposited on the electrode is proportional to their equivalent weight if the same amount of electricity is passed through different electrolytic solutions.

Faraday’s law : M ∝ E

M=nE

Where,

M is the mass of substance deposited

n is the number of moles

E is the equivalent weight of the substance

Q=It

where,

Q is the charge

I is the current and

t is time(in seconds)

Putting the values in the above formula , we get

1A x (10 x 60) sec = 600C

The electrochemical reactions for silver and copper are as given below:

Cu²⁺ + 2e^-→ Cu

Ag⁺ + e^-→ Ag

1 mole of electrons have a charge equal to 1 Faraday= 96500 C

For copper, we need 2 moles of electrons or (2F of charge) and for silver only one mole of the electron (1F of charge)

So, mass deposited

For silver,

If 1 F (96500 C) or 1 mole of electrons can deposit 1 mole i.e., 108 g of Ag, then

600 C can deposit,

g = 0.671 g of silver

For copper,

If 2 Faraday(96500 C) or 2 moles of electrons can deposit 1 mole i.e., 63g of Cu, then

600 C can deposit,

g = 0.195 g of copper

The highlighted fractions represent the equivalent weight.

Equivalent mass of Cu²⁺ is different from the equivalent mass of Ag⁺ so obviously the mass of copper deposited will not be the same as the mass of silver deposited.

Faraday’s second law states that the mass of substances deposited on the electrode is proportional to their equivalent weight if the same amount of electricity is passed through different electrolytic solutions.

Faraday’s law : M ∝ E

M=nE

Where,

M is the mass of substance deposited

n is the number of moles

E is the equivalent weight of the substance

Q=It

where,

Q is the charge

I is the current and

t is time(in seconds)

Putting the values in the above formula , we get

1A x (10 x 60) sec = 600C

The electrochemical reactions for silver and copper are as given below:

Cu²⁺ + 2e^-→ Cu

Ag⁺ + e^-→ Ag

1 mole of electrons have a charge equal to 1 Faraday= 96500 C

For copper, we need 2 moles of electrons or (2F of charge) and for silver only one mole of the electron (1F of charge)

So, mass deposited

For silver,

If 1 F (96500 C) or 1 mole of electrons can deposit 1 mole i.e., 108 g of Ag, then

600 C can deposit,

g = 0.671 g of silver

For copper,

If 2 Faraday(96500 C) or 2 moles of electrons can deposit 1 mole i.e., 63g of Cu, then

600 C can deposit,

g = 0.195 g of copper

The highlighted fractions represent the equivalent weight.

Equivalent mass of Cu²⁺ is different from the equivalent mass of Ag⁺ so obviously the mass of copper deposited will not be the same as the mass of silver deposited.

Oxidation takes place at the anode and reduction takes place at the cathode.

Anode- Oxidation half-cell: Cu → Cu²⁺ + 2e^-

Cathode- Reduction half-cell: 2Ag⁺ + 2e^- →2Ag

Overall reaction : Cu + 2Ag⁺→ Cu²⁺ + 2Ag

By convention while representing a galvanic cell, the following steps have to be followed:

1) The anode is always placed to the left and cathode to the right

2) The metal and electrolyte of the same half-cell are separated by a vertical line

3) A double vertical line which represents the salt bridge is placed in between the two half-cells to complete the notation of the galvanic cell.

Galvanic cell notation: Cu(s)| Cu²⁺(aq)|| Ag⁺(aq)| Ag(s)

Oxidation takes place at the anode and reduction takes place at the cathode.

Anode- Oxidation half-cell: Cu → Cu²⁺ + 2e^-

Cathode- Reduction half-cell: 2Ag⁺ + 2e^- →2Ag

Overall reaction : Cu + 2Ag⁺→ Cu²⁺ + 2Ag

By convention while representing a galvanic cell, the following steps have to be followed:

1) The anode is always placed to the left and cathode to the right

2) The metal and electrolyte of the same half-cell are separated by a vertical line

3) A double vertical line which represents the salt bridge is placed in between the two half-cells to complete the notation of the galvanic cell.

Galvanic cell notation: Cu(s)| Cu²⁺(aq)|| Ag⁺(aq)| Ag(s)

A positive value for the oxidation of Cl^- ions means the reduced form Cl^- is more stable than Cl₂ but still, oxidation of Cl^- takes place.

Cl^– (aq) → 1/2 Cl₂ (g) + e^– E_cell = 1.36 V ---(1)

2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e^– E_cell = 1.23 V ---(2)

Reaction (2) should take place at anode instead of Reaction (1) due to lower reduction potential.

In reaction (2), the oxidation of water to oxygen is kinetically unfavourable and requires excess potential called over-potential. Over-potential is the excess potential required to drive a reaction at a particular rate.

Therefore, oxidation of water does not take place at anode.

A positive value for the oxidation of Cl^- ions means the reduced form Cl^- is more stable than Cl₂ but still, oxidation of Cl^- takes place.

Cl^– (aq) → 1/2 Cl₂ (g) + e^– E_cell = 1.36 V ---(1)

2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e^– E_cell = 1.23 V ---(2)

Reaction (2) should take place at anode instead of Reaction (1) due to lower reduction potential.

In reaction (2), the oxidation of water to oxygen is kinetically unfavourable and requires excess potential called over-potential. Over-potential is the excess potential required to drive a reaction at a particular rate.

Therefore, oxidation of water does not take place at anode.

The potential difference developed between an electrode and electrolyte is called the electrode potential. When the potential is measured under standard conditions, i.e. when the concentration of all species in a half-cell is unity, then it is called standard electrode potential.

By convention, the standard electrode potential is always the reduction potential.

The potential difference developed between an electrode and electrolyte is called the electrode potential. When the potential is measured under standard conditions, i.e. when the concentration of all species in a half-cell is unity, then it is called standard electrode potential.

By convention, the standard electrode potential is always the reduction potential.

By convention, anode is placed on left and cathode is placed on the right. Oxidation occurs at anode and reduction occurs at the cathode.

Redox reaction for the cell is

Zn(s) + Cu²⁺→Zn²⁺(aq) + Cu(s)

Anode- oxidation half-cell: Zn(s) → Zn²⁺ (aq)+ 2e^-

Cathode- reduction half-cell: Cu²⁺ (aq)+ 2e^-→ Cu(s)

Zinc goes into solution as Zn²⁺ and leaves behind the electrons on Electrode A making it negatively charged. Cu²⁺ from the solution deposits on Electrode B making it positively charged.

Electrode A polarity- negative

Electrode B polarity- positive

By convention, anode is placed on left and cathode is placed on the right. Oxidation occurs at anode and reduction occurs at the cathode.

Redox reaction for the cell is

Zn(s) + Cu²⁺→Zn²⁺(aq) + Cu(s)

Anode- oxidation half-cell: Zn(s) → Zn²⁺ (aq)+ 2e^-

Cathode- reduction half-cell: Cu²⁺ (aq)+ 2e^-→ Cu(s)

Zinc goes into solution as Zn²⁺ and leaves behind the electrons on Electrode A making it negatively charged. Cu²⁺ from the solution deposits on Electrode B making it positively charged.

Electrode A polarity- negative

Electrode B polarity- positive

Direct current changes the concentration of ions in solution. Alternating current stops electrolysis from happening and keeps the concentration of ions constant so that proper measurement of resistance can be made.

Direct current changes the concentration of ions in solution. Alternating current stops electrolysis from happening and keeps the concentration of ions constant so that proper measurement of resistance can be made.

A galvanic cell has an electrical potential of 1.1V if an opposing potential of 1.1V is applied to this cell.

If the opposing potential equals the electric potential of the galvanic cell, then the flow of electrons stops.

When the reaction stops altogether, and no current flows through the cell.

Any further increase in the external potential again starts the reaction but in the opposite direction.

A galvanic cell has an electrical potential of 1.1V if an opposing potential of 1.1V is applied to this cell.

If the opposing potential equals the electric potential of the galvanic cell, then the flow of electrons stops.

When the reaction stops altogether, and no current flows through the cell.

Any further increase in the external potential again starts the reaction but in the opposite direction.

NaCl (aq)→Na⁺ (aq) + Cl^– (aq)

Cathode: H₂O(l ) + e^–→ 1/2 H₂(g) + OH^– (aq)

Anode: Cl^– (aq) → 1/2 Cl₂(g) + e^–

Overall reaction: NaCl(aq)+H₂O(l)→NaOH(aq)+H₂(g)+Cl₂(g)

The solution contains Na⁺ and OH^- which combine to form NaOH.

Since NaOH is a strong base. It will turn the brine solution basic, and the pH of the solution will increase.

NaCl (aq)→Na⁺ (aq) + Cl^– (aq)

Cathode: H₂O(l ) + e^–→ 1/2 H₂(g) + OH^– (aq)

Anode: Cl^– (aq) → 1/2 Cl₂(g) + e^–

Overall reaction: NaCl(aq)+H₂O(l)→NaOH(aq)+H₂(g)+Cl₂(g)

The solution contains Na⁺ and OH^- which combine to form NaOH.

Since NaOH is a strong base. It will turn the brine solution basic, and the pH of the solution will increase.

The following diagram represents a mercury cell.

The reactions take place inside a mercury cell are as follows:

Anode: Zn(Hg) + 2OH^–→ ZnO(s) + H₂O + 2e^–

Cathode: HgO + H₂O + 2e^–→ Hg(l) + 2OH^–

Galvanic cell equation: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

The above overall reaction contains no ions whose concentration can change over time. So, the cell potential remains constant throughout the mercury cell life.

The following diagram represents a mercury cell.

The reactions take place inside a mercury cell are as follows:

Anode: Zn(Hg) + 2OH^–→ ZnO(s) + H₂O + 2e^–

Cathode: HgO + H₂O + 2e^–→ Hg(l) + 2OH^–

Galvanic cell equation: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

The above overall reaction contains no ions whose concentration can change over time. So, the cell potential remains constant throughout the mercury cell life.

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing one mole of electrolyte kept between two electrodes with an area of cross section A and distance of unit length.

Molar conductivity, Λ_m= or Λ_m= κV

where

κ is the conductivity

A is the area of cross-section and

l is the distance between the electrodes.

For weak electrolytes the number of ions increases and the degree of dissociation increases with dilution and the increase is drastic. Hence, A has to be a weak electrolyte whereas B has to be a strong electrolyte because for strong electrolytes the number of ions does not change, but interionic distance decreases and therefore the increase is gradual.

Thus, electrolyte B is strong as on dilution the number of ions remains the same, only interionic attraction decreases; therefore, the increase in Λ_m is small.

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing one mole of electrolyte kept between two electrodes with an area of cross section A and distance of unit length.

Molar conductivity, Λ_m= or Λ_m= κV

where

κ is the conductivity

A is the area of cross-section and

l is the distance between the electrodes.

For weak electrolytes the number of ions increases and the degree of dissociation increases with dilution and the increase is drastic. Hence, A has to be a weak electrolyte whereas B has to be a strong electrolyte because for strong electrolytes the number of ions does not change, but interionic distance decreases and therefore the increase is gradual.

Thus, electrolyte B is strong as on dilution the number of ions remains the same, only interionic attraction decreases; therefore, the increase in Λ_m is small.

The following reactions take place when acidulated water is electrolysed.

Anode: 2H₂O(l)→ O₂(g) + 4H⁺ (aq) + 4e^–

Cathode: 4H⁺ + 4e^-→ 2H₂

Overall cell reaction: 2H₂O(l) → O₂(g) + 2H₂(g)

pH remains the same because concentration of H⁺ ions remains constant.

The following reactions take place when acidulated water is electrolysed.

Anode: 2H₂O(l)→ O₂(g) + 4H⁺ (aq) + 4e^–

Cathode: 4H⁺ + 4e^-→ 2H₂

Overall cell reaction: 2H₂O(l) → O₂(g) + 2H₂(g)

pH remains the same because concentration of H⁺ ions remains constant.

The conductivity of a solution by definition at any given concentration is the conductance of one unit volume of solution kept between two platinum electrodes with a unit area of cross-section and at a distance of unit length.

G=

Where

G is the conductance

κ is the conductivity

A is the area of cross-section

l is the distance between the platinum electrodes

Now, the addition of water dilutes the electrolyte. The number of ions in a given volume decreases and conductivity also reduces.

The conductivity of a solution by definition at any given concentration is the conductance of one unit volume of solution kept between two platinum electrodes with a unit area of cross-section and at a distance of unit length.

G=

Where

G is the conductance

κ is the conductivity

A is the area of cross-section

l is the distance between the platinum electrodes

Now, the addition of water dilutes the electrolyte. The number of ions in a given volume decreases and conductivity also reduces.

Standard hydrogen electrode (SHE) is used as a reference electrode. The potential of SHE is assigned the value zero. The electrode potential of other electrodes are measured with respect to this.

The standard hydrogen electrodes represented by

Pt(s)⎥ H₂(g)⎥ H⁺ (aq)

Standard hydrogen electrode (SHE) is used as a reference electrode. The potential of SHE is assigned the value zero. The electrode potential of other electrodes are measured with respect to this.

The standard hydrogen electrodes represented by

Pt(s)⎥ H₂(g)⎥ H⁺ (aq)

The oxidation takes place at the anode while the reduction takes place at the cathode.

By convention, while representing a galvanic cell, Anode is always placed to the left and cathode to the right.

Galvanic cell: Cu|Cu²⁺|| Cl^—|Cl₂, Pt

Anode- Oxidation half-cell: Cu(s) → Cu²⁺ + 2e^-

Cathode- Reduction half-cell: Cl₂(g)+ 2e^-→ 2Cl^-

Thus, Cu is anode as it is getting oxidised.

Cl₂ is cathode as it is getting reduced.

The oxidation takes place at the anode while the reduction takes place at the cathode.

By convention, while representing a galvanic cell, Anode is always placed to the left and cathode to the right.

Galvanic cell: Cu|Cu²⁺|| Cl^—|Cl₂, Pt

Anode- Oxidation half-cell: Cu(s) → Cu²⁺ + 2e^-

Cathode- Reduction half-cell: Cl₂(g)+ 2e^-→ 2Cl^-

Thus, Cu is anode as it is getting oxidised.

Cl₂ is cathode as it is getting reduced.

Galvanic cell reaction: Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s)

Anode- oxidation half-cell: Zn(s) → Zn²⁺ (aq)+ 2e^-

Cathode- reduction half-cell: Cu²⁺ (aq)+ 2e^-→ Cu(s)

Overall reaction: Zn(s) + Cu²⁺ (aq) → Zn²⁺(aq) +Cu(s)

Nernst equation:

Where, E _cellis the electrode potential at any given concentration

E^°_cell is the standard electrode potential

R is the gas constant 8.314 JK^-1mol^-1

F is the Faraday constant (96500 C mol^-1)

T is the temperature in Kelvin

[Zn²⁺] and [Cu²⁺] are the concentration of the Zn²⁺ and Cu²⁺ ion in the solution.

When we substitute the value for R, F and T(298K), we can reduce the equation to :

Galvanic cell reaction: Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s)

Anode- oxidation half-cell: Zn(s) → Zn²⁺ (aq)+ 2e^-

Cathode- reduction half-cell: Cu²⁺ (aq)+ 2e^-→ Cu(s)

Overall reaction: Zn(s) + Cu²⁺ (aq) → Zn²⁺(aq) +Cu(s)

Nernst equation:

Where, E _cellis the electrode potential at any given concentration

E^°_cell is the standard electrode potential

R is the gas constant 8.314 JK^-1mol^-1

F is the Faraday constant (96500 C mol^-1)

T is the temperature in Kelvin

[Zn²⁺] and [Cu²⁺] are the concentration of the Zn²⁺ and Cu²⁺ ion in the solution.

When we substitute the value for R, F and T(298K), we can reduce the equation to :

Fuel cells are cells that convert the energy of combustion of fuels like hydrogen, methanol into electrical energy. Fuel cells run continuously as long as reactants are supplied. Primary batteries discharge and are one-time use only, and secondary batteries can be recharged, but recharging takes a lot of time.

Thus, the fuel cell has scalability advantages over the primary and secondary battery.

Fuel cells are cells that convert the energy of combustion of fuels like hydrogen, methanol into electrical energy. Fuel cells run continuously as long as reactants are supplied. Primary batteries discharge and are one-time use only, and secondary batteries can be recharged, but recharging takes a lot of time.

Thus, the fuel cell has scalability advantages over the primary and secondary battery.

The cell reaction of lead storage battery when it is discharged is give below:

At anode: Pb(s) + SO₄^2–(aq) → PbSO₄ (s) + 2e^–

At cathode: PbO₂(s)+ SO₄^2–(aq)+4H⁺(aq) +2e^–→PbSO₄(s)+2H₂O (l)

Overall reaction:

Pb(s)+PbO₂(s)+2H₂SO₄(aq)→ 2PbSO₄(s) + 2H₂O(l)

The density is high when the concentration of H₂SO₄ is high, but during discharge this sulphuric acid gets consumed and becomes diluted with water. So, density decreases as the product water form dilute the H₂SO₄ concentration during the discharge of the battery.

The cell reaction of lead storage battery when it is discharged is give below:

At anode: Pb(s) + SO₄^2–(aq) → PbSO₄ (s) + 2e^–

At cathode: PbO₂(s)+ SO₄^2–(aq)+4H⁺(aq) +2e^–→PbSO₄(s)+2H₂O (l)

Overall reaction:

Pb(s)+PbO₂(s)+2H₂SO₄(aq)→ 2PbSO₄(s) + 2H₂O(l)

The density is high when the concentration of H₂SO₄ is high, but during discharge this sulphuric acid gets consumed and becomes diluted with water. So, density decreases as the product water form dilute the H₂SO₄ concentration during the discharge of the battery.

CH₃COOH is a weak electrolyte and CH₃COONa is a strong electrolyte. A weak electrolyte has a lower degree of dissociation at higher concentration but upon dilution, the degree of dissociation increases, the number of ions per unit volume increases and this lead to an increase in _{.So, the increase in drastic}

_{For strong electrolytes upon dilution, the number of ions doesn’t change,but interionic attraction decreases and hence the increase is gradual.}

CH₃COOH is a weak electrolyte and CH₃COONa is a strong electrolyte. A weak electrolyte has a lower degree of dissociation at higher concentration but upon dilution, the degree of dissociation increases, the number of ions per unit volume increases and this lead to an increase in _{.So, the increase in drastic}

_{For strong electrolytes upon dilution, the number of ions doesn’t change,but interionic attraction decreases and hence the increase is gradual.}

(i) Λ_m → (c) S cm² mol^-1

Here, Λ_m represents the molar conductivity.

Molar conductivity is expressed as Conductivity of the solution divided by its concentration.

Thus,

Λ_m = = = S cm² mol^-1

(ii) E_cell → (d) V

Here, E_cell represents the emf (electromotive force) of the cell and is expressed as the difference of the potential of the two half cells i.e., cathode and anode. Thus, the unit of E_cell is equalto ‘Volt’ and is denoted as ‘V’.

(iii) κ → (a) S cm^-1

Here, κ (kappa) denotes the conductivity or the specific conductance of the cell.

We know, Resistance (R) = ρ =

Or, κ = G = = S cm^-1

Where, ρ = Resistivity or specific resistance =

L= Distance between the electrodes = length

A= Area of cross section of the electrode

G= = Conductance

S= Siemens= SI unit of conductance

(iv) G^* → (b) m^-1

Here, G^* represents the cell constant.

We know, G^* =

Or, G^*= Conductivity x Resistance = x Ω

= S m^-1 x S^-1 (Since, Ω = S^-1)

= m^-1

Where, Ω is the SI unit of resistance.

(i) Λ_m → (c) S cm² mol^-1

Here, Λ_m represents the molar conductivity.

Molar conductivity is expressed as Conductivity of the solution divided by its concentration.

Thus,

Λ_m = = = S cm² mol^-1

(ii) E_cell → (d) V

Here, E_cell represents the emf (electromotive force) of the cell and is expressed as the difference of the potential of the two half cells i.e., cathode and anode. Thus, the unit of E_cell is equalto ‘Volt’ and is denoted as ‘V’.

(iii) κ → (a) S cm^-1

Here, κ (kappa) denotes the conductivity or the specific conductance of the cell.

We know, Resistance (R) = ρ =

Or, κ = G = = S cm^-1

Where, ρ = Resistivity or specific resistance =

L= Distance between the electrodes = length

A= Area of cross section of the electrode

G= = Conductance

S= Siemens= SI unit of conductance

(iv) G^* → (b) m^-1

Here, G^* represents the cell constant.

We know, G^* =

Or, G^*= Conductivity x Resistance = x Ω

= S m^-1 x S^-1 (Since, Ω = S^-1)

= m^-1

Where, Ω is the SI unit of resistance.

(i) Λ_m → (d) increases with dilution

(ii) → (a) intensive property

(iii) Κ → (b) depends on number of ions/volume

(iv) Δ_rG_cell → (c) extensive property

(i) Molar conductivity of the solution, Λ_m = = κ V

(Electrodes are at unit distance)

Where, V= volume of the solution containing 1 mole of electrolyte.

In other words, molar conductivity of the solution is directly proportional to the volume of the solution. Thus, with dilution (increase in volume of the solution), Λ_m increases.

(ii) Intensive property is the property of the system that does not depend of the mass of the system.

is an intensive property as it does not depend on the amount or size of the system. In fact, it depends on energy per coulomb of charge transferred.

(iii) The conductivity or the specific conductance of electrolyte depends on the number of ions per unit volume.

(iv) Δ_rG_cell = -nFE_cell

Where, Δ_rG_cell is the Gibbs free energy of the reaction.

F= Faraday constant

nF = Amount of charge passed

n= number of electron transferred in the radox reaction.

Since, Δ_rG_cell depends on the n, it is an extensive property.

(i) Λ_m → (d) increases with dilution

(ii) → (a) intensive property

(iii) Κ → (b) depends on number of ions/volume

(iv) Δ_rG_cell → (c) extensive property

(i) Molar conductivity of the solution, Λ_m = = κ V

(Electrodes are at unit distance)

Where, V= volume of the solution containing 1 mole of electrolyte.

In other words, molar conductivity of the solution is directly proportional to the volume of the solution. Thus, with dilution (increase in volume of the solution), Λ_m increases.

(ii) Intensive property is the property of the system that does not depend of the mass of the system.

is an intensive property as it does not depend on the amount or size of the system. In fact, it depends on energy per coulomb of charge transferred.

(iii) The conductivity or the specific conductance of electrolyte depends on the number of ions per unit volume.

(iv) Δ_rG_cell = -nFE_cell

Where, Δ_rG_cell is the Gibbs free energy of the reaction.

F= Faraday constant

nF = Amount of charge passed

n= number of electron transferred in the radox reaction.

Since, Δ_rG_cell depends on the n, it is an extensive property.

(i) Lead storage battery → (d) Pb is anode, PbO₂ is cathode

(ii) Mercury cell → (c) gives steady potential

(iii) Fuel cell → (a) maximum efficiency

(iv) Rusting → (b) prevented by galvanization

(i) Lead storage battery is an example of secondary battery. It consists of a Pb anode and cathode made up of Lead dioxide.

(ii) The overall reaction of Mercury cell is represented as

Zn(Hg) + HgO_(S) → ZnO_(S) + Hg_(l)

Mercury cell is a primary cell and the cell potential value is ~1.35V and remains constant throughout its life as they do not involve any ion whose concentration varies during its lifetime. Thus, gives a steady potential throughout its life.

(iii) Fuel cell uses Hydrogen and Oxygen to produce water molecules. The overall reaction is represented as

2H_2(g) + O_2(g)→ 2 H₂O_(l)

These cells run continuously until the reactants are depleted. They produce electricity with a higher efficiency value as compared to that by the thermal plants.

(iv) Rusting is basically the corrosion of iron in presence of air and water, where Iron is oxidized by loss of electrons to O_2(g) and forms Iron oxide. But rusting can be prevented by galvanization, by applying a protective Zinc coating on the Iron.

(i) Lead storage battery → (d) Pb is anode, PbO₂ is cathode

(ii) Mercury cell → (c) gives steady potential

(iii) Fuel cell → (a) maximum efficiency

(iv) Rusting → (b) prevented by galvanization

(i) Lead storage battery is an example of secondary battery. It consists of a Pb anode and cathode made up of Lead dioxide.

(ii) The overall reaction of Mercury cell is represented as

Zn(Hg) + HgO_(S) → ZnO_(S) + Hg_(l)

Mercury cell is a primary cell and the cell potential value is ~1.35V and remains constant throughout its life as they do not involve any ion whose concentration varies during its lifetime. Thus, gives a steady potential throughout its life.

(iii) Fuel cell uses Hydrogen and Oxygen to produce water molecules. The overall reaction is represented as

2H_2(g) + O_2(g)→ 2 H₂O_(l)

These cells run continuously until the reactants are depleted. They produce electricity with a higher efficiency value as compared to that by the thermal plants.

(iv) Rusting is basically the corrosion of iron in presence of air and water, where Iron is oxidized by loss of electrons to O_2(g) and forms Iron oxide. But rusting can be prevented by galvanization, by applying a protective Zinc coating on the Iron.

(i) K → (d)

(ii) Λ_m → (c)

(iii) ɑ →(b)

(iv) Q → (a) I × t

(i) We know, Resistance (R) = ρ =

Or, R κ = = G*

Or, κ=

Where, ρ = Resistivity or specific resistance =

L= Distance between the electrodes = length

A= Area of cross section of the electrode

G*= = Cell constant

(ii) Molar conductivity, Λ_m is defined as the conductivity of electrolyte divided by the molar concentration of the electrolyte.

Mathematically,

Λ_m =

Where, κ= Conductivity of the solution

C= Concentration

(iii) Degree of dissociation, α is defined as the ratio of the molar conductivity at any concentration to the limiting molar conductivity of an electrolyte.

Mathematically,

α =

(iv) Electric charge, Q is defined as the product of the electrical current passed and the duration.

Mathematically, Q = I x T

(i) K → (d)

(ii) Λ_m → (c)

(iii) ɑ →(b)

(iv) Q → (a) I × t

(i) We know, Resistance (R) = ρ =

Or, R κ = = G*

Or, κ=

Where, ρ = Resistivity or specific resistance =

L= Distance between the electrodes = length

A= Area of cross section of the electrode

G*= = Cell constant

(ii) Molar conductivity, Λ_m is defined as the conductivity of electrolyte divided by the molar concentration of the electrolyte.

Mathematically,

Λ_m =

Where, κ= Conductivity of the solution

C= Concentration

(iii) Degree of dissociation, α is defined as the ratio of the molar conductivity at any concentration to the limiting molar conductivity of an electrolyte.

Mathematically,

α =

(iv) Electric charge, Q is defined as the product of the electrical current passed and the duration.

Mathematically, Q = I x T

(i) Lechlanche cell → (d) reaction at anode, Zn → Zn²⁺ + 2e^–

(ii) Ni–Cd cell → (c) rechargeable

(iii) Fuel cell → (a) cell reaction 2H₂ + O₂→ 2H₂O ,

(e) converts energy of combustion into electrical energy

(iv) Mercury cell → (b) does not involve any ion in solution and is used in hearing aids.

(i) Lechlanche cell consists of a Zinc anode. The chemical reaction involved at anode is represented as:

Zn_(S) → Zn²⁺ + 2e^–

And the chemical reaction involved at cathode is represented as:

MnO_{2 +} NH₄⁺ + e^-→ MnO(OH) + NH₃

(ii) Ni-Cd cell is a type of secondary cell and is discharged after some time. But the good thing is that it can be charged again and therefore, they have a longer life than the lead storage cell. Thus, they are rechargeable.

(iii) Fuel cell uses Hydrogen and Oxygen to produce water molecules. The overall reaction is represented as:

2H_2(g) + O_2(g)→ 2 H₂O_(l)

It converts the energy of combustion into electrical energy. Oxygen is reduced at Cathode and the reaction involved is represented as:

O_2(g) + 2H₂O_(l) → 4OH^-_(aq)

(iv) Mercury cell is a type of primary cell. The overall reaction of Mercury cell is represented as

Zn(Hg) + HgO_(S) → ZnO_(S) + Hg_(l)

They do not involve any ion whose concentration varies during its lifetime and thus, the cell remains constant throughout its life. They are used for low current devices like in hearing aids, watches, etc.

(i) Lechlanche cell → (d) reaction at anode, Zn → Zn²⁺ + 2e^–

(ii) Ni–Cd cell → (c) rechargeable

(iii) Fuel cell → (a) cell reaction 2H₂ + O₂→ 2H₂O ,

(e) converts energy of combustion into electrical energy

(iv) Mercury cell → (b) does not involve any ion in solution and is used in hearing aids.

(i) Lechlanche cell consists of a Zinc anode. The chemical reaction involved at anode is represented as:

Zn_(S) → Zn²⁺ + 2e^–

And the chemical reaction involved at cathode is represented as:

MnO_{2 +} NH₄⁺ + e^-→ MnO(OH) + NH₃

(ii) Ni-Cd cell is a type of secondary cell and is discharged after some time. But the good thing is that it can be charged again and therefore, they have a longer life than the lead storage cell. Thus, they are rechargeable.

(iii) Fuel cell uses Hydrogen and Oxygen to produce water molecules. The overall reaction is represented as:

2H_2(g) + O_2(g)→ 2 H₂O_(l)

It converts the energy of combustion into electrical energy. Oxygen is reduced at Cathode and the reaction involved is represented as:

O_2(g) + 2H₂O_(l) → 4OH^-_(aq)

(iv) Mercury cell is a type of primary cell. The overall reaction of Mercury cell is represented as

Zn(Hg) + HgO_(S) → ZnO_(S) + Hg_(l)

They do not involve any ion whose concentration varies during its lifetime and thus, the cell remains constant throughout its life. They are used for low current devices like in hearing aids, watches, etc.

(i) F₂ → (c) non-metal which is the best oxidizing agent

(ii) Li → (a) metal is the strongest reducing agent

(iii) Au³⁺ → (g) metal ion which is an oxidizing agent

(iv) Br^– → (e) anion that can be oxidized by Au³⁺

(v) Au → (d) unreactive metal

(vi) Li ⁺→ (b) metal ion which is the weakest oxidizing agent

(vii) F^– → (f) anion which is the weakest reducing Agent

(i) We know, Fluoride is a non-metal and the standard electrode potential is +2.87V. High positive value means they are good oxidizing agent. Thus, F₂ get easily reduced to F^-.

(ii) Li⁺/Li have a standard electrode potential of -3.5V. Thus, this radox couple is a stronger reducing agent than the H⁺/H₂ radox couple. In other words, they have the most negative potential value among the above chemical elements which makes Li metal a strongest reducing agent.

(iii) The standard electrode potential for Au³⁺/Au is + 1.4V. Thus, they are a weaker reducing agent than the H⁺/H₂ radox couple. Positive standard electrode potential makes Au³⁺ a good oxidizing agent.

(iv) The standard electrode potential for this radox couple is +1.09V. Clearly, one can say that the standard electrode potential of this radox couple is less than that of Au³⁺/Au. Thus, Br^- is weak oxidizing agent and stronger reducing agent than Au³⁺. Therefore, Br^- anion can be oxidized by Au³⁺.

(v) Au³⁺ + 3e^-→ Au_(S)

Since, Au is in the solid state. Thus, its standard electrode potential will be zero. So, Au is an unreactive metal.

(vi) Li⁺/Li have a standard electrode potential of -3.5V. Thus, this radox couple is a stronger reducing agent than the H⁺/H₂ radox couple. In other words, Li⁺ metal ion is the weakest oxidizing agent.

(vii) F₂/F^- have the standard electrode potential equal to +2.87V. High positive value of standard radox potential indicates that F₂ is the best oxidizing agent, whereas, F^- is the weakest reducing Agent.

(i) F₂ → (c) non-metal which is the best oxidizing agent

(ii) Li → (a) metal is the strongest reducing agent

(iii) Au³⁺ → (g) metal ion which is an oxidizing agent

(iv) Br^– → (e) anion that can be oxidized by Au³⁺

(v) Au → (d) unreactive metal

(vi) Li ⁺→ (b) metal ion which is the weakest oxidizing agent

(vii) F^– → (f) anion which is the weakest reducing Agent

(i) We know, Fluoride is a non-metal and the standard electrode potential is +2.87V. High positive value means they are good oxidizing agent. Thus, F₂ get easily reduced to F^-.

(ii) Li⁺/Li have a standard electrode potential of -3.5V. Thus, this radox couple is a stronger reducing agent than the H⁺/H₂ radox couple. In other words, they have the most negative potential value among the above chemical elements which makes Li metal a strongest reducing agent.

(iii) The standard electrode potential for Au³⁺/Au is + 1.4V. Thus, they are a weaker reducing agent than the H⁺/H₂ radox couple. Positive standard electrode potential makes Au³⁺ a good oxidizing agent.

(iv) The standard electrode potential for this radox couple is +1.09V. Clearly, one can say that the standard electrode potential of this radox couple is less than that of Au³⁺/Au. Thus, Br^- is weak oxidizing agent and stronger reducing agent than Au³⁺. Therefore, Br^- anion can be oxidized by Au³⁺.

(v) Au³⁺ + 3e^-→ Au_(S)

Since, Au is in the solid state. Thus, its standard electrode potential will be zero. So, Au is an unreactive metal.

(vi) Li⁺/Li have a standard electrode potential of -3.5V. Thus, this radox couple is a stronger reducing agent than the H⁺/H₂ radox couple. In other words, Li⁺ metal ion is the weakest oxidizing agent.

(vii) F₂/F^- have the standard electrode potential equal to +2.87V. High positive value of standard radox potential indicates that F₂ is the best oxidizing agent, whereas, F^- is the weakest reducing Agent.

(iii) Assertion is true but the reason is false.

=+0.34V

= 0.00 V

Since, this radox couple have positive standard electrode potential, they are a weak reducing agent and therefore, Cu is less reactive than Hydrogen.

Hence, Cu is less reactive than Hydrogen as is positive.

=+0.34V

= 0.00 V

Since, this radox couple have positive standard electrode potential, they are a weak reducing agent and therefore, Cu is less reactive than Hydrogen.

Hence, Cu is less reactive than Hydrogen as is positive.

(iii) Assertion is true but the reason is false.

E_cell should have a positive value for the cell to function. As,we know,

Δ_rG_cell = -nFE_cell

Where, Δ_rG_cell = Gibbs free energy of the reaction.

For the proper functioning of the cell, the reaction must be spontaneous, i.e., Δ_rG_cell should have a negative value and this is true only when E_cell have a positive value.

We know, E_cell = E_cathode – E_anode

For reaction to be spontaneous, E_cell should have a positive value. This is true only when E_cathode > E_anode.

E_cellshouldhave a positive value for the cell to function. As,we know,

Δ_rG_cell = -nFE_cell

Where, Δ_rG_cell= Gibbs free energy of the reaction.

For the proper functioning of the cell, the reaction must be spontaneous, i.e., Δ_rG_cellshould have a negative value and this is true only when E_cellhave a positive value.

We know, E_cell= E_cathode – E_anode

For reaction to be spontaneous, E_cellshould have a positive value. This is true only when E_cathode>E_anode.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.

Both the statements are true and the reason is also true. Conductivity of all electrolytes (both, weak and strong electrolytes) decreases on dilution as number of ions per unit volume decreases with dilution. Thus, the amount of current carried by the ions per unit volume also decreases.

Hence, Conductivity of all electrolytes decreases on dilution because with dilution, the number of ions per unit volume decreases.

Both the statements are true and the reason is also true. Conductivity of all electrolytes (both, weak and strong electrolytes) decreases on dilution as number of ions per unit volume decreases with dilution. Thus, the amount of current carried by the ions perunit volume also decreases.

Hence, Conductivity of all electrolytes decreases on dilution because with dilution, the number of ions per unit volume decreases.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.

Λ_mfor weak electrolytes shows a sharp increase when the electrolytic solution is diluted as degree of dissociation and the number of ions in total volume of solution that contains 1 mol of electrolyte increases with dilution of solution.

Where, C= Concentration

Hence, Λ_m for weak electrolytes shows a sharp increase when the electrolytic solution is diluted because degree of dissociation increases with dilution of solution.

Λ_mfor weak electrolytes shows a sharp increase when the electrolytic solution is diluted as degree of dissociation and the number of ions in total volume of solution that contains 1 mol of electrolyte increases with dilution of solution.

Where, C= Concentration

Hence, Λ_mfor weak electrolytes shows a sharp increase when the electrolytic solution is diluted because degree of dissociation increases with dilution of solution.

(v) Assertion is false but reason is true.

The overall reaction of Mercury cell is represented as

Zn(Hg) + HgO_(S) → ZnO_(S) + Hg_(l)

The cell potential value is ~1.35V and remains constant throughout its life as they do not involve any ion whose concentration varies during its lifetime. Thus, gives a steady potential throughout its life.

The overall reaction of Mercury cell is represented as

Zn(Hg) + HgO_(S) →ZnO_(S) + Hg_(l)

The cell potential value is ~1.35V and remains constant throughout its life as they do not involve any ion whose concentration varies during its lifetime. Thus, gives a steady potential throughout its life.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.

In the electrolysis of NaCl solution, Chlorine is produced at anode instead of O_2. The two half-cell reactions are:

2Cl^-→ Cl_2(g) + 2e^- E^° = -1.36V

2H₂O → O_2(g) + 4H⁺ + 4e^- E^° = -1.23V

Generally, the one with the lower value is preferred but in this case Chlorine is produced at anode. This is because formation of oxygen at anode requires overvoltage.

In the electrolysis of NaCl solution, Chlorine is produced at anode instead of O_2. The two half-cell reactions are:

2Cl^-→ Cl_2(g) + 2e^- E^° = -1.36V

2H₂O →O_2(g) + 4H⁺ + 4e^- E^° = -1.23V

Generally, the one with the lower value is preferred but in this case Chlorine is produced at anode. This is because formation of oxygen at anode requires overvoltage.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.

For measuring resistance of an ionic solution an AC (Alternating current) source is used because concentration of ionic solution will change if DC (Direct current) source is used. DC current can change the composition of the ionic solution.

For measuring resistance of an ionic solution an AC (Alternating current) source is used because concentration of ionic solution will change if DC (Direct current) source is used. DC current can change the composition of the ionic solution.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.

We know, E_cell = E_cathode – E_anode

For reaction to be spontaneous, E_cell should have a positive value and Δ_rG_cell negative.

At equilibrium, E_cell = 0, as current stops flowing.

Or, E_cathode = E_anode

Hence, Current stops flowing when E_Cell = 0 because equilibrium of the cell reaction is attained.

We know, E_cell= E_cathode – E_anode

For reaction to be spontaneous, E_cellshould have a positive value and Δ_rG_cellnegative.

At equilibrium, E_cell = 0, as current stops flowing.

Or, E_cathode = E_anode

Hence, Current stops flowing when E_Cell = 0 because equilibrium of the cell reaction is attained.

(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.

Ag⁺_(aq) + e^- → Ag_(S) E° = + 0.80 V, i.e., positive radox potential

According to the Nernst equation,

At 25°C,

E_cell = - 0.059 log (For, n=1)

= - 0.059 log

(Since, molar concentration of an element in the solid state is equal to one, because the ratio of their density to their molar mass remains the same throughout the reaction)

= + 0.059 log[Ag⁺]

We know, for Ag⁺/Ag couple is fixed. Thus, with increase in Ag⁺ concentration, i.e., [Ag⁺], log[Ag⁺] increases. Thus, E_cell value also increases.

Hence, E_Ag+/Ag increases with increase in concentration of Ag⁺ ions.

Ag⁺_(aq) + e^- → Ag_(S) E° = + 0.80 V, i.e., positive radox potential

According to the Nernst equation,

At 25°C,

E_cell= - 0.059 log (For, n=1)

= - 0.059 log

(Since, molar concentration of an element in the solid state is equal to one, because the ratio of their density to their molar mass remains the same throughout the reaction)

= + 0.059 log[Ag⁺]

We know, for Ag⁺/Ag couple is fixed. Thus, with increase in Ag⁺ concentration, i.e., [Ag⁺], log[Ag⁺] increases. Thus, E_cell value also increases.

Hence, E_Ag+/Ag increases with increase in concentration of Ag⁺ ions.

Cu²⁺ + 2e^-→ Cu_(S) E° = 0.34V

Zn²⁺ + 2e^-→ Zn_(S) E° =-0.76V

Overall reaction,

Cu²⁺ + Zn_(S) → Cu_(S) + Zn²⁺ ; E° = 0.34V – (-0.76V) = + 1.1V

Copper (Cu²⁺) sulphate can’t be stored in zinc vessel as Zinc is more reactive than Copper. Clearly, one can say that Zinc is a stronger reducing agent than Copper. Thus, Zinc donate electrons to Cu²⁺ and the above radox reaction occurs.

(iv) Both assertion and reason are false.

Cu²⁺ + 2e^-→ Cu_(S) E° = 0.34V

Zn²⁺ + 2e^-→ Zn_(S) E° = -0.76V

Overall reaction,

Cu²⁺ + Zn_(S) → Cu_(S) + Zn²⁺ ; E° = 0.34V – (-0.76V) = + 1.1V

Copper (Cu²⁺) sulphate can’t be stored in zinc vessel as Zinc is more reactive than Copper. Clearly, one can say that Zinc is a stronger reducing agent than Copper. Thus, Zinc donate electrons to Cu²⁺ and the above radox reaction occurs.

(i) Given,

E_Cell(A) = 2V

E_Cell(B) = 1.1V

• The electrolytic cell with small E_cell will act as an electrolytic cell. Therefore, Cell B with lower emf value will act as an electrolytic cell.

• The half-cell reactions are:

• Anode: Cu_(S)→ Cu²⁺ + 2e^-

• Cathode: Zn²⁺ + 2e^-→ Zn_(S)

(ii) Given,

E_Cell(A) = 0.5V

E_Cell(B) = 1.1V

• Cell B having higher emf value will behave as a galvanic cell.

• Thus, new half-cell reactions are:

• Anode: Zn_(S)→ Zn²⁺ + 2e^-

• Cathode: Cu²⁺ + 2e^-→ Cu_(S)

(i) Given,

E_Cell(A) = 2V

E_Cell(B) = 1.1V

• The electrolytic cell with small E_cell will act as an electrolytic cell. Therefore, Cell B with lower emf value will act as an electrolytic cell.

• The half-cell reactions are:

• Anode: Cu_(S)→ Cu²⁺ + 2e^-

• Cathode: Zn²⁺ + 2e^-→ Zn_(S)

(ii) Given,

E_Cell(A) = 0.5V

E_Cell(B) = 1.1V

• Cell B having higher emf value will behave as a galvanic cell.

• Thus, new half-cell reactions are:

• Anode: Zn_(S)→ Zn²⁺ + 2e^-

• Cathode: Cu²⁺ + 2e^-→ Cu_(S)

(i) Here, the electrons will move from Zn to Ag.

(ii) The silver plate is Cathode, the one on the right hand side (RHS).

(iii) Salt bridge maintains the electrical neutrality and connects the two half-cells.

If the salt bridge is removed, the cell will stop working as electrons will not be able to flow from one side to another. Radox reaction will not occur.

(iv) The cell will stop functioning when equilibrium is attained. In other words, When E_cell = 0

Or, E_cathode = E_anode

(v) When cell is functioning, the concentration of Zn²⁺ions will increase and concentration of Ag⁺ ions will decrease because of the radox reaction between these two chemical species.

(vi) No change in the concentration of Zn²⁺ and Ag⁺ as the cell is dead, i.e., E_cell = 0.

According to the Nernst equation,

E_cell = - log

Therefore, = log (E_cell = 0)

We know, value is fixed at a fixed temperature, thus, the concentration of Zn²⁺ and Ag⁺ remains the same.

(i) Here, the electrons will move from Zn to Ag.

(ii) The silver plate is Cathode, the one on the right hand side (RHS).

(iii) Salt bridge maintains the electrical neutrality and connects the two half-cells.

If the salt bridge is removed, the cell will stop working as electrons will not be able to flow from one side to another.Radox reaction will not occur.

(iv) The cell will stop functioning when equilibrium is attained. In other words, When E_cell= 0

Or, E_cathode = E_anode

(v) When cell is functioning, the concentration of Zn²⁺ions will increase and concentration of Ag⁺ions will decrease because of the radox reaction between these two chemical species.

(vi) No change in the concentration of Zn²⁺ and Ag⁺ as the cell is dead, i.e., E_cell = 0.

According to the Nernst equation,

E_cell= - log

Therefore, = log (E_cell = 0)

We know, value is fixed ata fixed temperature, thus, the concentrationof Zn²⁺ and Ag⁺ remains the same.

• The relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell:

Δ_rG_cell = -nFE_cell

Where, Δ_rG_cell= Gibbs free energy of the reaction.

F= Faraday constant

E_cell = Emf of the cell

n= Number of moles of electron transferred in the radox reaction.

• At standard condition,

When, concentration of all reacting species is unity, T= 298 K

E_cell=

Or, ΔG° = -nF

• The maximum work can be obtained from a galvanic cell when the charge is passed reversibly.

• The reversible work done by the galvanic cell is written as,

Δ_rG_cell = -nFE_cell

• The relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell:

Δ_rG_cell = -nFE_cell

Where, Δ_rG_cell = Gibbs free energy of the reaction.

F= Faraday constant

E_cell = Emf of the cell

n= Number of moles of electron transferred in the radox reaction.

• At standard condition,

When, concentration of all reacting species is unity, T= 298 K

E_cell =

Or, ΔG° = -nF

• The maximum work can be obtained from a galvanic cell when the charge is passed reversibly.

• The reversible work done by the galvanic cell is written as,

Δ_rG_cell = -nFE_cell