Electrochemistry

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: (i) The diagram is as follows;

(ii) Agis cathode where the reduction process is taking place where Ag⁺ takes electrons and deposits them at the cathode

(iii) Potential is zero when the salt bridge is suddenly removed.

(iv) Cell will stop functioning at discharging position when the cell potential is zero

(v) The concentration of Zn²⁺ ions will increase and the concentration of Ag⁺ ions will decrease due to conversion in oxidized and reduced forms.

(vi) When the cell is dead, the potential is zero and at equilibrium condition. Thus, the concentration of Zn²⁺ and Ag⁺ will not change.

Given -

(i) All the ions are in aqueous state.

Reaction in solution:

AgNO_{3 (s)} + aq → Ag ⁺ + NO^3-

H₂O → H ⁺ + OH ^-

At cathode:

Ag + (aq) + e - →Ag (s)

Ag ⁺ ions have lower discharge potential than H ⁺ ions. Hence, Ag ⁺ ions get deposited as Ag in preference to H ⁺ ions.

At anode:

Ag (s)→ Ag + (aq) + e -

As Ag anode is attacked by NO^3- ions, Ag of the anode will dissolve to form Ag ⁺ ions in the aqueous solution.

(ii) Reaction in solution:

AgNO_{3 (s)} + aq → Ag ⁺ + NO^3-

H₂O óH ⁺ + OH ^-

At cathode:

2Ag ⁺ _(aq) + 2e ^- →2Ag _(s)

Ag ⁺ ions have lower discharge potential than H ⁺ ions. Hence, Ag ⁺ ions get deposited as Ag in preference to H ⁺ ions.

At anode:

2OH^- (aq) → O₂ (g) + 2H⁺ (aq) + 4e -

As anode is not attackable, out of OH ^- and NO^3- ions, OH ^- having lower discharge potential, will be discharged in preference to NO^3- ions. These then decompose to give out O₂.

(iii) Reaction in solution

H₂SO₄ (aq) → 2H⁺ (aq) + SO^2-₄  (aq)

At cathode:

2H + (aq) + 2e - →H₂ (g)

At anode:

2OH^-  (aq) → O₂ (g) + 2H⁺ (aq) + 4e -

∴ H₂ gas is evolved at cathode and O_{2 (g)} is evolved at anode.

(iv) Reaction in solution: CuCl₂ (s) + aq → Cu²⁺ (aq) + Cl^- (aq)

H₂O óH ⁺ + OH ^-

At cathode:

Cu²⁺ (aq) + 2e - →Cu (g)

At anode:

2Cl^- (aq) - 2e^- → Cl₂ (g)

∴ Cu will be deposited at cathode and Cl₂ gas will be liberated at anode.

The electrode reaction is written as,

2Fe³⁺ + 2I ^- → 2Fe²⁺ + I₂

= 0.54V - 0.77V

∴ E⁰_cell = - 0.23 V

It is not feasible, as E⁰_cell is negative, ∴ ?G⁰ is positive.

• The electrode reaction is written as,
• 2Ag⁺ (aq) + Cu(s)→ Cu²⁺ (aq) + Ag(s)

= + 0.80V - 0.34V

∴ E⁰_cell = 0.46V

It is feasible, as E_cell ⁰ is positive, ∴ ?G⁰ is negative.

• (iii) The electrode reaction is written as,
• 2Fe³⁺ (aq) + 2Br^- (aq)→ 2Fe²⁺ (aq) + Br₂

k

= 0.77V - 1.09V

∴ E⁰_cell = - 0.32 V

It is not feasible, as E⁰_cell is negative, ∴ ?G⁰ is positive.

• (iv) The electrode reaction is written as,
• Ag(s) + Fe³⁺ (aq) → Fe²⁺ (aq) + Ag⁺ (aq)

= 0.77V - 0.80V

∴ E⁰_cell = - 0.03

It is not feasible, as E⁰_cell is negative, ∴ ?G⁰ is positive.

• (v) The electrode reaction is written as,
• Br₂ + 2Fe²⁺ (aq) → 2Br^- (aq) + 2Fe³⁺ (aq)

= 1.09V - 0.77V

∴ E⁰_cell = 0.32 V

It is feasible, as E_cell⁰ is positive, ∴ ?G⁰ is negative.