NCERT solution Class 12 electrochemistry

NCERT Solutions

Class 12 Chemistry

Electrochemistry

NCERT Questions

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Q.18

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO ₃ with silver electrodes.

(ii) An aqueous solution of AgNO ₃with platinum electrodes.

(iii) A dilute solution of H ₂SO₄ with platinum electrodes.

(iv) An aqueous solution of CuCl₂ with platinum electrodes.

At cathode:

The following reduction reactions compete to take place at the cathode.

Ag⁺ (aq) + e^- à Ag(s); E⁰ = +0.80V

H⁺ (aq) + e^- à (1/2) H₂ (g); E⁰= 0.00V

The reaction with a higher value of E⁰takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

The Ag anode is attacked by NO₃^- ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag⁺.

At cathode:

The following reduction reactions compete to take place at the cathode.

Ag⁺ (aq) + e^- à Ag(s); E⁰ = +0.80V

H⁺ (aq) + e^- à (1/2) H₂ (g); E⁰= 0.00V

The reaction with a higher value of E ⁰takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

Since Pt electrodes are inert, the anode is not attacked by NO ₃^-ions. Therefore, OH⁻ or NO₃^- ions can be oxidized at the anode. But OH ⁻ ions having a lower discharge potential and get preference and decompose to liberate O₂.

OH^-  OH + e^- 4OH^-  H₂O + O₂

(iii) At the cathode, the following reduction reaction occurs to produce H ₂

gas.

H⁺ (aq) + e^- à (1/2) H₂ (g)

At the anode, the following processes are possible.

For dilute sulphuric acid, reaction (i) is preferred to produce O₂ gas. But for concentrated sulphuric acid, reaction (ii) occurs.

(iv) At cathode:

The following reduction reactions compete to take place at the cathode.

Cu²⁺ (aq) + 2e^- à Cu(s); E⁰ = +0.34V

H⁺ (aq) + e^-à (1/2) H₂ (g); E⁰= 0.00V

The reaction with a higher value of E ⁰takes place at the cathode. Therefore, deposition of copper will take place at the cathode.

At anode:

The following oxidation reactions are possible at the anode.

Cl^-(aq) à (1/2) Cl₂ (g) +e^-1; E⁰ = 1.36 V

2H₂O (l) à O₂ (g) + 4H⁺ (aq) + 4e^- ; E⁰ =+1.23 V

At the anode, the reaction with a lower value of E ⁰is preferred. But due to the over-potential of oxygen, Cl⁻ gets oxidized at the anode to produce Cl ₂ gas.

Video solution:

Electrochemistry Class 12 Chemistry NCERT Chapter 2 NCERT Solutions

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