Ncert Exercises & Solutions

3.18 Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO₃ with silver electrodes.

(ii) An aqueous solution of AgNO₃ with platinum electrodes.

(iii) A dilute solution of H₂SO₄ with platinum electrodes.

(iv) An aqueous solution of CuCl₂with platinum electrodes.

(i) At cathode:

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

${Ag}^{+}_{(aq)} + e^{-} \to {Ag}_{(s)}; E ° = 0.80 V$

${H^{+}}_{(aq)} + e^{-} \to \frac{1}{2} H_{2 (g)}; E ° = 0.00 V$

The reaction with a higher value of

$E °$ takes place at the cathode. therefore, deposition of silver will take place at the cathode.

At anode:

The Ag anode is attacked by

${NO}_{3}^{-}$ ions. therefoe, the silver electrode at the anode dissolves in the solution to form

${Ag}^{+}$ .

(ii) At cathode:

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

${Ag}^{+}_{(aq)} + e^{-} \to {Ag}_{(s)}; E ° = 0.80 V$

${H^{+}}_{(aq)} + e^{-} \to \frac{1}{2} H_{2 (g)}; E ° = 0.00 V$

The reaction with a higher value of

$E °$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode;

Since Pt electrodes are inert, the anode is attacked by

${NO}_{3}^{-}$ ions. Therefore,

${OH}^{-}$ or

${NO}_{3}^{-}$ ions can be oxidized at the anode. But

${OH}^{-}$ ions have a lower discharge potential and get preference and decompose to liberate

$O_{2}$ .
4OH^-→2H₂O+O₂ + 4e^-

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

$H_{2}$ gas.

${H^{+}}_{(aq)} + e^{-} \to \frac{1}{2} H_{2 (g)}$

At the anode, the following processes are possible.

$2 H_{2} O_{(t)} \to O_{2 (g)} + 4 {H^{+}}_{(aq)} + 4 e^{-}; E ° = + 1.23 V (i)$

$2 {SO}_{4 (aq)}^{2 -} \to S_{2} O_{6 (aq)}^{2 -} + 2 e^{-}; E ° = + 1.96 V (ii)$

For dilute sulphuric acid, reaction (i) is preferred to produce

$O_{2}$ gas. But for concentrated sulphuric acid, reaction (ii) occurs.

(iv) At cathode:

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

${Cu}^{2 +}_{(aq)} + 2 e^{-} \to {Cu}_{(s)}; E ° = 0.34 V$

${H^{+}}_{(aq)} + e^{-} \to 1 / 2 H_{2 (g)}; E ° = 0.00 V$

$E °$ The reaction with a higher value takes 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.

At the anode, the reaction with a lower value of

${Cl}^{-}_{(aq)} \to 1 / 2 {Cl}_{2 (g)} + e^{- 1}; E ° = 1.36 V$

$2 H_{2} O_{(t)} \to O_{2 (g)} + 4 {H^{+}}_{(aq)} + 4 e^{-}; E ° = + 1.23 V$

$E °$ is preferred. But due to the overpotential of oxygen,

${Cl}^{-}$ gets oxidized at the anode to produce

${Cl}_{2}$ gas.

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