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

Question:

Predict the products of electrolysis in each of the following:

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

(ii) An aqueous solution of AgNO3with platinum electrodes.

(iii) A dilute solution of H2SO4with platinum electrodes.

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

Solution:

(i) At cathode:

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

$\mathrm{Ag}_{(a q)}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}_{(s)} ; E^{0}=0.80 \mathrm{~V}$

$\mathrm{H}^{+}{ }_{(\alpha)}^{+}\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2(g)} ; \quad E^{o}=0.00 \mathrm{~V}$

The reaction with a higher value of $E^{\circ}$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

The $\mathrm{Ag}$ anode is attacked by $\mathrm{NO}_{3}^{-}$ions. Therefore, the silver electrode at the anode dissolves in the solution to form $\mathrm{Ag}^{+}$.

(ii) At cathode:

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

$\mathrm{Ag}_{(a q)}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}_{(s)} ; \quad E^{0}=0.80 \mathrm{~V}$

$\mathrm{H}^{+}{ }_{(a q)}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2(g)} ; \quad E^{\circ}=0.00 \mathrm{~V}$

The reaction with a higher value of $E^{\circ}$ 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 not attacked by $\mathrm{NO}_{3}^{-}$ions. Therefore, $\mathrm{OH}^{-}$or $\mathrm{NO}_{3}^{-}$ions can be oxidized at the anode. But $\mathrm{OH}^{-}$ions having a lower discharge potential and get preference and decompose to liberate $\mathrm{O}_{2}$.

$\mathrm{OH}^{-} \longrightarrow \mathrm{OH}+\mathrm{e}^{-}$

$4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$

(iii) At the cathode, the following reduction reaction occurs to produce H2 gas.

$\mathrm{H}^{+}{ }_{(a q)}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2(g)}$

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.

The reaction with a higher value of $E^{\circ}$ 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 $E^{\circ}$ is preferred. But due to the over-potential of oxygen, $\mathrm{Cl}^{-}$gets oxidized at the anode to produce $\mathrm{Cl}_{2}$ gas.