Question

Write equations for the reactions taking place at the two electrodes (mentionting clearly the name of the electrode) during the electrolysis of:
(i) Acidified copper sulphate solution with copper electrodes.

(ii) Molten lead bromide with inert electrodes.

Answer 1

(i) Dissociation reaction :

${\mathrm{CuSO}}_4\left(aq\right)\to {\mathrm{Cu}}^{2+}\left(aq\right)+{\mathrm{SO}}_4^{2-}\left(aq\right)$

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right) \\ \left(\mathrm{Cations}\right)\left(\mathrm{Anions}\right) \end{gathered}$$

At the cathode:

${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2e^{-}\mathit{}\to \mathrm{Cu}\left(\mathrm{s}\right)$

Copper $\left(\mathrm{Cu}\right)$ metal gets converted into ions.

At the anode:

$Cu\left(s\right)-2e^{-}\mathit{}\mathit{}\to {\mathrm{Cu}}^{2+}\left(aq\right)$

(ii) Dissociation reaction :

$$\begin{gathered} {PbBr}_2\left(\mathrm{l}\right)\to {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right) \\ \left(\mathrm{Cations}\right)\left(\mathrm{Anions}\right) \end{gathered}$$

The cations get attracted to the negative rod, Cathode.

The anions get attracted to the positive rod, Anode.

At the cathode:

$$\begin{gathered} {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2e^{-}\mathit{}\mathit{}\mathit{}\mathit{}\to \mathrm{Pb}\left(\mathrm{l}\right) \\ \mathrm{Lead} \end{gathered}$$

Then, the Lead metal is obtained in a molten state

At the anode:

$2{\mathrm{Br}}^{-}\left(aq\right)-2e^{-}\mathit{}\mathit{}\to {\mathrm{Br}}_2\left(g\right)$

Then, reddish-brown vapors of bromine are formed.