Predict the products of electrolysis in each of the following:
(iii) A dilute solution of H_{2} SO_{4} with platinum electrodes.
(iv) An aqueous solution of CuCl _{2} with platinum electrodes.
Predict the products of electrolysis in each of the following:
(iii) A dilute solution of H_{2} SO_{4} with platinum electrodes.
(iv) An aqueous solution of CuCl _{2} with platinum electrodes.

Solution:

(iii) At the cathode, the following reduction reaction occurs to produce H_{2} gas.

    \[H{(a q)}^{+}+e^{-} \rightarrow \frac{1}{2} H_{2(g)}\]


At the anode, the following processes are possible.

    \[2 H_{2} O_{(l)} \rightarrow O_{2(g)}+4 H_{(a q)}^{+}+4 e^{-} ; E ^{0}=+1.23 V \quad--(i)\]


    \[2 S O_{4(a q)}^{2-} \rightarrow S_{2} O_{6(a q)}^{2-}+2 e^{-} ; E ^{0}=+1.96 V \quad--( ii )\]

When dealing with dilute sulphuric acid, reaction 1 is the favored method of producing oxygen gas. However, reaction (ii) occurs when concentrated sulphuric acid is used.

(iv) At the cathode: The following reduction reactions are in competition with one another for space at the cathode.

    \[\begin{aligned}&Cu_{(a q)}^{2+}+2 e^{-} \rightarrow C u_{(s)} ; E ^{0}=0.34 V \&H_{(a q)}^{+}+e^{-} \rightarrow \frac{1}{2} H_{2(g)} ; E ^{0}=0.00 V\end{aligned}\]


The cathode is the location where the reaction with a greater value will take place. As a result, copper will be deposited at the cathode during the reaction.
At anode:
The following oxidation reactions are possible at the anode.

    \[Cl_{(a q)}^{-} \rightarrow \frac{1}{2} C l_{2(g)}+e^{-} ; E ^{0}=1.36 V\]


    \[2 H_{2} O {(l)} \rightarrow O_{2(g)}+4 H_{(a q)}^{+}+e^{-} ; E ^{0}=+1.23 V\]


At the anode, the reaction with a lower value of E^{0} is preferred. But due to the over potential of oxygen, Cl^-1 gets oxidized at the anode to produce Cl _{2} gas.