Question

What will happen during the electrolysis of aqueous solution of $CuS{O}_4$ by using platinum electrodes?

• A. Copper will deposit at cathode
• B. Oxygen will be released at anode
• C. Copper will deposit at anode
• D. Copper will dissolve at anode

Answer 1

Answer: (A), (B)

Oxygen will be released at anode

Redox Reactions at cathode and anode:

In aqueous solution of $CuS{O}_4$, $CuS{O}_4\text{and}{H}_{2}O$ are present.

Hence
$CuS{O}_4\rightleftharpoons C{u}^{2+}+S{O}_4^{2-}$
$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

At cathode, reduction reaction takes place and at anode, oxidation reaction takes place.

At cathode:

Between $C{u}^{2+}\text{and}{H}_{2}O$, $C{u}^{2+}$ will get reduced at cathode to $Cu$ because $C{u}^{2+}/Cu$ has higher standard reduction potential as compared to $H^{+}/\frac{1}{2}{H}_2$.

Hence copper gets deposited at cathode.
$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$

At anode:

Between $H_{2}O\text{and}S{O}_4^{2-}$, $H_{2}O$ will oxidize to $O_2$ because $H_{2}O$ has a greater tendency to get oxidized.

Thus reactions taking place at anode is
$2H_{2}O\to O_2+4H^{+}+4e^{-}$

Therefore, copper will deposit at cathode and oxygen gas will be released at anode.

Hence, the correct options are (a) and (c).