Question

Two platinum electrodes were immersed in a solution of $CuS{O}_4$ and electric current was passed through the solution. After some time, it was found that colour of $CuS{O}_4$ disappeared with evolution of gas at the electrode. The colorless solution contains:

• A. $CuS{O}_4$
• B. $H_{2}S{O}_4$
• C. both A & B
• D. None of the above

Answer 1

The correct option is B $H_{2}S{O}_4$

$CuS{O}_4\left(aq\right)\stackrel{Electrolysis}{\to }C{u}^{2+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)$

At cathode: $C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(reduction\right)$

The blue color of $CuS{O}_4$ disappears due to the deposition of Cu on Pt electrode.

At anode: $H_{2}O\to 2H^{\oplus }+2e^{-}+\frac{1}{2}{O}_2\left(g\right)$

Since oxidation potential of $H_{2}O$ > oxidation potential of $S{O}_4^{2-}$, so oxidation of $H_{2}O$ occurs and $O_2\left(g\right)$ is evolved at anode.

The colourless solution is due to the formation of $H_{2}S{O}_4$ as follows:

$2H^{\oplus }\left(fromanode\right)+S{O}_4^{2-}\to H_{2}S{O}_4$