Question

During the electrolysis of dilute sulphuric acid, the following process is possible at anode.

• A. $2H_{2}O_{\left(l\right)} \to O_{2\left(g\right)}+4H^{+}_{\left(aq\right)}+4e^{-}$
• B. $2SO^{2-}_{4\left(aq\right)} \to S_{2}O^{2-}_{8\left(aq\right)}+2e^{-}$
• C. $H_{2}O_{\left(l\right)} \to H^{+}_{\left(aq\right)}+OH^{-}_{\left(aq\right)}$
• D. $H_{2}O_{\left(l\right)}+e^{-} \to\frac{1}{2} H_{2\left(g\right)}+OH^{-}_{\left(aq\right)}$

Answer 1

Answer: (A)

During electrolysis of $H _{2} SO _{4}$
$H _{2}$ is evolved at cathode.
$2 H ^{+}+2 e ^{-} \longrightarrow H _{2}$
$O _{2}$ is evolved at anode,
$2 H _{2} O \longrightarrow O _{2}+4 H ^{+}+4 e ^{-}$