Question

In electrolysis of dilute $ H_2SO_4$, what is liberated at anode?

• A. $H_{2}$
• B. $ SO_{4}^{2-} $
• C. $SO_2$
• D. $O_2$

Answer 1

The Correct Option is D

Pure water does not conduct electricity. But when small amount of acid say $H _{2} SO _{4}$ is added to it, water ionises. On passing electricity, it decomposes. $H _{2} SO _{4}$ being a strong electrolyte ionise completely whereas water is feebly ionised $H _{2} SO _{4} \longrightarrow 2 H ^{+}+ SO _{4}^{2-}$ $H _{2} O \rightleftharpoons H ^{+}+ OH ^{-}$ During electrolysis, the hydrogen ions migrate towards the cathode and discharge here in the form of hydrogen gas $2 H ^{+}+2 e^{-} \longrightarrow H _{2} \uparrow$ At anode, the concentration of $OH ^{-}$ions is too low to maintain a reaction and sulphate ions are not oxidised but remain in solution. Thus water molecules must be the species reacting at anode $2 H _{2} O \longrightarrow O _{2}+4 H ^{+}+4 e^{-}$ The overall reactions are : At cathode $2 H ^{+}+2 e^{-} \longrightarrow H _{2}$ $4 H ^{+}+4 e^{-} \longrightarrow 2 H _{2}$ At anode $2 H _{2} O \longrightarrow O _{2}+4 H ^{+}+4 e^{-}$ Overall cell reaction : $4 H ^{+}+2 H _{2} O \rightleftharpoons 2 H _{2}+ O _{2}+4 H ^{+}$ as we see that acid used is regenerated at the end. Therefore, the whole electrolysis reaction is the dissociation of water to give oxygen at anode and hydrogen at cathode catalysed by $H _{2} SO _{4}$.