Question

In electrolysis of dilute $H_{2}S{O}_4$, what is liberated at anode?

• A. $H_2$
• B. $S{O}_4^{2-}$
• C. $S{O}_2$
• D. $O_2$

Answer 1

Answer: (D)

Pure water does not conduct electricity. But when small amount of acid say $H_{2}S{O}_4$ is added to it, water ionises. On passing electricity, it decomposes. $H_{2}S{O}_4$ being a strong electrolyte ionise completely whereas water is feebly ionised
$H_{2}S{O}_4⟶2H^{+}+S{O}_4^{2-}$
$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$
During electrolysis, the hydrogen ions migrate towards the cathode and discharge here in the form of hydrogen gas
$2H^{+}+2e^{-}⟶H_2\uparrow$
At anode, the concentration of $O{H}^{-}$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
$2H_{2}O⟶O_2+4H^{+}+4e^{-}$
The overall reactions are :
At cathode
$2H^{+}+2e^{-}⟶H_2$
$4H^{+}+4e^{-}⟶2H_2$
At anode
$2H_{2}O⟶O_2+4H^{+}+4e^{-}$
Overall cell reaction :
$4H^{+}+2H_{2}O\rightleftharpoons 2H_2+O_2+4H^{+}$
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 $H_{2}S{O}_4$.