Question

On electrolysis of dilute $H_{2}SO_{4}$ using platinum electrodes, the gas evolved at the anode is:

• A. $O_{2}$
• B. $H_{2}$
• C. $ SO_{2}$
• D. $ SO_{3}$

Answer 1

Answer: (A)

During electrolysis of dilute $H _{2} SO _{4}$ using platinum electrodes, the oxygen gas is evolved at the anode and hydrogen is liberated at cathode.
At cathode: $H^{+}+e^{-} \rightarrow \frac{1}{2} H_{2}$
At anode: $2 H_{2} O(l) \rightarrow O_{2}(g)+4 e^{-} ; \Delta E^{o}=+1.23\, V(I)$
$2 S O_{4}^{2-}(aq) \rightarrow S_{2} O_{8}^{2-}+2 e^{-} ; E=+1.96\, V(II)$
For dilute solution reaction (l) is preferred and
$O_{2}$ liberates at anode but at higher concentration reaction (II) is preferred.