Question

When conc. $H_{2}S{O}_4$ is electrolysed with high current using Pt electrodes, the product obtained at anode is:

• A. $S{O}_2$
• B. $S{O}_3$
• C. $O_2$
• D. $H_{2}S{O}_4$

Answer 1

The correct option is C $O_2$

$O{H}^{-}$ ions having a lower discharge potential than $S{O}_4^{2-}$. $O{H}^{-}$ will be liberated at anode in preference to $S{O}_4^{2-}$ ions.

$O{H}^{-}\to OH+e^{-}$

$4OH\to 2H_{2}O+O_2$