The product obtained at anode when $50 \%$ $\mathrm{H}_2 \mathrm{SO}_4$ aqueous solution is electrolysed…

The product obtained at anode when $50 \%$ $\mathrm{H}_2 \mathrm{SO}_4$ aqueous solution is electrolysed using platinum electrodes is :
  1. $\mathrm{H}_2 \mathrm{SO}_3$
  2. $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$
  3. $\mathrm{O}_2$
  4. $\mathrm{H}_2$

Solution

$50 \% \mathrm{H}_2 \mathrm{SO}_4$ aqueous solution can be electrolysed by using Pt electrodes as $2 \mathrm{H}_2 \mathrm{SO}_4 2 \mathrm{HSO}_4^{-}+2 \mathrm{H}^{+}$ $2 \mathrm{HSO}_4^{-} \longrightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8+2 e^{-}$(at anode)

Asked in: AP EAMCET 2003

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