Question

During the electrolysis of conc $H_{2}S{O}_4$, it was found that $H_2{S}_2{O}_8$ and $O_2$ were liberated in a molar ratio of 3:1. How many moles of $H_2$ were found is terms of moles of $H_2{S}_2{O}_8$ ?
[Express your answer as : $3\times$ moles of $H_2$ ; integer answer is between 0 and 15].

Answer 1

Let x mol of $O_2$ is liberated and 3x mol of $H_2{S}_2{O}_8$ is formed.
Reactions at cathode (reduction) :
$2H_{2}O+2e^{-}\to H_2+2\stackrel{\ominus }{O}H$

Reactions at anode (oxidation) :
(i) $2H_{2}O\to O_2+4H^{\oplus }+4e^{-}$ $\left[\begin{array}{ccc}1mol{O}_2& \equiv & 4F\\ xmol{O}_2& =& 4xF\end{array}\right]$
(II) $2S{O}_4^{2-}\to S_2{O}_8^{2-}+2e^{-}$ $\left[\begin{array}{ccc}1mol{S}_2{O}_8^{2-}& =& 2F\\ 3xmol{S}_2{O}_8^{2-}& =& 6xF\end{array}\right]$

Total faradays at anode = (4x + 6x)F = 10x F.
Total Faradays at acthode = 2F = 1 mol $H_2$
10x F $\equiv$ Total Faradays at cathode = Total faradays at anode

$\therefore 2Fcathode\equiv 1molof{H}_2$.
10x F at cathode $\equiv \frac{1}{2F}\times 10xF=5xmolof{H}_2$

$Ratio=\frac{Molesof,H_{2}at,cathode}{Molesof{H}_2{S}_2{O}_{8}atanode}=\frac{5x}{3x}=\frac{5}{3}$

Number of moles of $H_2=3\times \frac{5}{3}=5$