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Faraday's Second Law

During the el...

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].

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Solution

Let x mol of $O_{2}$ is liberated and 3x mol of $H_{2} S_{2} O_{8}$ is formed.
Reactions at cathode (reduction) :
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 ⊖ O H$
Reactions at anode (oxidation) :
(i) $2 H_{2} O \to O_{2} + 4 H^{\oplus} + 4 e^{-}$ $[\begin{matrix} 1 m o l O_{2} & \equiv & 4 F x m o l O_{2} & = & 4 x F \end{matrix}]$
(II) $2 S O_{4}^{2 -} \to S_{2} O_{8}^{2 -} + 2 e^{-}$ $[\begin{matrix} 1 m o l S_{2} O_{8}^{2 -} & = & 2 F 3 x m o l S_{2} O_{8}^{2 -} & = & 6 x F \end{matrix}]$
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
$∴ 2 F c a t h o d e \equiv 1 m o l o f H_{2}$ .
10x F at cathode $\equiv \frac{1}{2 F} \times 10 x F = 5 x m o l o f H_{2}$
$R a t i o = \frac{M o l e s o f, H_{2} a t, c a t h o d e}{M o l e s o f H_{2} S_{2} O_{8} a t a n o d e} = \frac{5 x}{3 x} = \frac{5}{3}$
Number of moles of $H_{2} = 3 \times \frac{5}{3} = 5$

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