Question

During an electrolysis of conc. $H_{2}S{O}_4$, perdisulphuric acid $\left(H_2{S}_2{O}_8\right)$ and $O_2$ form in equimolar amount. The amount of $H_2$ that will form simultaneously will be: $(2H_{2}S{O}_4\to H_2{S}_2{O}_8+2H^{+}+2e^{-})$

• A. Thrice that of O${}_2$ in moles
• B. Twice that of O${}_2$ in moles
• C. Equal to that of O${}_2$ in moles
• D. Half of that of O${}_2$ in moles

Answer 1

The correct option is A Thrice that of O${}_2$ in moles

Reaction of process:
$Anode\left\{\begin{array}{c}2H_{2}S{O}_4\to H_2{S}_2{O}_8+2H^{+}+2e^{-}\\ 2H_{2}O\to O_2+4H^{+}+4e^{-}\end{array}\right\}$

$Cathode\{2H_{2}O\to H_2+2O{H}^{-}-2e^{-}\}\times 3$
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Net: $2H_{2}S{O}_4+8H_{2}O\to H_2{S}_2{O}_8+O_2+3H_2+6H^{+}+6O{H}^{-}$

Hence ratio of $n_{O_2}$ and $n_{H_2}$ is 1 : 3.