ElectrochemistryHard
Question
H2O2 can be prepared by successive reactions:
2NH4HSO4 → H2 + (NH4)2S2O8
(NH4)2S2O8 + 2H2O → 2NH4HSO4 + H2O2
The first reaction is an electrolytic reaction and second is steam distillation. What amount of current would have to be used in first reaction to produce enough intermediate to yield 102 g pure H2O2 per hour. Assume current efficiency 50%.
Options
A.643.33 A
B.321.67 A
C.160.83 A
D.1286.67 A
Solution
$n_{H_{2}O_{2}\text{ formed}} = n_{\left( NH_{4} \right)_{2}S_{2}O_{8}} = n\text{ and }n_{eq}\left( NH_{4} \right)_{2}S_{2}O_{8} = \frac{Q}{F}$
$\Rightarrow n \times 2 = \frac{102}{34} \times 2 = \frac{i \times 3600 \times 0.5}{96500} $$$\left\lbrack 2SO_{4}^{2 -} \rightarrow S_{2}O_{8}^{2 -} + 2e^{-} \right\rbrack \Rightarrow i = 321.67\text{ A}$$
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