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Electrolytes- Strong,Weak

The electrosy...

Question

The electrosynthesis of $M n O_{2}$ is carried out from a solution of $M n S O_{4}$ in $H_{2} S O_{4}$ (aq). If a current of 25.5 ampere is used with a current efficiency of 85%, how long would it take to produce 1 kg of $M n O_{2}$ ?

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Solution

The overall reaction :
${M n}^{2 +} (a q) + 2 H_{2} O (l) ⟶ {M n O}_{2} (s) + 2 H^{+} (a q) + H_{2} (g)$
The reaction involves ${2 e}^{-}$ transfer.
Wt of ${M n O}_{2}$ deposited, $W = 1 k g = 1000 g m$
Electrochemical equivalent, $Z = \frac{E q . w t . o f {M n O}_{2}}{96500} = \frac{M o l . w t . o f {M n O}_{2}}{2 \times 96500} = \frac{86.9}{2 \times 96500} = 0.00045$
Current, $I = 25.5 \times \frac{85}{100} = 21.675 A$
Know, time $t = \frac{W}{Z \times I} = \frac{1000}{0.00045 \times 21.675} = 1.025 \times 10^{5}$ seconds.

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