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Application of Electrolysis

Electrolysis ...

Question

Electrolysis of a solution of $M n S O_{4}$ in aqueous sulphuric acid is a method for the preparation of $M n O_{2}$ . Passing a current of $27 A$ for $24$ hours gives $1 k g$ of $M n O_{2}$ . The current efficiency in this process is:

100

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95.185

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80

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82.951

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Solution

The correct option is B $95.185$ %
Solution:- (B) $95.185 %$
${M n}^{2 +}_{(a q .)} + {H_{2} O}_{(l)} ⟶ {M n O_{2}}_{(s)} + {H^{+}}_{(a q .)} + {H_{2}}_{(g)}$
No. of electrons transferred $= 2$
Mol. wt. of $M n O_{2} = 87 g$
$∴$ Eq. wt of $M n O_{2} = \frac{87}{2} = 43.5 g$
As we know that,
$w = \frac{E . I . t}{F}$
where as,
$w =$ weight of $M n O_{2} = 1 k g = 1000 g$
$E =$ Equivalent wt. of $M n O_{2} = 43.5 g$
$I =$ Current drawn $= ?$
$t =$ time $= 24 h r s . = 24 \times 60 \times 60 = 86400 s e c$
$F =$ Faraday's constant $= 96500 C$
$∴ 1000 = \frac{43.5 \times I \times 86400}{96500}$
$\Rightarrow I = 25.676 A$
But given that the current of $27 A$ is passed.
$∴$ Current efficiency $= \frac{25.676}{27} \times 100 = 95.1 %$

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