In a zinc manganese dioxide dry cell, the anode is made up of zinc and cathode of a carbon rod surrounded by a mixture of $M n O_{2}, c a r b o n, N H_{4} C l$ and $Z n C l_{2}$ in aqueous base.
The cathodic reaction may be represented as:
$2 M n O_{2} (s) + Z n^{2 +} + 2 e^{-} \to Z n M n_{2} O_{4} (s)$
Let there be $8 g M n O_{2}$ in the cathodic compartment. How many days will the dry cell continue to give a current of $4 \times 10^{- 3}$ ampere?

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Solution

When $M n O_{2}$ will be used up in cathodic process, the dry cell will stop to produce current.
Cathodic process:
$+ 4 2 M n O_{2} (s) + Z n^{2 +} + 2 e^{-} \to + 3 Z n M n_{2} O_{4}$
Equivalent mass of $M n O_{2} = \frac{Molecular mass}{Change in oxidation state}$
$= \frac{87}{1} = 87$
From first law of electrolysis,
$W = \frac{I t E}{96500}$
$8 = \frac{4 \times 10^{- 3} \times t \times 87}{96500}$
$t = 2218390.8 s e c o n d$
$= \frac{2218390.8}{3600 \times 24} = 25.675 d a y s$ .

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