Electrolysis of a solution of `MnSO_(4)` in aqueous sulphuric acid is a method for the preparation of `MnO_(2)` as per reaction,
`Mn_((aq.))^(2+)+2H_(2)O rarr MnO_(2(s))+2H_((aq))^(+)+H_(2(g))`
Passing a current of 27 ampere for 24 hour gives one of `MnO_(2)`. Waht is the value of current efficiency ? Write the reaction taking place at the cathode the anode.

Explanation

To solve the problem of determining the current efficiency during the electrolysis of a solution of `MnSO4` in aqueous sulfuric acid, we will follow these steps:

### Step 1: Identify the Reaction and N-Factor
The given reaction is:
\[ \text{Mn}^{2+}_{(aq)} + 2 \text{H}_2\text{O} \rightarrow \text{MnO}_2_{(s)} + 2 \text{H}^+_{(aq)} + \text{H}_2_{(g)} \]

In this reaction, manganese is oxidized from +2 to +4. The change in oxidation state (N-factor) for manganese is:
\[ \text{N-factor} = \text{Change in oxidation state} \times \text{Number of Mn atoms} = (4 - 2) \times 1 = 2 \]
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