In acid medium, `MnO_(4)^(c-)` is an oxidizing agent.
`MnO_(4)^(c-)+8H^(o+)+5e^(-) rarr Mn^(2+)+4H_(2)O`
If `H^(o+)` ion concentration is doubled, electrode potential of the half cell `MnO_(4)^(c-), Mn^(2+)|Pt` will

Options

A. Increase by 28.36 mV
B. Decrease by 28.36mV
C. Increase by 14.23 mV
D. Decrease by 142 .23mV

Explanation

To solve the problem, we need to determine how the electrode potential of the half-cell reaction involving `MnO4^(-)` and `Mn^(2+)` changes when the concentration of `H^+` ions is doubled.

### Step-by-Step Solution:

1. **Identify the Half-Cell Reaction**:
The half-cell reaction given is:
\[ \text{MnO}_4^{-} + 8 \text{H}^{+} + 5 e^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

2. **Use the Nernst Equation**:
The Nernst equation for the half-cell can be expressed as:
\[ E = E^{\circ} - \frac{0.0591}{n} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^{-}][\text{H}^{+}]^8} \right) \]
where \( n \) is the number of electrons transferred (which is 5 in this case).

3. **Substituting Values**:
Substitute \( n = 5 \) into the equation:
\[ E = E^{\circ} - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^{-}][\text{H}^{+}]^8} \right) \]

4. **Effect of Doubling \( H^+ \) Concentration**:
If the concentration of \( H^+ \) ions is doubled, the new concentration becomes \( 2[\text{H}^{+}] \). Therefore, the term \( [\text{H}^{+}]^8 \) in the denominator of the logarithm changes to:
\[ (2[\text{H}^{+}])^8 = 2^8 [\text{H}^{+}]^8 = 256 [\text{H}^{+}]^8 \]

5. **Revising the Nernst Equation**:
The Nernst equation now becomes:
\[ E = E^{\circ} - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^{-}] \cdot 256[\text{H}^{+}]^8} \right) \]

6. **Logarithmic Simplification**:
This can be simplified using logarithmic properties:
\[ E = E^{\circ} - \frac{0.0591}{5} \left( \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^{-}][\text{H}^{+}]^8} \right) - \log(256) \right) \]
Since \( \log(256) = 8 \log(2) \approx 8 \times 0.301 = 2.408 \).

7. **Final Expression**:
Thus, the equation becomes:
\[ E = E^{\circ} - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^{-}][\text{H}^{+}]^8} \right) + \frac{0.0591}{5} \cdot 2.408 \]
This indicates that the electrode potential \( E \) increases due to the additional positive term.

8. **Calculating the Change**:
The change in potential due to the increase is:
\[ \Delta E = \frac{0.0591}{5} \cdot 2.408 \approx 0.02836 \text{ V} \text{ or } 28.36 \text{ mV} \]

### Conclusion:
When the concentration of \( H^+ \) ions is doubled, the electrode potential of the half-cell increases by approximately 28.36 mV.