`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H-(2)O`,
If `H^(+)` concentration is decreased from 1 M to `10^(-4)` M at `25^(@)C`, whereas concentration of `Mn^(2+)` and `MnO_(4)^(-)` remains 1M, then:

Options

A. The potential decreases by 0.38 V with decrease in oxidising power
B. The potential increases by 0.38 V with increase in oxidising power
C. The potential decreases by 0.25 V with decrease inn oxidising power
D. the potential decreases by 0.38 V without affecting oxidising power

Explanation

To solve the problem, we need to analyze the given half-reaction and how the change in the concentration of \( H^+ \) affects the cell potential (\( E_{cell} \)).

### Step-by-Step Solution:

1. **Identify the Reaction and Standard Conditions**:
The half-reaction is:
\[ \text{MnO}_4^{-} + 8H^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4H_2O \]
At standard conditions, the concentrations of \( \text{Mn}^{2+} \) and \( \text{MnO}_4^{-} \) are both 1 M.

2. **Use the Nernst Equation**:
The Nernst equation is given by:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]
where \( n \) is the number of electrons transferred in the reaction. Here, \( n = 5 \).

3. **Calculate \( E_{cell} \) at Initial Conditions**:
When \( [H^+] = 1 \, M \):
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^{-}][H^+]^8} \right) \]
Since both \( [\text{Mn}^{2+}] \) and \( [\text{MnO}_4^{-}] \) are 1 M, the equation simplifies to:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{5} \log \left( \frac{1}{1 \cdot 1^8} \right) = E^{\circ}_{cell} - 0 \]
Thus, \( E_{cell} = E^{\circ}_{cell} \).

4. **Calculate \( E_{cell} \) at New Conditions**:
When \( [H^+] = 10^{-4} \, M \):
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^{-}][H^+]^8} \right) \]
Plugging in the values:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{5} \log \left( \frac{1}{1 \cdot (10^{-4})^8} \right) \]
This simplifies to:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{5} \log \left( 10^{32} \right) \]
Since \( \log(10^{32}) = 32 \):
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059 \times 32}{5} \]
\[ E_{cell} = E^{\circ}_{cell} - 0.3776 \approx E^{\circ}_{cell} - 0.38 \, V \]

5. **Conclusion**:
The potential decreases by approximately 0.38 V when the concentration of \( H^+ \) is decreased from 1 M to \( 10^{-4} \, M \). This indicates a decrease in the oxidizing power of the solution.

### Final Answer:
The potential decreases by 0.38 V with a decrease in oxidizing power.