If the `H^(+)` concentration is decreased from 1 M to `0^(-4)M` at `25^(@)C` for the couple `MnO_(4)^(-) ` then the oxidation power of the `MnO_(4)^(-)/ Mn^(2+)` couple decrease by :

Options

A. `(0.18V)`
B. `(0.18V)`
C. `(0.38)`
D. `(0.38V)`

Explanation

To solve the problem, we need to determine how the oxidation power of the MnO4^-/Mn^2+ couple changes when the concentration of H^+ is decreased from 1 M to 10^(-4) M at 25°C. We will use the Nernst equation to find the change in cell potential (E).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The half-reaction for the MnO4^-/Mn^2+ couple in acidic medium is:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

2. **Determine the Number of Electrons (n)**:
From the balanced equation, we see that 5 electrons are involved in the reaction. Thus, \( n = 5 \).

3. **Write the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^0 - \frac{0.0591}{n} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][\text{H}^+]^8} \right) \]
where \( E^0 \) is the standard reduction potential.

4. **Calculate E1 (when [H^+] = 1 M)**:
When the concentration of H^+ is 1 M, we can express E1 as:
\[ E_1 = E^0 - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][1]^8} \right) \]

5. **Calculate E2 (when [H^+] = 10^(-4) M)**:
When the concentration of H^+ is decreased to 10^(-4) M, we express E2 as:
\[ E_2 = E^0 - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][10^{-4}]^8} \right) \]

6. **Find the Change in Potential (E2 - E1)**:
Now, we calculate the difference:
\[ E_2 - E_1 = \left( E^0 - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][10^{-4}]^8} \right) \right) - \left( E^0 - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][1]^8} \right) \right) \]
Simplifying this gives:
\[ E_2 - E_1 = -\frac{0.0591}{5} \left( \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][10^{-4}]^8} \right) - \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][1]^8} \right) \right) \]
This can be further simplified using the properties of logarithms:
\[ E_2 - E_1 = -\frac{0.0591}{5} \left( \log \left( \frac{10^{-32}}{1} \right) \right) = -\frac{0.0591}{5} \times (-32) \]

7. **Calculate the Final Value**:
\[ E_2 - E_1 = \frac{0.0591 \times 32}{5} = \frac{1.8912}{5} \approx 0.37824 \text{ volts} \]
Thus, the decrease in oxidation power is approximately:
\[ \Delta E \approx -0.38 \text{ volts} \]

### Final Answer:
The oxidation power of the MnO4^-/Mn^2+ couple decreases by approximately **0.38 volts**.