The magnitude of the change in oxidising power of the `MnO_4^(-)//Mn^(2+)` couple is ` x xx 10^(-4)V`, if the `H^(+)` concentration is decreased from 1M to ` 10^(-4)` M at ` 25^(@)C`. (Assume concentration of `MnO_4^(-) and Mn^(2+)` to be same on change in `H^(+)` concentration ) . The value of x is ___________. (Rounded off to the nearest integer )
[Given : `(2.303 RT)/(F) = 0.059`]

Answer

To solve the problem, we need to determine the change in the oxidizing power of the `MnO4^(-)//Mn^(2+)` couple when the concentration of `H^+` ions is decreased from 1 M to `10^(-4)` M. We will use the Nernst equation for this calculation.

### Step-by-Step Solution:

1. **Write the half-reaction**:
The half-reaction for the reduction of permanganate ion (`MnO4^(-)`) is:
\[ MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O \]

2. **Apply the Nernst equation**:
The Nernst equation for the half-reaction can be expressed as:
\[ E = E^{\circ} - \frac{0.059}{n} \log \left( \frac{[Mn^{2+}]}{[H^{+}]^8} \right) \]
where \( n = 5 \) (the number of electrons transferred).

3. **Calculate the initial and final cell potentials**:
- **Initial condition** (when `[H^+] = 1 M`):
\[ E_1 = E^{\circ} - \frac{0.059}{5} \log \left( \frac{[Mn^{2+}]}{1^8} \right) = E^{\circ} - \frac{0.059}{5} \log \left( [Mn^{2+}] \right) \]

- **Final condition** (when `[H^+] = 10^{-4} M`):
\[ E_2 = E^{\circ} - \frac{0.059}{5} \log \left( \frac{[Mn^{2+}]}{(10^{-4})^8} \right) = E^{\circ} - \frac{0.059}{5} \log \left( [Mn^{2+}] \cdot 10^{32} \right) \]

4. **Subtract the two equations**:
To find the change in potential:
\[ \Delta E = E_2 - E_1 \]
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \Delta E = \left( E^{\circ} - \frac{0.059}{5} \log \left( [Mn^{2+}] \cdot 10^{32} \right) \right) - \left( E^{\circ} - \frac{0.059}{5} \log \left( [Mn^{2+}] \right) \right) \]
Simplifying this gives:
\[ \Delta E = -\frac{0.059}{5} \left( \log \left( [Mn^{2+}] \cdot 10^{32} \right) - \log \left( [Mn^{2+}] \right) \right) \]
\[ \Delta E = -\frac{0.059}{5} \log \left( 10^{32} \right) = -\frac{0.059}{5} \cdot 32 \]

5. **Calculate \(\Delta E\)**:
\[ \Delta E = -\frac{0.059 \cdot 32}{5} = -0.3776 \text{ V} \]

6. **Express \(\Delta E\) in the required format**:
The problem states that the change in oxidizing power is \( x \times 10^{-4} V \). Thus:
\[ -0.3776 = x \times 10^{-4} \]
Rearranging gives:
\[ x = -0.3776 \times 10^{4} = -3776 \]
Since we need the magnitude, we take the absolute value:
\[ x = 3776 \]

### Final Answer:
The value of \( x \) is **3776**.