By how much is the oxidizing power of the `MnO_4//Mn^(+2)` couple decreased if the `H^+` concentration is decreased by 100 at `25^@C?` Assume other species have no change in concentration.

Answer

To solve the problem, we will use the Nernst equation to determine how the oxidizing power of the `MnO4^- / Mn^(2+)` couple changes when the concentration of `H^+` ions is decreased.

### Step-by-Step Solution:

1. **Understand the Reaction**: The half-reaction for the reduction of permanganate ion (`MnO4^-`) to manganese ion (`Mn^(2+)`) in acidic medium is:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

2. **Nernst Equation**: The Nernst equation for this reaction can be expressed as:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][\text{H}^+]^8} \right) \]
where \( n = 5 \) (the number of electrons transferred).

3. **Initial Conditions**: Let the initial concentration of `H^+` be \( [H^+] = X \). The initial reduction potential \( E_{initial} \) can be expressed as:
\[ E_{initial} = E^\circ - \frac{0.059}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-] \cdot X^8} \right) \]

4. **Final Conditions**: If the concentration of `H^+` is decreased by a factor of 100, the final concentration becomes:
\[ [H^+]_{final} = \frac{X}{100} \]
The final reduction potential \( E_{final} \) can be expressed as:
\[ E_{final} = E^\circ - \frac{0.059}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-] \cdot \left(\frac{X}{100}\right)^8} \right) \]

5. **Calculate the Change in Potential**: The change in potential \( \Delta E \) can be calculated as:
\[ \Delta E = E_{final} - E_{initial} \]
Substituting the expressions for \( E_{final} \) and \( E_{initial} \):
\[ \Delta E = \left( E^\circ - \frac{0.059}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-] \cdot \left(\frac{X}{100}\right)^8} \right) \right) - \left( E^\circ - \frac{0.059}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-] \cdot X^8} \right) \right) \]
This simplifies to:
\[ \Delta E = -\frac{0.059}{5} \left( \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-] \cdot \left(\frac{X}{100}\right)^8} \right) - \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-] \cdot X^8} \right) \right) \]

6. **Use Logarithm Properties**: Using the property of logarithms:
\[ \Delta E = -\frac{0.059}{5} \left( \log \left( \frac{100^8}{1} \right) \right) \]
Since \( 100 = 10^2 \), we have:
\[ \Delta E = -\frac{0.059}{5} \cdot 8 \cdot 2 = -\frac{0.059 \cdot 16}{5} \]

7. **Calculate the Value**:
\[ \Delta E = -\frac{0.944}{5} = -0.1888 \text{ V} \approx -0.189 \text{ V} \]

8. **Conclusion**: The oxidizing power of the `MnO4^- / Mn^(2+)` couple decreases by approximately **0.189 V** when the concentration of `H^+` is decreased by a factor of 100.