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 of how the oxidation power of the MnO4^−/Mn^2+ couple decreases when the H^+ concentration is decreased from 1 M to 10^(-4) M at 25°C, we will use the Nernst equation. Here’s a step-by-step breakdown of the solution:

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

### Step 2: Identify the standard reduction potential
The standard reduction potential (E°) for this half-reaction is typically known from standard tables. For MnO4^− to Mn^2+, E° is approximately +1.51 V.

### Step 3: Apply the Nernst equation
The Nernst equation is given by:
\[ E = E° - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Where:
- E° is the standard reduction potential,
- n is the number of electrons transferred (which is 5 in this case),
- [products] and [reactants] are the concentrations of the products and reactants.

### Step 4: Substitute the concentrations into the Nernst equation
Initially, when [H^+] = 1 M:
\[ E_1 = E° - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][\text{H}^+]^8} \right) \]

When [H^+] is decreased to 10^(-4) M:
\[ E_2 = E° - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][10^{-4}]^8} \right) \]

### Step 5: Calculate the change in potential
The change in potential (ΔE) can be calculated as:
\[ \Delta E = E_2 - E_1 \]

Substituting the values:
\[ E_1 = E° - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][1]^8} \right) \]
\[ E_2 = E° - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-}[10^{-32}]} \right) \]

Now, we can simplify:
\[ \Delta E = \left( E° - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-}[10^{-32}]} \right) \right) - \left( E° - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][1]} \right) \right) \]

The E° terms cancel out, and we are left with:
\[ \Delta E = -\frac{0.0591}{5} \left( \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-}[10^{-32}]} \right) - \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][1]} \right) \right) \]

This simplifies to:
\[ \Delta E = -\frac{0.0591}{5} \left( \log(10^{32}) \right) \]
\[ \Delta E = -\frac{0.0591}{5} \times 32 \]

### Step 6: Calculate the numerical value
Calculating this gives:
\[ \Delta E = -0.37872 \text{ V} \]

### Conclusion
Thus, the oxidation power of the MnO4^−/Mn^2+ couple decreases by approximately 0.38 V when the H^+ concentration is decreased from 1 M to 10^(-4) M.

---