At 298 K, the standard reduction potentials are 1.51 V for `MnO_(4)^(-) | Mn^(2+), 1.36 V` for `Cl^(2) | Cl^(-)`, 1.07 V for `Br_(2)|Br^(-),` and 0.54 V for `I_(2)|I^(-)`. At pH=3, permanganate is expected to oxidize `((RT)/(F) = 0.059V)`:

Options

A. `Cl^(-)` and `Br^(-)`
B. `Cl^(-)`,`Br^(-)` and `I^(-)`
C. `Br^(-)` and `I^(-)`
D. `I^(-)` only

Answer

To solve the problem, we need to determine whether permanganate ion (MnO₄⁻) can oxidize the other species at pH 3. We will use the Nernst equation to adjust the standard reduction potential for the conditions given.

### Step-by-Step Solution:

1. **Identify the Standard Reduction Potentials:**
- For MnO₄⁻ to Mn²⁺: E° = 1.51 V
- For Cl₂ to Cl⁻: E° = 1.36 V
- For Br₂ to Br⁻: E° = 1.07 V
- For I₂ to I⁻: E° = 0.54 V

2. **Adjust the Standard Reduction Potential for pH = 3:**
- The Nernst equation is given by:
\[ E = E° - \frac{RT}{nF} \ln Q \]
- At 298 K, \( \frac{RT}{F} = 0.059 \) V (for n = 1).
- The concentration of H⁺ at pH 3 is:
\[ [H^+] = 10^{-3} \text{ M} \]

3. **Calculate the adjusted potential for MnO₄⁻:**
- The half-reaction for MnO₄⁻ is:
\[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \]
- Here, n = 5 (number of electrons transferred).
- Adjusting the potential:
\[ E = 1.51 - \frac{0.059}{5} \log \left(\frac{1}{[H^+]^8}\right) \]
- Substitute [H⁺] = 0.001 M:
\[ E = 1.51 - \frac{0.059}{5} \log \left(\frac{1}{(0.001)^8}\right) \]
- Simplifying:
\[ E = 1.51 - \frac{0.059}{5} \log(10^8) \]
- Since \(\log(10^8) = 8\):
\[ E = 1.51 - \frac{0.059 \times 8}{5} \]
- Calculate:
\[ E = 1.51 - 0.0944 = 1.4156 \text{ V} \]

4. **Compare with Other Reduction Potentials:**
- Now we compare this adjusted potential of 1.4156 V with the reduction potentials of Cl₂, Br₂, and I₂.
- Since 1.4156 V > 1.36 V (Cl₂), 1.4156 V > 1.07 V (Br₂), and 1.4156 V > 0.54 V (I₂), MnO₄⁻ can oxidize all three species.

### Conclusion:
At pH 3, permanganate (MnO₄⁻) is expected to oxidize Cl⁻, Br⁻, and I⁻ due to its higher adjusted reduction potential.

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