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. `C1^(-),Br^(-)and I^(-)`
B. `C1^(-) and Br^(-)`
C. `Br^(-)and I^(-)`
D. `I(-) only`

Explanation

To solve the problem step by step, we will analyze the given standard reduction potentials and determine which species can be oxidized by permanganate ion (MnO4^-) at pH 3.

### Step 1: Write the half-reaction for permanganate reduction
The half-reaction for the reduction of permanganate ion (MnO4^-) to manganese ion (Mn2+) in acidic medium is:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
The standard reduction potential (E°) for this reaction is given as 1.51 V.

### Step 2: Use the Nernst equation
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