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 $$Br_2|Br^-$$, 0.54 V for $$I_2|I^-$$. At pH = 3, permanganate is expected to oxidize: $$\left(\frac{RT}{F} = 0.059\right)$$

We begin by noting the given standard reduction potentials at 298 K:

$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O,\qquad E^\circ = 1.51\;{\text V}$$

$$Cl_2 + 2e^- \rightarrow 2Cl^-,\qquad E^\circ = 1.36\;{\text V}$$

$$Br_2 + 2e^- \rightarrow 2Br^-,\qquad E^\circ = 1.07\;{\text V}$$

$$I_2 + 2e^- \rightarrow 2I^-,\qquad E^\circ = 0.54\;{\text V}$$

The permanganate reaction involves protons, so its potential changes with pH. The Nernst equation for any redox half-reaction is first stated:

$$E = E^\circ - \frac{0.059}{n}\log Q,$$

where $$n$$ is the number of electrons and $$Q$$ is the reaction quotient. For the permanganate half-reaction we have $$n = 5$$ and

$$Q = \frac{[Mn^{2+}]}{[MnO_4^-]\,[H^+]^{8}}.$$

We take the standard state concentrations for $$Mn^{2+}$$ and $$MnO_4^-$$ as $$1\;\text M$$, so $$Q = 1/[H^+]^{8}$$. Substituting this into the Nernst equation gives

$$E = 1.51 - \frac{0.059}{5}\log\!\left(\frac{1}{[H^+]^{8}}\right).$$

Because $$\log\!\left(\frac{1}{[H^+]^{8}}\right)= -\log[H^+]^{8} = -8\log[H^+] = 8\,\text{pH},$$ we obtain

$$E = 1.51 - \frac{0.059}{5}\,(8\,\text{pH}).$$

At the specified pH of $$3$$ we set $$\text{pH}=3$$:

$$E = 1.51 - \frac{0.059}{5}\,(8\times 3).$$

Evaluate the coefficient first:

$$\frac{0.059}{5}=0.0118.$$

Now multiply by $$8\times 3 = 24$$:

$$0.0118 \times 24 = 0.2832.$$

Finally, subtract this from the standard potential:

$$E = 1.51 - 0.2832 = 1.2268 \text{ V} \approx 1.23 \text{ V}.$$

This is the actual reduction potential of $$MnO_4^-$$ to $$Mn^{2+}$$ at pH 3.

Next we compare this value with the standard (pH-independent) reduction potentials of the halogen couples:

$$E_{MnO_4^-/Mn^{2+}}(pH=3)=1.23\;\text V,$$

$$E^\circ_{Cl_2/Cl^-}=1.36\;\text V,$$

$$E^\circ_{Br_2/Br^-}=1.07\;\text V,$$

$$E^\circ_{I_2/I^-}=0.54\;\text V.$$

If the permanganate potential is greater than the halogen potential, it can oxidize the corresponding halide ion. We therefore compare:

For chloride: $$1.23\;\text V \lt 1.36\;\text V$$  ⇒  permanganate is weaker than needed ⇒ no oxidation of $$Cl^-$$.

For bromide: $$1.23\;\text V \gt 1.07\;\text V$$  ⇒  permanganate is strong enough ⇒ $$Br^-$$ will be oxidized to $$Br_2$$.

For iodide: $$1.23\;\text V \gt 0.54\;\text V$$  ⇒  permanganate is also strong enough ⇒ $$I^-$$ will be oxidized to $$I_2$$.

Thus, at pH 3, permanganate can oxidize bromide and iodide ions, but not chloride.

Hence, the correct answer is Option B.