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(B) The reduction half-reaction for permanganate is: $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$. Using the Nernst equation: $E = E^\circ - \frac{0.05

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$Br^{-}$ and $I^{-}$

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(B) The reduction half-reaction for permanganate is: $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$. Using the Nernst equation: $E = E^\circ - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$. Assuming standard concentrations for $Mn^{2+}$ and $MnO_4^-$ $(1 \ M)$,the potential at $pH = 3$ is: $E = 1.51 - \frac{0.059}{5} 	imes \log \frac{1}{(10^{-3})^8} = 1.51 - \frac{0.059 	imes 24}{5} = 1.51 - 0.2832 = 1.2268 \ V$. Permanganate can oxidize species whose standard reduction potential is lower than its own potential $(1.2268 \ V)$. Comparing this with the given potentials: $Cl_2/Cl^- = 1.36 \ V$,$Br_2/Br^- = 1.07 \ V$,and $I_2/I^- = 0.54 \ V$. Since $1.2268 \ V > 1.07 \ V$ and $1.2268 \ V > 0.54 \ V$,but $1.2268 \ V < 1.36 \ V$,permanganate will oxidize $Br^-$ and $I^-$.

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 $\left( \frac{RT}{F} = 0.059 \ V \right)$

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