$E^{\circ} (MnO X_{4} X^{-}, H X^{+} / Mn X^{2 +}) = 1.51 V$ and $E^{\circ} (Br X_{2} / Br X^{-}) = 1.07 V$ at 298 K. With $[MnO X_{4} X^{-}] = [Mn X^{2 +}] = [Br X^{-}] = 1 M$ and $Br X_{2}$ as pure liquid, the pH above which permanganate can no longer oxidise bromide (both potentials become equal) is (two decimals; take $2.303 RT / F = 0.0591 V$ ).

Numerical answer type, tolerance 0.02.

Answer: 4.65

The

Br X_{2} / Br X^{-}

potential is pH independent and stays at 1.07 V. For permanganate at unit concentrations,

E = 1.51 - \frac{8 \times 0.0591}{5} pH = 1.51 - 0.0946 pH

. Setting

1.51 - 0.0946 pH = 1.07

gives

pH = 0.44/0.0946 = 4.65

. Above this pH,

E (MnO X_{4} X^{-}) < E (Br X_{2})

and the oxidation of

Br X^{-}

is no longer spontaneous.