We are asked to **calculate the equilibrium constant for each of the following reaction at 298 K**

**10Br ^{-}_{ (aq)} + 2MnO_{4} ^{-}_{ (aq)} + 16H^{+} _{ (aq)} → 2Mn^{2 + } _{ (aq)} + 8H_{2} O_{ (l)} + 5Br_{2}**

We can solve this using the formula below:

$\overline{){\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}\mathbf{}\mathbf{V}}{\mathbf{n}}{\mathbf{log}}{\mathbf{}}{\mathbf{K}}}$

E_{cell} = cell potential under non-standard conditions

E°_{cell} = standard cell potential

n = number of e^{-} transferred

K = equilibrium constant = [products]/[reactants]

Using the standard reduction potentials listed in Appendix E in the textbook, calculate the equilibrium constant for each of the following reactions at 298 K.

10Br^{-} (aq) + 2MnO_{4} ^{-} (aq) + 16H^{+} (aq) → 2Mn^{2 + } (aq) + 8H_{2} O(l) + 5Br_{2} (l)