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)

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Cell Potential concept. If you need more Cell Potential practice, you can also practice Cell Potential practice problems.