Question

Calculate the cell potential for a cell operating with the following reaction at 25 degrees Celsius, in which [MnO4^1-] = .01M, [Br^1-] = .01M, [Mn^2+] = .15M, and [H^1+] = 1M. The reaction is 2 MnO4^1-(aq) + 10 Br^1-(aq) + 16 H^1+(aq) --> 2 Mn^2+(aq) + 5 Br2(l) + 8 H2O(l)

Answer 1

The two half cell reactions are

Oxidation reaction

10Br- = 5Br2 + 10e-

Eox = - 1.07 V

Reduction reaction

2MnO4- + 16H+ + 10e- = 2Mn2+ + 8H2O

Ered = 1.51 V

Overall cell reaction

10Br- + 2MnO4- + 16H+ = 5Br2 + 2Mn2+ + 8H2O

E°cell = Eox + Ered

= - 1.07 + 1.51

= 0.44 V

From the Nernst equation

E = E°cell - (0.0592/2) log ( [products] / [reactants])

E = E°cell - (0.0592/2) log ( [Mn2+]² / [H+]¹⁶ [MnO4-]² [Br-]¹⁰)

E = 0.44 - (0.0592/2) log ( [0.15]² / [1]¹⁶ [0.01]² [0.01]¹⁰)

E = 0.44 - (0.0592/2) log (2.25*10^22)

E = - 0.222 V