Question

Consider the following redox problem:

MnO4 + 8H+ + e- ----> Mn2+ + 4H2O E0 =1.51V

a) Calculate the redox potential (non-standard conditions) if the concentration of MnO4 is

0.1M and for Mn2+ is 0.001M, and the pH is 5.

b) For a concentration of MnO4 0.1M and Mn2+ 0.0001M, what must be the pH of the solution to give a value of E = 1.15V?

Answer 1

a)

E cell = E0 cell - 2.303RT/nF log[Mn+2]*[H2O]/[MnO4]*[H+]

we have [MnO4] = 0.1 and [Mn+2] = 0.001

pH = 5

-log[H+] = 5

[H+} = 10^-5

R = 8.314 and T = 298K

n = 5 and F = 96500 coloumbs

E cell = 1.51 - ((2.303*8.314*298/(5*96500))*log(0.001/(0.1 * 10^-5)))

E cell = 1.4745 V

b)

E cell = E0 cell - 2.303RT/nF log[Mn+2]*[H2O]/[MnO4]*[H+]

1.15 = 1.51 - ((2.303*8.314*298/(5*96500))*log(0.0001/(0.1 * [H+]))

[H+] = 7.1123 * 10^-11

pH = -log[H+] = -log(7.1123 * 10^-11)

pH = 10.148