For the galvanic (voltaic) cell Fe(s) + Mn²⁺(aq) ⟶ Fe²⁺(aq) + Mn(s) (E° = 0.77 V at 25 °C), what is [Fe²⁺] if [Mn²⁺] = 0.040 M and E = 0.78 V? Assume T is 298 K

Answer 1

Fe(s) + Mn²⁺(aq) ==> Fe²⁺(aq) + Mn(s) .. Eº = 0.77

When conditions are anything but normal (1 M Fe²⁺ and 1 M Mn²⁺ @ 298K) we use the Arrhenius equation:

Ecell = Eº - RT / nF ln Q and at 298K and substituting for log_10, we have

Ecell = Eº - 0.0592 / n log Q

Ecell = 0.78

Eº = 0.77

n = 2 mols electrons

Q = [Fe²⁺] / [Mn²⁺] = ?

0.78 = 0.77 - 0.0296 log Q

log Q = -0.3378

Q = [Fe²⁺] / [Mn²⁺] = 0.459

[Fe²⁺] / [0.04] = 0.459

[Fe²⁺] = 0.018 M

(be sure to check all of the math)