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⁺] = 0.035 M and E = 0.78 V? Assume T is 298 K

Question

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⁺] = 0.035 M and E = 0.78 V? Assume T is 298 K

Answer 1

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

We are asked to determine the [Fe²⁺] under non-standard conditions (i.e. when Fe and Mn are not @ 1M). To do this we will want to employ the Nernst equation:

E_cell = Eº_cell - RT/nF ln Q

and at 25ºC (298K) and converting natural log to log base 10, this equation becomes

E_cell = Eº_cell - 0.0591/n log Q

E_cell = 0.78 V

Eº_cell = 0.77 V

n = moles electrons transferred = 2

Q = [Fe²⁺] / [Mn²⁺] = [Fe²⁺] /0.035

Solving for [Fe²⁺].....

0.78 = 0.77 - 0.0591/2 log [Fe²⁺] /0.035

0.78 = 0.77 - 0.02955 log [Fe²⁺] /0.035

0.01 = - 0.02955 log [Fe²⁺] /0.035

-0.3384 = log [Fe²⁺] /0.035

0.459 = [Fe²⁺] /0.035

[Fe²⁺] = 0.0161 M