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⁺] = 0.050 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⁺] = 0.050 M and E = 0.78 V? Assume T is 298 K

Answer 1

For this problem, we will need to use the Arrhenius equation:

E_cell = Eº_cell - RT / nF ln Q and since the cell is @298K this reduces to

E_cell = Eº_cell - 0.0592 / n log Q

E_cell = 0.78 V

Eº_cell = 0.77 V

n = 2 moles of electrons

Q = [Fe²⁺] / [Mn²⁺]

Plugging in values and solving for Q:

0.78 = 0.77 - 0.0592 / 2 log Q

0.78 = .077 - 0.0296 log Q

log Q = -0.3378

Q = 0.4594 = [Fe²⁺] / [Mn²⁺]

Solving for [Fe²⁺]

0.4594 = [Fe²⁺] / [Mn²⁺]

0.4594 = [Fe²⁺] / [0.05 M]

[Fe²⁺] = 0.023 M

(be sure to check all of the math)