For the galvanic (voltaic) cell Cd²⁺(aq) + Fe(s) ⟶ Cd(s) + Fe²⁺(aq) (E° = 0.0400 V), what is the ratio [Fe²⁺]/[Cd²⁺] when E = 0.007 V? Assume T is 298 K

Question

Explain please.

Answer 1

At 298K we can write the Nernst equation as Ecell = Eº - 0.0592/n log Q

Ecell = 0.010

Eº = 0.007

n = 2 moles electrons

Q = [Fe²⁺] / [Cd²⁺] = ?

0.007= 0.0400 - 0.0296 log Q

-0.033 = -0.0296 log Q

log Q = 1.115

Q = 13.0 = [Fe²⁺] / [Cd²⁺]