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 = 0.010 V? Assume T is 298 K

Answer 1

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

Ecell = 0.010

Eº = 0.0400

n = 2 moles electrons

Q = [Fe²⁺] / [Cd²⁺] = ?

0.010 = 0.0400 - 0.0296 log Q

-0.03 = -0.0296 log Q

log Q = 1.0135

Q = 0.0969 = [Fe²⁺] / [Cd²⁺]

(be sure to check the math)