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 = 0.002 V? Assume T is 298 K

Answer 1

Cd²⁺(aq) + Fe(s) ⟶ Cd(s) + Fe²⁺(aq) ... Eº = 0.0400 V

Nernst equation when T = 298K:

Ecell = Eº - 0.0592/n log Q

Ecell = 0.002 V

Eº = 0.0400 V

n = 2

Q = [Fe²⁺] / [Cd²⁺] = ?

0.002 = 0.0400 - (0.0592/2) log [Fe²⁺] / [Cd²⁺]

0.038 = 0.0296 log [Fe²⁺] / [Cd²⁺]

log [Fe²⁺] / [Cd²⁺] = 1.284

[Fe²⁺] / [Cd²⁺] = 19.2