For the electrochemical cell 2 Al(s) + 3 Mn²⁺(aq) ⟶ 2 Al³⁺(aq) + 3 Mn(s) (E° = 0.48 V, [Al³⁺] = 1.0 M), what is the value of E when [ Mn²⁺] = 0.072 M? Assume T is 298 K

Answer 1

2 Al(s) + 3 Mn²⁺(aq) ⟶ 2 Al³⁺(aq) + 3 Mn(s) ... Eº = 0.48 V

Nernst equation (when T = 298K):

Ecell = Eº - 0.0592/n log Q

Eº = 0.48

n = 6

Q = [Al³⁺]² / [Mn²⁺]³ = (1)² / (0.072)³ = 2679

Ecell = 0.48 - (0.0592/6) log 2679 = 0.48 - 0.034

Ecell = 0.446 = 0.45 V