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⁺] = 0.050 M? Assume T is 298 K

Question

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⁺] = 0.050 M? Assume T is 298 K

Answer 1

Ecell = Eºcell - 0.0592/n log Q

Ecell = 0.48 - 0.0592/6 log [Al³⁺]² / [Mn²⁺]³

Ecell = 0.48 - 0.00987 log (1)² / (0.05)³

Ecell = 0.48 - (0.00987)(3.90)

Ecell = 0.48 - 0.038

Ecell = 0.44 V