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⁺] = 0.022 M? Assume T is 298 K

Answer 1

Nernst equation:

Ecell = Eºcell - RT/nF ln Q and at 298K and converting to log₁₀ it is...

Ecell = Eºcell - 0.0591 / n log Q

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

Ecell = ?

Eºcell = 0.48

n = 6 (since 6 mols of electrons were transferred)

Q = [Al³⁺]² / [Mn²⁺]³

Ecell = Eºcell - 0.0591 / n log Q

Ecell = 0.48 - 0.0591/6 log (1² / 0.022³ = 0.48 - (0.00985 x 4.97)

Ecell = 0.48 - 0.05

Ecell = 0.43 V

(be sure to check for any math errors)