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⁺] = 0.046 M? Assume T is 298 K

Question

I got -0.0108 using Nernst equation but got it wrong.

Answer 1

2Al(s) + 3Mn²⁺(aq) ==> 2Al³⁺(aq) + 3Mn(s)

Nernst equation: E_cell = Eº - RT/nF lnQ and when T=298 we can reduce this to

E_cell = Eº - 0.0592/n log Q

E_cell = ?

n = 3 moles electrons

Q = [Al³⁺]² / [Mn²⁺]³ = (1.0) / (0.046)³ = 10,274

E_cell = 0.48 - 0.0592/3 log 10,274

E_cell = 0.48 - (0.0197 x 4.01)

E_cell = 0.48 - 0.079

E_cell = 0.40

(be sure to check the math)