Calculate the cell potential for the reaction as written at 25.00 °C , given that [Zn2+]=0.805 M and [Fe2+]=0.0180 M . Use the standard reduction potentials in this table.

Question

Zn(s)+Fe2+(aq)↽−−⇀ Zn2+(aq)+Fe(s)

Answer 1

I'm using the following standard reduction potentials.

Zn²⁺ + 2e- ==> Zn(s) Eº = -0.76 V (anode, oxidation)

Fe²⁺ + 2e- ==> Fe(s) Eº = -0.44 V (cathode, reduction)[

Eº = -0.44 -(-0.76) = 0.32

Use the Nernst equation to correct the cell potential for the given conditions. Since the system is @ 25ºC, we can use

E_cell = Eº - 0.0592/n log Q

n = 2 moles electrons

Q = [Zn²⁺] / [Fe²⁺] = 0.805 / 0.0180 = 44.7

Ecell = 0.32 - 0.0592/2 log 44.7 = 0.32 - 0.0488

Ecell = 0.27 V