What is the cell potential for the following cell at 25°C?

Question

Pt(s)|Sn2+(aq)(0.001M),Sn4+(aq)(0.005M)||Ag+(aq)(0.500M)|Ag(s)

Answer 1

Looking at the two reduction reactions and the associated reduction potentials:

Sn⁴⁺(aq) + 2e- ==> Sn²⁺(aq) Eº = +0.14 V (anode, oxidation)

Ag⁺(aq) + e- ==> Ag(s) Eº = +0.80 V (cathode, reduction)

Overall redox reaction: 2Ag⁺(aq) + Sn²⁺(aq) ==> 2Ag(s) + Sn⁴⁺(aq) with transfer of 2 mols electrons

Eº_cell = 0.80 - 0.14 = 0.66 V

Since the concentrations of the ions are not standard (1 M), we must use the Nernst equation to find the cell potential.

Nernst Equation

E_cell = Eº_cell - RT / nF ln Q

Since the temperature is 25ºC and R and F are constants we can write this as

E_cell = Eº_cell - 0.0592 / n log Q (also have changed ln Q to log Q)

Q = [Sn4+] / [Ag+]² [Sn2+] = (0.005) / (0.500)²(0.001) = 20

E_cell = 0.66 - (0.0592/2) log 20

E_cell = 0.66 - (0.0296)(1.30) = 0.66 - 0.0385

E_cell = 0.62 V

(be sure to check the math)