Use the Nernst Equation to calculate the cell potentials of the following cells at 298 K

Question

Use the Nernst Equation to calculate the cell potentials of the following cells at 298 K

Answer 1

E_cell = E_ºcell -RT/nF ln Q

At 298K this can be written as

E_cell = Eº_cell - 0.0592/n log Q

I'll do (a) and hopefully you'll be able to do part (b) in a similar fashion.

We need to find Eº, by using standard reduction potentials. We also need to find Q and log Q (see below)

Pb²⁺ + 2e- ==> Pb Eº = -0.126 V

Ag⁺ + e- ==> +0.7996 V

Eº = 0.7996 + 0.126 = 0.926 V

Q = reaction quotient = [Pb²⁺] / [Ag⁺]² = (0.30 M) / (0.20 M)² =7.5

E_cell = Eº_cell - 0.0592/n log Q where n is moles of electrons = 2

Ecell = 0.926 - (0.0592/2) log 7.5

Ecell = 0.926 - 0.0259

Ecell = 0.900 V