Answers to problems in "Electrode Potentials"

Question 2

The cell diagram is:

Pb(s) | Pb2+(aq), (a=1) || Ag+(aq), (a=1) | Ag(s)

The EMF if E0cell = +0.912 V.

Potential determining equilibria:

at RH Electrode: Ag+(aq) + e- ⇌ Ag(s) E0Ag|Ag+ = ?
at LH Electrode: ½Pb2+(aq) + e- ⇌ ½Pb(s) E0Pb|Pb2+ = -0.130V

Formal Cell Reaction:

Ag+(aq) + ½Pb(s) ⇌ ½Pb2+(aq) + Ag(s)

Nernst Equation: Ecell = E0cell + {RT/F}ln{aAg+ aPb½/aAg aPb2+½}

Lead and silver are both pure solids at room temperature (298K), and therefore the activities of these species are unity.

Hence, 0.912 = E0cell + (8.313 x 298/ 96 490)ln{0.5/1½}

ie., E0cell = E0RHE - E0LHE = E0Ag|Ag+ + 0.130 = +0.930V

Therefore, E0Ag|Ag+ = +0.800V

This result is consistent with data given in Table 1.1 on page 21.