Answers to problems in "Electrode Potentials"
Question 2
The cell diagram is:
The EMF if E^{0}_{cell} = +0.912 V.
Potential determining equilibria:
at RH Electrode: | Ag^{+}(aq) + e^{-} ⇌ Ag(s) | E^{0}_{Ag|Ag+} = ? |
at LH Electrode: | ½Pb^{2+}(aq) + e^{-} ⇌ ½Pb(s) | E^{0}_{Pb|Pb2+} = -0.130V |
Formal Cell Reaction:
Ag^{+}(aq) + ½Pb(s) ⇌ ½Pb^{2+}(aq) + Ag(s) |
Nernst Equation: E_{cell} = E^{0}_{cell }+ {RT/F}ln{a_{Ag+} a_{Pb}^{½}/a_{Ag} a_{Pb2+}^{½}}
Lead and silver are both pure solids at room temperature (298K), and therefore the activities of these species are unity.
Hence, 0.912 = E^{0}_{cell} + (8.313 x 298/ 96 490)ln{0.5/1^{½}}
ie., E^{0}_{cell} = E^{0}_{RHE} - E^{0}_{LHE} = E^{0}_{Ag|Ag+ }+ 0.130 = +0.930V
Therefore, E^{0}_{Ag|Ag+
}= +0.800V
This result is consistent with data given in Table 1.1 on page 21.