Answers to problems in "Electrode Potentials"

Question 1

The cell diagram is:

Zn(s) | Zn2+(aq), (a=1) || Cu2+(aq), (a=1) | Cu(s)

This is the Daniel cell.

Potential determining equilibria:

at RH Electrode: ½Cu2+(aq) + e- ⇌ ½ Cu(s) E0Cu|Cu2+ = +0.34V
at LH Electrode: ½Zn2+(aq) + e- ⇌ ½ Zn(s) E0Zn|Zn2+ = -0.76V

Formal Cell Reaction: ½Cu2+(aq) + ½Zn(s) ⇌ ½Zn2+(aq) + ½Cu(s)

E0cell = E0Cu|Cu2+ - E0Zn|Zn2+ - {RT/2F} ln(aZn2+ /aCu2+)

E0cell = 0.34 - (-0.76) = +1.10V

E0cell > 0 Zinc reduces a copper(II) solution to metallic copper.