Answers to problems in "Electrode Potentials"
Question 1
The cell diagram is:
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.