CHEM1012 Worksheet 8 โ Answers to Critical Thinking Questions
The worksheets are available in the tutorials and form an integral part of the learning outcomes and
experience for this unit.
Model 1: Reduction Potentials
1. Oxidising agents are themselves reduced โ the strongest oxidising agent is the most easily reduced. This is
Ag+(aq) as it has the most positive Eredยฐ value (it has the strongest attraction to electrons).
2. Reaction (1) will remain a reduction. Reaction (2) will reverse to become an oxidation, as Ag+(aq) is the
strongest oxidising agent.
3. Reaction (3) will remain a reduction. Reaction 4 will reverse to become an oxidation as Zn2+(aq) is the
stronger reducing agent. It does not matter that they are both negative as it is the difference between the two
E0red values which determines the reaction.
Model 2: Voltaic Cells
1. The Zn/Zn2+ half reaction is proceeding as an oxidation as it has a lower E0red value than that for Cu2+/Cu.
When we flip a reduction to an oxidation, we reverse the sign of the potential.
2. The Zn electrode will erode and the Cu electrode will gain mass.
3. Oxidation takes place at anode. Reduction takes place at cathode
4. Electrons flow from the anode, the Zn metal, to the cathode. As the Zn atoms lose electrons to form Zn2+ ions,
the electrons flow through the wire and to the cathode where they combine with Cu2+ to form Cu.
5. Anode is negative, cathode is positive
6. The overall charge in each compartment must be equalised. Cations are being made in the anode department
and lost in the cathode compartment, so to balance the charges cations move from anode to cathode and
anions in the other direction. You can also think of it as โcompleting the circuitโ โ electrons flow from anode
to cathode through the wire, then the negative anions flow form cathode to anode through the solution.
7. Cathode - reduction: Ag+(aq) + eโโAg(s). Anode - oxidation: Cu(s) โ Cu2+ (aq) + 2eโ
E0cell = [0.80 +(-0.34)] V = 0.46 V
8. Cathode - reduction: Cu2+(aq) + 2eโโCu(s). Anode - oxidation: Ni(s) โ Ni(aq) + 2eโ
E0cell = [0.34 + 0.25] V = 0.59 V
9. Cathode - reduction: Ni2+(aq) + 2eโโNi(s). Anode -oxidation: ๐๐ โ ๐๐2+ + 2๐โ
E0cell = [-0.25 + 0.76] V = 0.51 V
10. Ag+(aq) + eโโAg(s) and Zn(s) โ Zn2+(aq) + 2eโ; E0cell = 1.56 V
Model 3: Nernst Equation
1. 1.0 V
2. The voltmeter reading goes to zero: the battery is dead.
3. Put a current through it.
Ag+(aq) + eโ โ Ag(s)
Cu2+(aq) + 2eโ โ Cu(s)
Cu(s) โ Cu2+(aq) + 2eโ
Ni(s) โ Ni(aq) + 2eโ
Ni2+(aq) + 2eโ โ Ni(s)
๐๐ โ ๐๐2+ + 2๐โ
E0cell = [0.80 +(-0.34)] V = 0.46 V
E0cell = [0.34 + 0.25] V = 0.59 V
E0cell = [-0.25 + 0.76] V = 0.51 V
