Assume that you are provided with the following materials: • Strips of metallic zinc, metallic copper, metallic iron
• 1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine(I2)
• Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos, identify 4 different combinations of Voltaic cells that are possible to be created with the above materials. For each cell created, include the following details.
A) Which electrode was the anode, and which was the Cathode?
B) The anode and cathode half reactions.
C) Balanced equation for each cell you propose to construct.
D) Calculated Eocelle Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed).

See explanation

Explanation:

First voltaic cell;

Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Zinc

Cathode;

Copper

Oxidation half equation;

Zn(s)> Zn^2+(aq) + 2e

Reduction half equation;

Cu^2+(aq) +2e > Cu(s)

Overall; Zn(s) + Cu^2+(aq) > Zn^2+(aq) + Cu(s)

E°cell = 0.34 -(-0.76) =1.1 V

Second voltaic cell;

Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)

Anode;

Zinc

Cathode;

Iron

Oxidation half equation;

Zn(s)> Zn^2+(aq) + 2e

Reduction half equation;

Fe^2+(aq) +2e > Fe(s)

Overall; Zn(s) + Fe^2+(aq) > Zn^2+(aq) + Fe(s)

E°cell = (-0.44) -(-0.76) = 0.32 V

Third voltaic cell;

Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Iron

Cathode;

Copper

Oxidation half equation;

Fe(s)> Fe^2+(aq) + 2e

Reduction half equation;

Cu^2+(aq) +2e > Cu(s)

Overall; Fe(s) + Cu^2+(aq) > Fe^2+(aq) + Cu(s)

E°cell = 0.34 -(-0.44) = 0.78 V

Fourth voltaic cell

Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)

Anode;

Copper

Cathode;

Graphite rod

Oxidation half equation;

Cu(s)> Cu^2+(aq) + 2e

Reduction half equation;

I2(aq) +2e > 2I^-(aq)

Overall; Cu(s) + I2(aq) > Cu^2+(aq) + 2I^-(aq)

E°cell = 0.54 -0.34 = 0.20 V

fe 55.85 g/mol and h2o 18.02 g/mol.