Experiment: Part I: Voltaic Cell
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.
Which electrode was the anode and which was the Cathode?
The anode and cathode half reactions.
Balanced equation for each cell you propose to construct.
Calculated Eocell
Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed)

Here are four possible voltaic cells.  

Explanation:

1. Standard reduction potentials

                                         E°/V

I₂(s) + 2e⁻ ⟶ 2I⁻(aq);        0.54

Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);   0.34

Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);   -0.41

Zn²⁺(aq) + 2e⁻ ⟶ Zn(s);   -0.76

2. Possible Voltaic cells

(a) Zn/I₂

                                                                       E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                 0.76

Cathode:  I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 1.30

Zn(s)|Zn²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Zn is the anode; graphite is the cathode.

(b) Zn/Cu²⁺

                                                                          E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                    0.76

Cathode:  Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);                  0.34

Cell:          Zn(s) +  Cu²⁺(s) ⟶  Zn²⁺(aq) + Cu(s); 1.10

Zn(s)|Zn²⁺(aq)∥Cu²⁺(aq)|Cu(s)

Zn is the anode; Cu is the cathode.

(c) Zn/Fe²⁺

                                                                            E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                     0.76

Cathode:  Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);                    -0.41

Cell:          Zn(s) +  Fe²⁺(s) ⟶  Zn²⁺(aq) + Fe(s);  0.35

Zn(s)|Zn²⁺(aq)∥Fe²⁺(aq)|Fe(s)

Zn is the anode; Fe is the cathode.

(d) Fe/I₂

                                                                         E°/V

Anode:     Fe(s) ⟶ Fe²⁺(aq) + 2e⁻;                   0.41

Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 0.95

Fe(s)|Fe²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Fe is the anode; graphite is the cathode.

i just got this problem. it's 90.3

b.

the accuracy of the data.