Answer:
Here are four possible voltaic cells.
Explanation:
1. Standard reduction potentials
<u> E°/V
</u>
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
<u> E°/V
</u>
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
<u>Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq); </u> <u> 0.54
</u>
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⁺
<u> E°/V
</u>
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: <u>Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); </u> <u>0.34
</u>
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⁺
<u> E°/V</u>
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: <u>Fe²⁺(aq) + 2e⁻ ⟶ Fe(s); </u> <u>-0.41
</u>
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
<u> E°/V
</u>
Anode: Fe(s) ⟶ Fe²⁺(aq) + 2e⁻; 0.41
Cathode: <u> I₂(s) + 2e⁻ ⟶ 2I⁻(aq); </u> <u> 0.54</u>
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.