Commercial Cells
Useful Half-Equations
| Half-Equation | Eo (V) |
|---|---|
| MnO₂(s) + H₂O(l) + e⁻ → MnO(OH)(s) + OH⁻(aq) | +0.74 V |
| Li⁺ + CoO₂ + e⁻ → LiCoO₂ | +0.60 V |
| NiO(OH)(s) + H₂O(l) + e⁻ → Ni(OH)₂(s) + OH⁻(aq) | +0.52 V |
| Ag₂O(s) + H₂O(l) + 2e⁻ → 2Ag(s) + 2OH⁻(aq) | +0.34 V |
| Li⁺ + MnO₂ + e⁻ → LiMnO₂ | -0.15 V |
| Zn²⁺(aq) + 2e⁻ → Zn(s) | -0.76 V |
| [Zn(NH₃)₂]²⁺ + 2e⁻ → Zn + 2NH₃ | -0.80 V |
| Cd(OH)₂(s) + 2e⁻ → Cd(s) + 2OH⁻(aq) | -0.88 V |
| ZnO(s) + H₂O(l) + 2e⁻ → Zn(s) + 2OH⁻(aq) | -1.26 V |
| Li⁺ + e⁻ → Li | -3.04 V |
(Derived from: Cathode: 2MnO₂ + 2H₂O + 2e⁻ → 2MnO(OH) + 2OH⁻ and Anode: Zn + 2OH⁻ → ZnO + H₂O + 2e⁻).
EMF = E(cathode) – E(anode)
= +0.74 – (-1.26) = 2.00 V
Cathode (Reduction): Manganese oxide half-cell.
Anode (Oxidation): Zinc half-cell.
+0.60 – (-3.04) = 3.64 V
+3
(Li = +1, O = -2 x 2 = -4. Total charge is 0, so Co must be +3).
The electricity used to charge them can be produced from renewable sources (like wind or solar) that do not release carbon dioxide, unlike burning petrol/diesel.
-0.15 – (-3.04) = 2.89 V
Full Notation: Li(s) | Li⁺(aq) || Li⁺(aq), MnO₂(s), LiMnO₂(s) | Pt(s)
Simplified: Li | Li⁺ || Li⁺, MnO₂, LiMnO₂ | Pt
(Note: An inert electrode like Pt or Graphite is required on the right because the reaction involves solids and ions, but no conductive metal is produced/consumed).
+3
(OH is -1, O is -2. Ni + (-2) + (-1) = 0, so Ni = +3).
+0.52 – (-0.88) = 1.40 V