Redox and Electrode Potentials

Rules

1. Elements: oxidation state = 0
2. Simple ions: oxidation state = charge (Na⁺ = +1, Cl⁻ = −1)
3. In compounds, sum of oxidation states = 0
4. In polyatomic ions, sum = charge on ion
5. Specific rules: O = −2 (except peroxides = −1), H = +1 (except metal hydrides = −1), F = always −1

Examples

• $\text{MnO}_4^-$: Mn + 4(−2) = −1, so Mn = +7
• $\text{Cr}_2\text{O}_7^{2-}$: 2Cr + 7(−2) = −2, so Cr = +6
• $\text{H}_2\text{SO}_4$: S = +6

Identifying Redox

• Oxidation: increase in oxidation state (loss of electrons)
• Reduction: decrease in oxidation state (gain of electrons)
• Oxidising agent: causes oxidation; is itself reduced
• Reducing agent: causes reduction; is itself oxidised

Half-Equation Method

1. Write the oxidation and reduction half-equations separately
2. Balance atoms (use H₂O for O atoms, H⁺ for H atoms)
3. Balance charges using electrons
4. Multiply to equalise electrons
5. Add half-equations and cancel common species

Example: Fe²⁺ reacting with MnO₄⁻ in acidic solution

Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (×5)

Overall: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$

An electrochemical cell converts chemical energy into electrical energy (or vice versa).

Standard Electrode Potential ($E^\ominus$)

The standard electrode potential is the voltage of a half-cell measured relative to the standard hydrogen electrode (SHE) under standard conditions (298 K, 100 kPa, 1.00 mol dm⁻³).

The SHE is defined as $E^\ominus = 0.00$ V.

Using $E^\ominus$ Values

All half-equations are written as reductions:

$\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad E^\ominus = +0.34 \text{ V}$ $\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad E^\ominus = -0.76 \text{ V}$

Cell EMF

$E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ $= E^\ominus(\text{more positive}) - E^\ominus(\text{more negative})$

For a Zn/Cu cell: $E^\ominus_{\text{cell}} = +0.34 - (-0.76) = +1.10 \text{ V}$

The more negative half-cell is oxidised (anode); the more positive is reduced (cathode).

A reaction is thermodynamically feasible if: $E^\ominus_{\text{cell}} > 0$

The species with the more positive $E^\ominus$ will be reduced (gains electrons). The other will be oxidised.

Example

Will $\text{Fe}^{3+}$ oxidise $\text{I}^-$ to $\text{I}_2$?

$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ $E^\ominus = +0.77$ V $\text{I}_2 + 2e^- \rightarrow 2\text{I}^-$ $E^\ominus = +0.54$ V

$E_{\text{cell}} = 0.77 - 0.54 = +0.23$ V > 0 ✓

Yes, the reaction is feasible. Fe³⁺ is reduced; I⁻ is oxidised.

Limitations

• $E^\ominus$ predicts thermodynamic feasibility, not kinetic feasibility
• The reaction may be feasible but too slow (high activation energy)
• Non-standard conditions may change the actual cell potential

• Oxidation states: remember the rules; practice with complex ions
• Half-equations: balance atoms first, then charges with electrons
• $E^\ominus_{\text{cell}} = E^\ominus(\text{more positive}) - E^\ominus(\text{more negative})$
• Positive cell EMF = feasible reaction
• Always state that $E^\ominus$ predicts thermodynamics only, not kinetics

• Oxidation = loss of electrons / increase in oxidation state
• Reduction = gain of electrons / decrease in oxidation state
• $E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$
• Positive $E^\ominus_{\text{cell}}$ → reaction is thermodynamically feasible
• The more positive half-cell is reduced; the more negative is oxidised
• Feasibility ≠ rate — kinetics may prevent a feasible reaction from occurring