Nernst Equation – IIT‑JEE Chemistry Study Notes
A compact, self‑contained guide to the Nernst equation – its derivation, applications and typical JEE‑style problems.
1. Simple Explanation
The Nernst equation links the electrode potential of a redox couple to the activities (concentrations) of the reacting species.
It tells us how a change in concentration, pressure or temperature alters the electromotive force (EMF) of a galvanic or electrolytic cell.
Key idea:
When the reaction quotient (Q) differs from unity, the potential deviates from its standard value (E^\circ).
2. Key Points
| Item | Description |
|---|---|
| General form | (E = E^\circ - \dfrac{RT}{nF},\ln Q) |
| Simplified at 25 °C | (E = E^\circ - \dfrac{0.0592}{n}\log_{10} Q) |
| Variables | (E) – electrode potential (V) (E^\circ) – standard electrode potential (V) (R = 8.314,\text{J mol}^{-1}\text{K}^{-1}) (T) – temperature (K) (n) – moles of e⁻ transferred (F = 96485,\text{C mol}^{-1}) (Q) – reaction quotient |
| Reaction quotient (Q) | For (aA + bB \rightleftharpoons cC + dD): |
| (Q = \dfrac{a_C^c,a_D^d}{a_A^a,a_B^b}) (activities). For dilute solutions, (a_i \approx [i]) (mol L⁻¹). | |
| Temperature dependence | If (T) increases, (RT/nF) grows → larger deviation from (E^\circ). |
| Spontaneity | Cell potential (E_{\text{cell}}>0) → spontaneous galvanic cell. |
| Standard conditions | (T = 298;\text{K}), (P = 1;\text{atm}), (a_i = 1) for solids & liquids, (a_{\text{gas}} = 1) atm. |
| Applications | • Calculating cell potentials under non‑standard conditions. • Determining concentrations at which a cell is at equilibrium ( (E_{\text{cell}} = 0) ). • Predicting the effect of temperature on EMF. • Estimating equilibrium constants from measured potentials. |
3. Derivation (Brief)
- Gibbs free‑energy relation for a cell reaction
[ \Delta G = -nFE_{\text{cell}} ] - Non‑standard Gibbs free energy
[ \Delta G = \Delta G^\circ + RT\ln Q ] - Combine
[ -nFE_{\text{cell}} = \Delta G^\circ + RT\ln Q ] - Solve for (E_{\text{cell}})
[ E_{\text{cell}} = E^\circ - \frac{RT}{nF}\ln Q ] (where (E^\circ = -\Delta G^\circ/(nF))).
4. Worked Examples
Example 1 – Half‑Cell Potential
Problem:
Calculate the potential of the half‑cell ( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} ) at (25^\circ\text{C}) when ([\text{Cu}^{2+}] = 0.01;\text{M}).
(E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34;\text{V}).
Solution
- Identify (n = 2).
- Reaction quotient (