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Concept 02 / 07Electrochemistry

Nernst Equation, Gibbs Energy and the Equilibrium Constant

Dependence of electrode and cell potentials on concentration through the Nernst equation, its form at 298 K, concentration cells, relation between cell emf and Gibbs energy (Delta G = -nFE), and between standard emf and the equilibrium constant, with pH measurement using the hydrogen electrode.

Figure 1
Standard potentials refer to 1 M1\,\mathrm{M} solutions and 1 bar1\,\mathrm{bar} gases. For other concentrations, the Nernst equation applies. Electrode potential. For Mn+(aq)+ne−→M(s)\mathrm{M^{n+}(aq) + ne^{-} \rightarrow M(s)}: E=E∘−RTnFln⁡1[Mn+]=E∘−2.303RTnFlog⁡1[Mn+]E = E^{\circ} - \frac{RT}{nF}\ln\frac{1}{[\mathrm{M^{n+}}]} = E^{\circ} - \frac{2.303RT}{nF}\log\frac{1}{[\mathrm{M^{n+}}]}. At 298 K298\,\mathrm{K}, 2.303RTF=0.0591 V\frac{2.303RT}{F} = 0.0591\,\mathrm{V}. Cell emf. For a cell reaction aA+bB→cC+dDa\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D} transferring nn electrons: Ecell=Ecell∘−0.0591nlog⁡QE_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n}\log Q at 298 K298\,\mathrm{K}, with Q=[C]c[D]d[A]a[B]bQ = \frac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}} (pure solids and liquids omitted, gases as partial pressures in bar). For the Daniell cell: E=1.10−0.05912log⁡[Zn2+][Cu2+]E = 1.10 - \frac{0.0591}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} (see figure). Increasing [Cu2+][\mathrm{Cu^{2+}}] or decreasing [Zn2+][\mathrm{Zn^{2+}}] raises the emf. Emf and Gibbs energy. The maximum electrical work a cell can do equals the decrease in Gibbs energy: ΔrG=−nFEcell\Delta_{r}G = -nFE_{\text{cell}} and ΔrG∘=−nFEcell∘\Delta_{r}G^{\circ} = -nFE^{\circ}_{\text{cell}}, with F=96487 C mol−1F = 96487\,\mathrm{C\,mol^{-1}} (about 9650096500). A positive EE means negative ΔG\Delta G: spontaneous. ΔG\Delta G is extensive (it doubles if the equation is doubled, together with nn), but EE is intensive. Emf and equilibrium constant. At equilibrium, Ecell=0E_{\text{cell}} = 0 and Q=KQ = K: Ecell∘=0.0591nlog⁡KE^{\circ}_{\text{cell}} = \frac{0.0591}{n}\log K, or log⁡K=nEcell∘0.0591\log K = \frac{nE^{\circ}_{\text{cell}}}{0.0591}. Combined with ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT\ln K, this links electrochemistry, thermodynamics and equilibrium. Concentration cells. The same electrodes in solutions of different concentration, such as Cu∣Cu2+(c1)∣∣Cu2+(c2)∣Cu\mathrm{Cu | Cu^{2+}(c_{1}) || Cu^{2+}(c_{2}) | Cu} with c2>c1c_{2} > c_{1}: E∘=0E^{\circ} = 0 and E=0.0591nlog⁡c2c1E = \frac{0.0591}{n}\log\frac{c_{2}}{c_{1}}. Measuring pH. For the hydrogen electrode, H++e−→12H2\mathrm{H^{+} + e^{-} \rightarrow \frac{1}{2}H_{2}} at 1 bar1\,\mathrm{bar}: E=−0.0591 pHE = -0.0591\,\text{pH}. Key Takeaways: - Ecell=Ecell∘−0.0591nlog⁡QE_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n}\log Q at 298 K298\,\mathrm{K}. - ΔG=−nFE\Delta G = -nFE; ΔG∘=−nFE∘\Delta G^{\circ} = -nFE^{\circ}. - log⁡K=nE∘0.0591\log K = \frac{nE^{\circ}}{0.0591}. - Concentration cell: E=0.0591nlog⁡ccathodecanodeE = \frac{0.0591}{n}\log\frac{c_{\text{cathode}}}{c_{\text{anode}}}. - Pitfall: in QQ, products go in the numerator even though the Nernst equation has a minus sign; do not invert. Brain-Teaser 1: Find the emf of Mg∣Mg2+(0.001 M)∣∣Cu2+(0.0001 M)∣Cu\mathrm{Mg | Mg^{2+}(0.001\,M) || Cu^{2+}(0.0001\,M) | Cu} at 298 K298\,\mathrm{K} (E∘E^{\circ}: Mg −2.37-2.37, Cu +0.34 V+0.34\,\mathrm{V}). Solution: E∘=0.34+2.37=2.71 VE^{\circ} = 0.34 + 2.37 = 2.71\,\mathrm{V}. E=2.71−0.05912log⁡0.0010.0001=2.71−0.0296=2.68 VE = 2.71 - \frac{0.0591}{2}\log\frac{0.001}{0.0001} = 2.71 - 0.0296 = 2.68\,\mathrm{V}. Brain-Teaser 2: Calculate ΔrG∘\Delta_{r}G^{\circ} and KK for the Daniell cell at 298 K298\,\mathrm{K}. Solution: ΔrG∘=−2×96487×1.10=−212.3 kJ mol−1\Delta_{r}G^{\circ} = -2 \times 96487 \times 1.10 = -212.3\,\mathrm{kJ\,mol^{-1}}. log⁡K=2×1.100.0591=37.2\log K = \frac{2 \times 1.10}{0.0591} = 37.2, so K≈1.7×1037K \approx 1.7 \times 10^{37}: the reaction goes essentially to completion. Brain-Teaser 3: What is the potential of a hydrogen electrode dipped in a solution of pH 1010? And the emf of Ag∣Ag+(0.01 M)∣∣Ag+(0.1 M)∣Ag\mathrm{Ag | Ag^{+}(0.01\,M) || Ag^{+}(0.1\,M) | Ag}? Solution: E=−0.0591×10=−0.591 VE = -0.0591 \times 10 = -0.591\,\mathrm{V}. For the concentration cell, n=1n = 1: E=0.0591log⁡0.10.01=0.0591 VE = 0.0591\log\frac{0.1}{0.01} = 0.0591\,\mathrm{V}.

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