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Electrochemistry - Nernst equation

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell) to the standard electrode potential (E∘E^\circ), temperature, and activities (approximated by molar concentrations) of the chemical species.

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For a general redox reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD, the reaction quotient QQ is expressed as Q=[C]c[D]d[A]a[B]bQ = \frac{[C]^c [D]^d}{[A]^a [B]^b}.

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The value of the constant 2.303RTF\frac{2.303 RT}{F} is approximately 0.05910.0591 at 298 K298\text{ K}, which simplifies calculations for standard laboratory conditions.

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At equilibrium, the cell potential EcellE_{cell} becomes 00, as the battery is 'dead'. This state allows for the calculation of the equilibrium constant KcK_c using the standard cell potential.

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The relationship between the Gibbs free energy change and the cell potential is given by ΔG=−nFEcell\Delta G = -nFE_{cell}, where nn is the number of moles of electrons transferred and FF is Faraday's constant (96500 C mol−196500\text{ C mol}^{-1}).

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For a pure solid or liquid, the concentration is taken as unity (1) in the Nernst equation expression.

📐Formulae

Ecell=Ecell∘−RTnFln⁡QE_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q

Ecell=Ecell∘−0.0591nlog⁡10[Products][Reactants] (at 298 K)E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log_{10} \frac{[\text{Products}]}{[\text{Reactants}]}\text{ (at 298 K)}

Ecell∘=0.0591nlog⁡10Kc (at equilibrium, 298 K)E^\circ_{cell} = \frac{0.0591}{n} \log_{10} K_c\text{ (at equilibrium, 298 K)}

ΔG=−nFEcell\Delta G = -nFE_{cell}

ΔG∘=−nFEcell∘=−2.303RTlog⁡Kc\Delta G^\circ = -nFE^\circ_{cell} = -2.303 RT \log K_c

💡Examples

Problem 1:

Calculate the EMFEMF of the cell Mg(s)∣Mg2+(0.1 M)∣∣Cu2+(1×10−4 M)∣Cu(s)Mg(s) | Mg^{2+}(0.1\text{ M}) || Cu^{2+}(1 \times 10^{-4}\text{ M}) | Cu(s) at 298 K298\text{ K}. Given: EMg2+/Mg∘=−2.37 VE^\circ_{Mg^{2+}/Mg} = -2.37\text{ V} and ECu2+/Cu∘=+0.34 VE^\circ_{Cu^{2+}/Cu} = +0.34\text{ V}.

Solution:

  1. Calculate Ecell∘E^\circ_{cell}: Ecell∘=Ecathode∘−Eanode∘=0.34 V−(−2.37 V)=2.71 VE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34\text{ V} - (-2.37\text{ V}) = 2.71\text{ V}.
  2. Identify nn: For the reaction Mg+Cu2+→Mg2++CuMg + Cu^{2+} \rightarrow Mg^{2+} + Cu, n=2n = 2.
  3. Apply Nernst Equation: Ecell=2.71−0.05912log⁡[Mg2+][Cu2+]E_{cell} = 2.71 - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}.
  4. Substitute values: Ecell=2.71−0.02955log⁡0.110−4=2.71−0.02955log⁡(103)E_{cell} = 2.71 - 0.02955 \log \frac{0.1}{10^{-4}} = 2.71 - 0.02955 \log(10^3).
  5. Ecell=2.71−0.02955×3=2.71−0.08865=2.62135 VE_{cell} = 2.71 - 0.02955 \times 3 = 2.71 - 0.08865 = 2.62135\text{ V}.

Explanation:

The standard cell potential is first determined. Then, using the Nernst equation at 298 K298\text{ K}, we account for the non-standard concentrations of the ions to find the actual EMFEMF of the cell.

Problem 2:

Calculate the equilibrium constant KcK_c for the reaction Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) if Ecell∘=0.46 VE^\circ_{cell} = 0.46\text{ V} at 298 K298\text{ K}.

Solution:

  1. Use the formula: log⁡Kc=nEcell∘0.0591\log K_c = \frac{n E^\circ_{cell}}{0.0591}.
  2. For this reaction, n=2n = 2 electrons are transferred.
  3. log⁡Kc=2×0.460.0591=0.920.0591≈15.566\log K_c = \frac{2 \times 0.46}{0.0591} = \frac{0.92}{0.0591} \approx 15.566.
  4. Kc=antilog(15.566)≈3.68×1015K_c = \text{antilog}(15.566) \approx 3.68 \times 10^{15}.

Explanation:

At equilibrium, the concentration of reactants and products are related to the standard cell potential. A positive Ecell∘E^\circ_{cell} results in a very large KcK_c, indicating the reaction goes near completion.

Nernst equation - Revision Notes & Key Formulas | ICSE Class 12 Chemistry