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Electrochemistry - Galvanic Cells and Nernst Equation

Grade 12CBSEChemistry

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

🔑Concepts

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A Galvanic (Voltaic) cell is an electrochemical cell that converts chemical energy from spontaneous redox reactions into electrical energy. A common example is the Daniel Cell involving ZnZn and CuCu.

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The electrode where oxidation occurs is the Anode (negative polarity in galvanic cells), and the electrode where reduction occurs is the Cathode (positive polarity).

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Cell notation is represented as: Anode∣Anode Ion (c1)∣∣Cathode Ion (c2)∣CathodeAnode | Anode\ Ion \ (c_1) || Cathode\ Ion \ (c_2) | Cathode. The double vertical line ∣∣|| represents the salt bridge.

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The Electromotive Force (EMF) or Cell Potential (EcellE_{cell}) is calculated as: Ecell=Ecathode−EanodeE_{cell} = E_{cathode} - E_{anode} using standard reduction potentials.

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The Nernst Equation relates the cell potential to the concentration of electrolytes: Ecell=Ecell∘−RTnFln⁡QE_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln Q. At 298 K298\text{ K}, this simplifies using log⁡10\log_{10}.

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Gibbs Free Energy change (ΔG\Delta G) is related to cell potential by ΔG=−nFEcell\Delta G = -nFE_{cell}. For a reaction to be spontaneous, ΔG<0\Delta G < 0 and Ecell>0E_{cell} > 0.

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At equilibrium, Ecell=0E_{cell} = 0 and the reaction quotient QQ becomes the equilibrium constant KcK_c.

📐Formulae

Ecell∘=Ecathode∘−Eanode∘E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}

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

ΔG∘=−nFEcell∘\Delta G^{\circ} = -nFE^{\circ}_{cell}

log⁡Kc=nEcell∘0.0591 (at 298 K)\log K_c = \frac{nE^{\circ}_{cell}}{0.0591} \text{ (at 298 K)}

Wmax=ΔG∘ (Maximum electrical work done)W_{max} = \Delta G^{\circ} \text{ (Maximum electrical work done)}

💡Examples

Problem 1:

Calculate the emf of the cell in which the following reaction takes place: Ni(s)+2Ag+(0.002M)→Ni2+(0.160M)+2Ag(s)Ni(s) + 2Ag^{+}(0.002 M) \rightarrow Ni^{2+}(0.160 M) + 2Ag(s). Given that Ecell∘=1.05 VE^{\circ}_{cell} = 1.05\text{ V}.

Solution:

  1. Identify nn: The number of electrons transferred is n=2n = 2.
  2. Apply Nernst Equation: Ecell=Ecell∘−0.0591nlog⁡[Ni2+][Ag+]2E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^2}.
  3. Substitute values: Ecell=1.05−0.05912log⁡0.160(0.002)2E_{cell} = 1.05 - \frac{0.0591}{2} \log \frac{0.160}{(0.002)^2}.
  4. Calculate log term: 0.1600.000004=40,000\frac{0.160}{0.000004} = 40,000. log⁡(40,000)=4.602\log(40,000) = 4.602.
  5. Final calculation: Ecell=1.05−(0.02955×4.602)=1.05−0.136=0.914 VE_{cell} = 1.05 - (0.02955 \times 4.602) = 1.05 - 0.136 = 0.914\text{ V}.

Explanation:

The Nernst equation is used here to find the potential under non-standard concentrations. Note that the concentration of Ag+Ag^{+} is squared because its stoichiometric coefficient in the balanced equation is 2.

Problem 2:

Calculate the standard Gibbs energy (ΔG∘\Delta G^{\circ}) for the reaction: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s). Given EZn2+/Zn∘=−0.76 VE^{\circ}_{Zn^{2+}/Zn} = -0.76\text{ V}, ECu2+/Cu∘=+0.34 VE^{\circ}_{Cu^{2+}/Cu} = +0.34\text{ V}, and F=96500 C mol−1F = 96500\text{ C mol}^{-1}.

Solution:

  1. Calculate Ecell∘=Ecathode∘−Eanode∘=0.34−(−0.76)=1.10 VE^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 - (-0.76) = 1.10\text{ V}.
  2. Identify n=2n = 2 (since Zn→Zn2++2e−Zn \rightarrow Zn^{2+} + 2e^{-}).
  3. Use formula ΔG∘=−nFEcell∘\Delta G^{\circ} = -nFE^{\circ}_{cell}.
  4. ΔG∘=−(2)×(96500)×(1.10)=−212300 J mol−1=−212.3 kJ mol−1\Delta G^{\circ} = -(2) \times (96500) \times (1.10) = -212300\text{ J mol}^{-1} = -212.3\text{ kJ mol}^{-1}.

Explanation:

The negative value of ΔG∘\Delta G^{\circ} indicates that the reaction is thermodynamically spontaneous under standard conditions.