Electrochemistry, Corrosion & Colloidal (9 hrs) · Engineering Chemistry

Electrochemical Theory: EMF, electrode potential, Nernst equation; numerical problems

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EMF, Electrode Potential & Nernst Equation with Numericals

1. Electrochemical cell basics

A galvanic (voltaic) cell converts chemical energy to electrical energy: oxidation at anode (–ve), reduction at cathode (+ve), salt bridge completes circuit.

Zn∣Zn2+∣∣Cu2+∣Cu(Daniel cell)Zn | Zn^{2+} || Cu^{2+} | Cu \quad \text{(Daniel cell)}
  • Anode: Zn→Zn2++2e−Zn \to Zn^{2+} + 2e^-; cathode: Cu2++2e−→CuCu^{2+} + 2e^- \to Cu.
  • Cell representation: anode | anode-ion || cathode-ion | cathode (oxidation on left by convention).

2. Electrode potential and EMF

  • Standard electrode potential E∘E^\circ: potential of Mn+/MM^{n+}/M vs standard hydrogen electrode (SHE, E∘=0E^\circ = 0 by definition), all at 1 M, 25 °C, 1 atm.
  • EMF of cell: Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}.
  • Daniel cell: Ecell∘=(+0.34)−(−0.76)=1.10E^\circ_{cell} = (+0.34) - (-0.76) = 1.10 V.
  • Relation to free energy and equilibrium:
ΔG∘=−nFEcell∘,Ecell∘=0.0591nlog⁡10K(at 298 K)\Delta G^\circ = -nFE^\circ_{cell}, \qquad E^\circ_{cell} = \frac{0.0591}{n}\log_{10}K \quad \text{(at 298 K)}

Spontaneous if Ecell>0E_{cell} > 0 (ΔG<0\Delta G < 0).

3. Nernst equation

For aA+bB→cC+dDaA + bB \to cC + dD:

E=E∘−RTnFln⁡Q⇒E=E∘−0.0591nlog⁡10Q(at 25 °C)E = E^\circ - \frac{RT}{nF}\ln Q \quad \Rightarrow \quad E = E^\circ - \frac{0.0591}{n}\log_{10}Q \quad \text{(at 25 °C)}

For metal electrode Mn++ne−→MM^{n+} + ne^- \to M: E=E∘+0.0591nlog⁡10[Mn+]E = E^\circ + \frac{0.0591}{n}\log_{10}[M^{n+}].

  • Increasing reactant ion concentration raises EE; product build-up lowers it.
  • pH electrodes, ion-selective electrodes, and corrosion-potential shifts all follow Nernst behaviour.

4. Solved numericals

Ex 1: Ecell∘E^\circ_{cell} for Zn+Cu2+(1M)→Zn2+(1M)+CuZn + Cu^{2+}(1M) \to Zn^{2+}(1M) + Cu: E∘=0.34−(−0.76)=1.10E^\circ = 0.34 - (-0.76) = 1.10 V; ΔG∘=−2×96500×1.10=−212\Delta G^\circ = -2 \times 96500 \times 1.10 = -212 kJ/mol.
Ex 2: Zn electrode in 0.01 M Zn2+Zn^{2+}: E=−0.76+0.05912log⁡10(0.01)=−0.76−0.0591=−0.819E = -0.76 + \frac{0.0591}{2}\log_{10}(0.01) = -0.76 - 0.0591 = -0.819 V.
Ex 3: Cell with [Zn2+]=0.1[Zn^{2+}] = 0.1, [Cu2+]=0.001[Cu^{2+}] = 0.001: E=1.10−0.05912log⁡100.10.001=1.10−0.0591=1.04E = 1.10 - \frac{0.0591}{2}\log_{10}\frac{0.1}{0.001} = 1.10 - 0.0591 = 1.04 V.
Ex 4: Find KK if E∘=0.46E^\circ = 0.46 V, n=2n = 2: log⁡10K=2×0.46/0.0591=15.6\log_{10}K = 2 \times 0.46/0.0591 = 15.6 → K≈4×1015K \approx 4 \times 10^{15}.

Exam Points

  • Cell representation + anode/cathode reactions (2–3 marks).
  • Nernst equation statement, 298 K form, and one concentration-cell numerical (5 marks favourite); ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ conversion is a guaranteed sub-part.
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