Temperature coefficient of a Zn/Cu cell
A student sets up the cell Zn(s) | Zn(NO3)2(aq, 0.100 M) || Cu(NO3)2(aq, 0.100 M) | Cu(s) and measures the open-circuit potential with a high-impedance digital voltmeter at 298, 308, and 318 K. The cell is placed in a water bath, but readings are taken as soon as the thermometer shows the new temperature. Values: 1.083 V, 1.079 V, and 1.075 V. The student uses these to calculate ΔG°, ΔS°, and ΔH° for the cell reaction, assuming the measured potentials are standard. The salt bridge is 1.0 M KNO3, and the electrodes are polished but not calibrated. The literature E° for the Zn/Cu cell at 298 K is about 1.100 V. A second run at 298 K after 30 min gives 1.097 V.
1. Calculate ΔG°, ΔS°, and ΔH° at 298 K from the 298–318 K data. State the stoichiometry and sign conventions, and assess whether the assumption that E = E° is justified for this cell.
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Reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), n = 2. dE/dT = (1.075 − 1.083)/(318 − 298) = −4.0×10−4 V K−1. At 298 K, ΔG° = −nFE = −2(96485)(1.083) = −2.090×10^5 J mol−1 = −209.0 kJ mol−1. ΔS° = nF(dE/dT) = 2(96485)(−4.0×10−4) = −77.2 J K−1 mol−1. ΔH° = ΔG° + TΔS° = −209.0 + 298(−0.0772) = −232.0 kJ mol−1. Because [Zn2+] = [Cu2+] = 0.100 M, Q = 1 and E = E° if activity coefficients are equal; the assumption is approximate, not exact, because γ(Zn2+) and γ(Cu2+) differ and a liquid-junction potential may be present.
2. Critique the temperature-control and electrode practice, and propose improvements to obtain a reliable temperature coefficient.
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The temperature coefficient is the small difference between large numbers, so temperature control must be much better than the 5 K spacing. Taking a reading as soon as the thermometer changes does not mean the electrodes, salt bridge, and solution have equilibrated; thermal EMFs and slow electrode response can bias dE/dT. Thermostat the cell to ±0.1 K, stir, and wait until the potential drifts by less than about 0.1 mV per minute. Measure at each temperature both on heating and cooling, calibrate the voltmeter against a known cell, use clean electrodes and fresh solutions, and minimize/estimate the liquid-junction potential. The 1.097 V second run at 298 K suggests the first reading was not at equilibrium or the cell changed.
3. Calculate log K at 298 K from 1.083 V and compare it with the value from 1.100 V. Does the 17 mV difference change the thermodynamic conclusion?
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Using E = 1.083 V, log K = nE/0.05916 = 2(1.083)/0.05916 = 36.6, so K ≈ 4×10^36. Using 1.100 V gives log K = 37.2, K ≈ 1.5×10^37. The difference in ΔG is nF(0.017 V) = 2(96485)(0.017) = 3.3 kJ mol−1, small on the scale of the reaction free energy. Both values show the reaction is strongly product-favored; the 17 mV discrepancy is more important as a measurement problem than as a change in the thermodynamic conclusion.