Diagnosing a Galvanic Cell That Reads the Wrong Voltage
A student builds a cell: Zn(s) | Zn²⁺(1.0 mol/L) || Cu²⁺(1.0 mol/L) | Cu(s) at 298 K, using a KNO₃ salt bridge and a high-impedance voltmeter. Standard potentials: E°(Zn²⁺/Zn) = −0.76 V; E°(Cu²⁺/Cu) = +0.34 V. The measured EMF is 0.94 V, not the expected 1.10 V. Observations: (i) the zinc electrode loses mass; (ii) the copper electrode gains mass; (iii) the salt bridge becomes warm after several minutes; (iv) the voltmeter reading drifts downward over 10 minutes. The student also finds the Cu²⁺ solution was made from CuSO₄·5H₂O that had been left open to air and had partly dehydrated.
1. Write the cell reaction and calculate the expected EMF; then explain the 0.16 V shortfall using the Nernst equation.
Read the worked answer
Reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
EMF: E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V. The observations (Zn dissolves = anode; Cu deposits = cathode) confirm this direction.
Nernst: E = E° − (0.0592/2)·log([Zn²⁺]/[Cu²⁺]).
For E = 0.94 V: 0.94 = 1.10 − 0.0296·log(Q) → log(Q) = 0.16/0.0296 = 5.4 → Q = [Zn²⁺]/[Cu²⁺] ≈ 2.5 × 10⁵.
So the measured voltage is consistent with the Cu²⁺ concentration being far below 1.0 mol/L - exactly what partly dehydrated CuSO₄·5H₂O would cause (less Cu²⁺ than assumed) plus Zn²⁺ accumulation at the anode. Alternatively, if [Zn²⁺] rose to ~0.1 mol/L while [Cu²⁺] fell to ~4 × 10⁻⁷, the same Q results. The Nernst equation fully accounts for the shortfall; the cell is not faulty.
2. Explain observations (iii) and (iv) and state whether they indicate a fault.
Read the worked answer
(iii) Warming: the cell is short-circuited internally through the voltmeter (a real voltmeter draws a small current) and through ion migration; the reaction is spontaneous (ΔG° = −nFE° = −2 × 96500 × 1.10 = −212 kJ/mol), so chemical energy is released partly as heat. Warming is expected, not a fault.
(iv) Downward drift: as the cell discharges, [Zn²⁺] rises and [Cu²⁺] falls, so Q increases and E falls according to the Nernst equation. The drift is thermodynamically required, not an instrument error. A high-impedance voltmeter minimises but cannot eliminate this.
Conclusion: neither observation indicates a fault; both are predicted by thermodynamics and the Nernst equation.
3. The student wants to restore the reading to 1.10 V. Propose two chemically valid actions and one that would not work, with reasons.
Read the worked answer
Works (1): Replace the Cu²⁺ solution with freshly prepared 1.0 mol/L CuSO₄ so [Cu²⁺] = 1.0 mol/L and replace the Zn²⁺ solution with fresh 1.0 mol/L ZnSO₄, restoring Q = 1 and E = E° = 1.10 V.
Works (2): Increase [Cu²⁺] above 1.0 mol/L (or decrease [Zn²⁺] below 1.0 mol/L). Since E = E° − (0.0592/2)·log([Zn²⁺]/[Cu²⁺]), raising [Cu²⁺] raises E; e.g. [Cu²⁺] = 10 mol/L with [Zn²⁺] = 1 gives E = 1.10 + 0.0296 = 1.13 V, overshooting, so a moderate increase restores ~1.10 V.
Does not work: Reversing the electrodes or swapping the salt bridge for a wire. Reversing electrodes reverses the sign of E (the cell would read −1.10 V, not +1.10 V), and a wire replaces the salt bridge, causing junction-potential and mixing problems but not restoring the standard EMF. Changing the salt bridge cation (e.g. to KCl) does not change E° either, since the bridge only maintains electroneutrality.