Galvanic Cell Voltage Below the Nernst Prediction
Students assemble a galvanic cell from a copper strip in 0.0010 M Cu(NO3)2 and a silver strip in 1.00 M AgNO3, joined by a KNO3 agar salt bridge, and read 0.50 V on a digital voltmeter at 25 °C. One group predicts 0.46 V by subtracting the tabulated standard potentials; a second group predicts -0.46 V "because electrons flow from silver to copper." The silver strip was polished, but that beaker had last held a chloride solution and was rinsed only with tap water. The salt bridge is a narrow agar tube with appreciable resistance, and the reading drifts by about 0.02 V while the class watches. The instructor asks for the expected voltage at the stated concentrations, the identity of the cathode, and a defensible explanation of the gap between prediction and measurement.
1. Calculate the expected cell potential at the stated concentrations, identify the anode and cathode, and correct the sign error in the second group's prediction.
Read the worked answer
E°(Ag+/Ag) = +0.80 V and E°(Cu2+/Cu) = +0.34 V. The larger (more positive) reduction potential wins the reduction, so silver is the cathode and copper is the anode. The spontaneous cell reaction is Cu(s) + 2 Ag+ → Cu2+ + 2 Ag(s), and E°cell = 0.80 - 0.34 = 0.46 V.
The standard-potential subtraction applies only when both ion concentrations are 1 M, and the copper solution is 0.0010 M. The Nernst equation corrects for that. With n = 2:
Q = [Cu2+]/[Ag+]^2 = 0.0010/(1.00)^2 = 1.0 × 10^-3
E = 0.46 - (0.0592/2) log(1.0 × 10^-3) = 0.46 - (0.0296)(-3) = 0.46 + 0.089 = 0.55 V.
Diluting the copper ion, the product of the cell reaction, increases the driving force, so the cell voltage should be about 0.55 V, not 0.46 V.
The -0.46 V prediction confuses the direction of electron flow in the external wire with the sign convention for reduction potentials. Electrons do flow from copper to silver through the wire, and the cell potential is reported as a positive quantity for that spontaneous direction.
2. The measured 0.50 V is below the 0.55 V prediction. Rank the candidate explanations and describe the observation that would discriminate among them.
Read the worked answer
The cell is underperforming by about 0.05 V. Consider each candidate against the Nernst equation:
- Chloride contamination of the silver half-cell. A AgCl film lowers the free Ag+ activity at the electrode surface. Solving E = 0.46 - 0.0296 log(0.0010/[Ag+]^2) = 0.50 V gives an effective [Ag+] near 0.1 M rather than 1.0 M. A factor-of-ten drop in silver ion accounts for the entire discrepancy, which makes this the leading explanation given the history of the beaker.
- Uncompensated resistance in the narrow agar bridge. Drawing current through a high-resistance bridge lowers the terminal voltage, and the 0.02 V drift is consistent with a poorly stable junction. A high-impedance voltmeter would reduce this effect; a cheap meter would worsen it.
- Junction potentials at the bridge ends, typically millivolts, not tens of millivolts.
- Temperature differing from 25 °C, worth roughly 0.1 mV per degree - too small.
The discriminating test is simple: repolish the silver strip, use a clean, DI-rinsed beaker, and prepare fresh AgNO3. If the reading climbs to about 0.55 V, chloride passivation is confirmed. If it does not, suspect the bridge.
3. The instructor asks what would happen to the voltage if a few drops of Na2S were added to the copper half-cell, precipitating CuS. Predict the direction, estimate a value, and explain the limits of that estimate.
Read the worked answer
Removing a product of the cell reaction increases the driving force. Adding sulfide drops free Cu2+ to an extremely low value, since CuS is one of the least soluble sulfides. Taking [Cu2+] around 1 × 10^-20 M for illustration:
E = 0.46 - (0.0296) log(1 × 10^-20/1) = 0.46 + 0.59 = 1.05 V.
The direction is the reliable part of this answer: the voltage rises, and the increase is roughly 30 mV for each factor of ten by which the copper ion concentration falls, since copper contributes a two-electron couple.
The number needs hedging. The Nernst equation lets E grow without bound as [Cu2+] approaches zero, but a real cell cannot exceed the potential of whatever couple takes over at the electrode. Once copper ion is essentially gone, some other reducible species - dissolved oxygen, water, or the sulfide system itself - will define the potential at that electrode, and the measured voltage would plateau well below 1.05 V. A student who reports 1.05 V as a prediction without that caveat has applied the equation outside its physical range.