A Galvanic Cell That Reads Low
A group of students constructs a galvanic cell with a zinc electrode in 0.10 M Zn(NO3)2 and a copper electrode in 0.10 M Cu(NO3)2, connected by a potassium nitrate salt bridge. At 22 °C, the voltmeter reads 1.02 V. The students calculated the standard cell potential as 1.10 V and expected the measured value to be lower because 'the solutions are more dilute than 1 M.' Their lab manual gives E°(Cu2+/Cu) = +0.34 V and E°(Zn2+/Zn) = -0.76 V. The voltmeter has a high input impedance (10 MΩ). The salt bridge is freshly prepared. The copper electrode looks slightly tarnished. The students ask whether their Nernst-equation explanation is correct and what else could account for the 0.08 V difference. The teacher must help them evaluate competing hypotheses without simply giving the answer.
1. Evaluate the students' claim that the lower concentrations explain the 1.02 V reading. Use the Nernst equation and the actual concentrations.
Read the worked answer
Nernst equation: E = E° − (0.0592/n) log Q at 25 °C. For Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), n = 2 and Q = [Zn2+]/[Cu2+] = 0.10/0.10 = 1.00. log Q = 0, so E = E° = 1.10 V. The Nernst term is zero because both ion concentrations were lowered by the same factor. Therefore the students' explanation is incorrect: dilution of both half-cells equally does not lower the cell potential. The 1.02 V reading must come from something else, such as electrode surface condition, junction/contact resistance, or a concentration error in one half-cell.
2. Propose two plausible sources of the 0.08 V discrepancy and describe a specific test for each.
Read the worked answer
Two plausible sources: (1) The tarnished copper electrode has an oxide or sulfide layer that lowers the effective Cu2+ activity and adds resistance. Test: polish the copper with fine abrasive or dip briefly in dilute HCl, rinse, and remeasure. If the voltage rises toward 1.10 V, the surface was the problem. (2) The measured concentrations may not actually be equal; if the Cu2+ solution were more dilute than 0.10 M, Q would be larger and E lower. Test: re-prepare both solutions from fresh stock or measure Cu2+ by a complexometric or spectrophotometric method. A smaller possibility is a low-resistance voltmeter drawing current; test with a potentiometer or a meter with much higher input impedance. Temperature (22 °C vs 25 °C) changes E only by a few millivolts, not 0.08 V.
3. Predict the cell potential if the copper half-cell concentration is raised to 1.00 M while the zinc half-cell remains 0.10 M. Show the calculation and explain the direction of the change.
Read the worked answer
Q = [Zn2+]/[Cu2+] = 0.10/1.00 = 0.10. log Q = −1.00. E = 1.10 − (0.0592/2)(−1.00) = 1.10 + 0.0296 = 1.13 V. The potential increases because increasing [Cu2+] makes Q smaller, so the forward reaction is more favorable; the Nernst term becomes positive. This assumes the cell remains at 25 °C and activities can be approximated by concentrations.