Galvanic Cell Voltage: Ideal versus Real
Students build galvanic cells using 0.10 M solutions and metal electrodes. One group constructs Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s) and measures 0.46 V with a voltmeter, close to the standard value. A second group uses the same metals but prepares the silver half-cell with 0.010 M AgNO3 and the copper half-cell with 1.0 M Cu(NO3)2. Their voltmeter reads 0.34 V, and a student says the cell is “broken” because the voltage is not 0.46 V. Another student claims that electrons travel through the salt bridge to complete the circuit, and a third student wants to know why a salt bridge is needed at all if the wire already connects the electrodes. The teacher has standard reduction potentials available: Ag+ + e− → Ag, E° = +0.80 V; Cu2+ + 2e− → Cu, E° = +0.34 V.
1. Identify the anode and cathode, write the balanced overall cell reaction, and calculate E°cell for the Cu–Ag cell.
Read the worked answer
Ag+ has the more positive reduction potential (+0.80 V) than Cu2+ (+0.34 V), so Ag+ is reduced at the cathode: Ag+ + e− → Ag. Copper is oxidized at the anode: Cu → Cu2+ + 2e−. Balancing electrons gives Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s). E°cell = E°cathode − E°anode = 0.80 V − 0.34 V = 0.46 V. The cell notation is Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s), with the anode on the left and the cathode on the right. The positive E°cell means the reaction is spontaneous under standard conditions, consistent with the first group’s measurement.
2. Calculate the expected cell potential for the second group’s concentrations ([Ag+] = 0.010 M, [Cu2+] = 1.0 M) and explain whether the 0.34 V reading is reasonable.
Read the worked answer
The overall reaction is Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s), so n = 2 and Q = [Cu2+]/[Ag+]2 = 1.0/(0.010)2 = 1.0 × 10^4. The Nernst equation at 25 °C is E = E° − (0.0592/n) log Q. Thus E = 0.46 − (0.0592/2)(4.00) = 0.46 − 0.118 = 0.342 V. The measured 0.34 V is therefore reasonable, not evidence of a broken cell. Lowering [Ag+] reduces the driving force for silver reduction, so the cell potential falls. Students often expect the standard potential to apply regardless of concentration; this example shows that voltage depends on the reaction quotient and that a cell can be far from standard conditions even when the same metals are used.
3. Correct the student’s statement that electrons travel through the salt bridge, and explain the salt bridge’s actual function in the cell.
Read the worked answer
Electrons do not travel through the salt bridge. They flow through the external wire from the anode (copper) to the cathode (silver), where they are consumed by Ag+ ions. The salt bridge contains an electrolyte that allows ions to migrate: anions move toward the anode compartment to balance the positive charge building up as Cu2+ is produced, and cations move toward the cathode compartment to replace the positive charge lost as Ag+ is reduced. Without this ion migration, charge imbalance would stop the redox reaction almost immediately. The salt bridge completes the electrical circuit by conducting ions, not electrons, and it also helps maintain electrical neutrality in each half-cell.