Galvanic Cell Design Under Real Constraints
A student builds a galvanic cell from a magnesium electrode in 1.0 M Mg(NO₃)₂ and a copper electrode in 1.0 M Cu(NO₃)₂, connected by a salt bridge and a voltmeter. Standard reduction potentials: Mg²⁺ + 2e⁻ → Mg, E° = −2.37 V; Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V. The measured cell potential is 2.61 V, lower than the calculated 2.71 V. The student also notices the copper electrode gains mass and the magnesium electrode corrodes. A classmate suggests replacing the salt bridge with a wire to 'make the circuit simpler'.
1. Calculate the theoretical cell potential, write the overall equation, and identify the anode and cathode.
Read the worked answer
The cell potential is E°cell = E°cathode − E°anode. Copper has the more positive reduction potential, so it is reduced at the cathode: Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V. Magnesium is oxidised at the anode: Mg → Mg²⁺ + 2e⁻, E° = +2.37 V when written as oxidation.
E°cell = 0.34 − (−2.37) = 2.71 V.
Overall: Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s).
Anode (oxidation): magnesium. Cathode (reduction): copper. This matches the observations: magnesium corrodes (loses mass) and copper gains mass as Cu²⁺ is reduced onto it. The positive cell potential confirms the reaction is spontaneous under standard conditions.
2. Explain why the measured 2.61 V is lower than 2.71 V, referring to at least two realistic causes.
Read the worked answer
The calculated value assumes standard conditions (1.0 M solutions, 25 °C, 100 kPa) and ideal behaviour. Real causes of a lower measured potential include:
- Non-standard or changing concentrations. As the reaction proceeds, [Cu²⁺] falls and [Mg²⁺] rises, shifting the electrode potentials according to the Nernst relationship and reducing the cell potential.
- Internal resistance and salt-bridge limitations. The salt bridge has finite conductance; ion migration is slower than electron flow, causing a voltage drop when current is drawn. A high-resistance voltmeter draws little current, but some drop remains.
- Electrode polarisation and overpotential, plus surface oxide films on magnesium, which impede electron transfer.
- Junction potentials and imperfect salt-bridge contact.
The 0.10 V shortfall is consistent with these combined effects. The judgement is that E°cell is a limiting ideal value; measured potentials are lower because the cell is not at standard conditions and is not ideal.
3. Evaluate the classmate's suggestion to replace the salt bridge with a wire, and explain the role of the salt bridge.
Read the worked answer
The suggestion must be rejected. A wire between the two solutions would create a short circuit: electrons would flow directly through the wire rather than through the external circuit and voltmeter, and the two half-cells would no longer be electrically separated. More importantly, a metal wire cannot perform the salt bridge's function of allowing ions to migrate between solutions to maintain electrical neutrality.
The salt bridge contains an electrolyte (e.g. KNO₃) that lets ions move: anions migrate toward the anode compartment to balance the excess positive charge from Mg²⁺ produced, and cations migrate toward the cathode compartment to replace the positive charge lost as Cu²⁺ is reduced. Without this ion flow, charge builds up, the reaction quickly stops, and the cell cannot deliver a sustained current. A wire conducts electrons, not ions, so it cannot replace the salt bridge. The correct judgement is that the salt bridge is essential for completing the internal circuit and maintaining neutrality, and replacing it with a wire would prevent the cell from working as intended.