Tin–Silver Galvanic Cell: When the Voltmeter Disagrees with E°cell
Two students build a galvanic cell from a tin electrode in 1.00 mol/L Sn(NO₃)₂(aq) and a silver electrode in AgNO₃(aq), joined by a potassium nitrate salt bridge. The cell is kept in a dark cupboard. Standard reduction potentials: Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V; Sn²⁺(aq) + 2e⁻ → Sn(s), E° = −0.14 V. Before measuring, the students calculate the cell potential they expect. With a freshly prepared 1.00 mol/L AgNO₃ solution, the voltmeter reads +0.76 V rather than their predicted value. They recheck every connection and confirm the electrodes are clean and the voltmeter is calibrated. They must decide whether the cell is faulty or whether something about the silver half-cell has changed.
1. Identify the anode and cathode, write the balanced overall cell reaction and the cell notation, and calculate E°cell.
Read the worked answer
Electrode assignment
The half-reaction with the more positive reduction potential proceeds as a reduction (cathode):
- Cathode (+): Ag⁺ + e⁻ → Ag, E° = +0.80 V
- Anode (−): Sn → Sn²⁺ + 2e⁻, E° = +0.14 V (oxidation; sign flipped)
Tin is oxidized, so tin is the anode; silver ion is reduced at the silver cathode.
Overall reaction (multiply the silver half-reaction by 2 to balance electrons; E° is not multiplied):
Sn(s) + 2Ag⁺(aq) → Sn²⁺(aq) + 2Ag(s)
Standard cell potential
E°cell = E°cathode − E°anode = (+0.80 V) − (−0.14 V) = +0.94 V
Cell notation
Sn(s) | Sn²⁺(1.00 mol/L) ‖ Ag⁺(1.00 mol/L) | Ag(s)
The positive E°cell confirms the reaction is spontaneous as written.
2. Assuming the tin half-cell is behaving ideally at 1.00 mol/L, use the Nernst equation to calculate the [Ag⁺] consistent with the 0.76 V reading. Conclude whether the cell is faulty.
Read the worked answer
Nernst equation (n = 2 electrons transferred)
E = E° − (0.0592/n)·log Q, with Q = [Sn²⁺]/[Ag⁺]²
0.76 = 0.94 − (0.0592/2)·log(1.00/[Ag⁺]²)
0.18 = 0.0296·log(1.00/[Ag⁺]²)
log(1.00/[Ag⁺]²) = 6.08
1.00/[Ag⁺]² = 1.2 × 10⁶
[Ag⁺]² = 8.3 × 10⁻⁷ → [Ag⁺] ≈ 9 × 10⁻⁴ mol/L
Conclusion
The cell is not faulty - the wiring and electrodes are fine. The silver half-cell simply is not at 1.00 mol/L; the measured potential corresponds to a silver ion concentration roughly a thousand times lower than stated. A large drop in [Ag⁺] shifts the silver reduction equilibrium left, lowering the driving force. Plausible causes: the AgNO₃ solution was diluted or partly decomposed, Ag⁺ was reduced or adsorbed onto the electrode or container, AgCl precipitated (if chloride contamination occurred), or the solution was made up incorrectly. Light exposure is not a good explanation here, since the cell was kept in a dark cupboard.
3. The students later replace the salt bridge with a copper wire and the reading falls to 0.00 V. Explain this result, and explain why the voltage would also fall toward 0 V over many hours even with the salt bridge in place.
Read the worked answer
Short-circuit by the copper wire
A metal wire offers electrons a low-resistance path directly between the two half-cells. Electrons flow through the wire instead of through the external circuit, so no useful potential difference is measured across the voltmeter. At the same time, no ions migrate between the compartments, so each beaker cannot maintain electroneutrality: the silver compartment becomes deficient in positive charge and the tin compartment becomes excess positive. Charge separation builds up and the reaction quickly stalls, giving 0.00 V.
Why a working cell still dies
With the salt bridge, ions migrate and the reaction proceeds:
Sn + 2Ag⁺ → Sn²⁺ + 2Ag
As it proceeds, [Ag⁺] falls and [Sn²⁺] rises, so Q = [Sn²⁺]/[Ag⁺]² increases. Since E = E° − (0.0592/2)·log Q, the cell potential steadily decreases. When Q reaches K (or when Ag⁺ is essentially consumed), log Q = log K and E = 0 V - the cell has reached equilibrium and is 'dead'. Because the reaction products remain in contact, this is not a rechargeable setup as built.