Temperature Coefficient of a Zn–Ag₂O Cell: Is the Cold Chamber Really Needed?
A development team is qualifying an alkaline Zn–Ag₂O button cell for an outdoor sensor that must start at −20 °C. A technician reports open-circuit voltages measured with a handheld 10 MΩ meter after 30 s stabilisation: 1.606 V at 0 °C, 1.592 V at 25 °C and 1.578 V at 45 °C. The project lead objects: 'The cell reaction is strongly exothermic, so the voltage should rise with temperature - the technician must have the sign wrong. And a 30 s reading on a button cell is meaningless anyway.' The supplier's datasheet gives 1.55 V nominal at 25 °C, 180 mAh at a 1 mA drain and an operating range of −20 to 60 °C. The team must decide by Friday whether to fund a −20 °C cold-chamber qualification or switch to a lithium chemistry. The electronics draw 3 mA at start-up with a 1.40 V cut-off.
1. Who is right about the sign of dE/dT? Compute the entropy of reaction and the EMF at −20 °C, and state what the number does and does not tell the team.
Read the worked answer
The project lead is wrong, and the reasoning is the standard Gibbs–Helmholtz confusion. For a reversible cell at constant pressure, dE/dT = ΔS/nF. The temperature coefficient is set by the entropy of reaction, not the enthalpy. ΔH = ΔG + TΔS = −nFE + nFT(dE/dT); nothing forces the entropy and enthalpy terms to share a sign.
Numbers (n = 2 for Zn + Ag₂O + H₂O → Zn(OH)₂ + 2Ag). dE/dT = (1.578 − 1.606)/(45 − 0) = −6.22 × 10⁻⁴ V/K. ΔS = nF(dE/dT) = 2 × 96,485 × (−6.22 × 10⁻⁴) = −120 J K⁻¹ mol⁻¹ - entropy decreases, consistent with converting a free hydroxide/oxide into a bound hydroxide and consuming water. ΔG(298 K) = −nFE = −2 × 96,485 × 1.592 = −307 kJ/mol. ΔH = ΔG + TΔS = −307.2 + 298.15 × (−0.1200) = −343 kJ/mol (exothermic, as the lead said - and yet E falls with temperature).
Extrapolation. E(−20 °C) = 1.592 + (−6.22 × 10⁻⁴)(−45.15) ≈ 1.620 V, comfortably above the 1.40 V cut-off; the measured 0 °C value of 1.606 V agrees with the fit to within 2 mV.
What it does not tell you. Thermodynamics says the driving force improves in the cold. It says nothing about electrolyte conductivity, ion mobility, Zn passivation or Ag₂O reduction kinetics, all of which worsen sharply below −10 °C. The cold chamber is still needed - but to measure internal resistance and start-up voltage sag under a 3 mA pulse, not to check the open-circuit voltage.
2. Evaluate the quality of the measurement. What are the dominant error sources, and what protocol would you specify?
Read the worked answer
Is 30 s with a 10 MΩ meter acceptable? Largely yes, and here is why: an Ag₂O button cell has an internal resistance of a few ohms, so a 10 MΩ meter draws under a microamp and causes no meaningful IR drop; 30 s is enough for a low-impedance cell to settle to its open-circuit value. The lead's objection is not the main issue.
Dominant error sources.
- Resolution and accuracy of the meter - ±1 mV on a 1.6 V reading is 0.06%. The slope across only 45 K is therefore uncertain by about √2 × 1 mV / 45 K ≈ 3 × 10⁻⁵ V/K, i.e. ±5% of the slope; add meter accuracy and the slope is good to perhaps ±10–15%.
- Thermal equilibration. A button cell has enough mass that it must be soaked at each temperature for at least an hour inside the chamber; measuring a cell taken out of a fridge in ambient air gives a drifting reading and is the likely reason for the 'meaningless' impression.
- History dependence - a cell that has been discharged, even briefly, has a different OCV. Use fresh cells for each point, or one cell only after full recovery.
- Junction and lead effects - for a two-electrode cell (a battery, not an electrochemical cell with a reference electrode) thermoelectric EMFs in dissimilar leads contribute microvolts, negligible here.
Protocol. Three or more temperatures spanning at least 50 K, one fresh cell per temperature, ≥1 h soak in a chamber, triplicate readings after 60 s, meter calibrated against a 1.000 V reference, and the slope fitted by least squares with the standard error reported. That converts an argument about signs into a slope with a stated confidence interval.
3. The reversible heat is sometimes said to make a cell 'self-cool' on discharge. Compute it here and explain the practical consequence for the sensor's duty cycle.
Read the worked answer
Reversible heat. q_rev = TΔS = 298.15 × (−0.1200 kJ K⁻¹ mol⁻¹) = −35.8 kJ per mole of cell reaction, i.e. released, not absorbed. Because the sign of TΔS is negative, this cell heats slightly on discharge under reversible conditions. It is about 10% of ΔH (−343 kJ/mol).
Per unit charge. −35.8 kJ per 2 mol of electrons = 35.8/193 kJ per 96,485 C … more usefully, per coulomb: 35,800 J / 193,000 C ≈ 0.186 J/C = 0.186 W per ampere. At the sensor's 3 mA start-up pulse, the reversible heat is only 0.6 mW, against an irreversible I²R term of (0.003)² × ~5 Ω = 0.05 mW - both negligible. At a 1 mA continuous drain the reversible term is 0.19 mW and the irreversible term 0.005 mW, so the entropic term dominates the (still tiny) heat balance.
Why this matters.
- The biggest contribution is heat into the cell in the cold: at −20 °C, TΔS = 253.15 × (−0.1200) = −30.4 kJ/mol, so ΔG(−20 °C) = −343.0 + 30.4 = −312.6 kJ/mol and E = 312,600/192,970 = 1.620 V. The cell gives more energy per coulomb in the cold; the reason cold start fails, if it does, is kinetic (electrolyte conductivity, Zn passivation), not thermodynamic.
- Self-heating is irrelevant at these currents, so the cell will sit at ambient. That is exactly why the cold-chamber test must be run at a controlled −20 °C, and why a self-heating argument cannot be used to waive it.
- For the Friday decision: the EMF margin (1.62 V vs a 1.40 V cut-off) is not the risk; the risk is the IR drop at 3 mA start-up in the cold and the resulting transient dipping below cut-off. Test that first - it is far cheaper than either a full qualification or a chemistry change.