What you'll be able to do: Predict how temperature changes a cell potential using ΔH° and ΔS°, and identify which reactions can be made favorable by heating or cooling.
Best after: Cell Potential, Free Energy, and the Equilibrium Constant
Some cells get stronger when heated and others get weaker. Which way a cell moves is decided entirely by the sign of its entropy change, and the reasoning is the same thermodynamics you met with Gibbs free energy.
These are recommended, not required. You can start this lesson at any time.
For standard states, ΔG° = ΔH° − TΔS° = −nFE°cell. Over a range where ΔH° and ΔS° are approximately constant, E° is approximately linear in T, with intercept −ΔH°/(nF) and slope ΔS°/(nF).
These are standard-state quantities; actual mixtures require ΔG = ΔG° + RT ln Q and E = E° − RT ln Q/(nF).
E°cell = (-ΔH° + T ΔS°) / (n F)
If ΔS° is positive, the slope is positive: heating increases the cell potential. If ΔS° is negative, heating decreases it. The magnitude of the effect is usually modest, on the order of millivolts per kelvin, because nF is large.
The sign follows algebra: E° = (TΔrS° − ΔrH°)/(nF). A positive reaction entropy raises E° with T, while a negative entropy lowers it. If ΔrS° = +100 J/(mol·K) and n = 2, the slope is 100/[2(96485)] = 5.18 × 10⁻⁴ V/K. Heating by 20.0 K gives ΔE° ≈ +0.0104 V over a range where the slope is approximately constant. A temperature difference can be given in K or °C here; the absolute T in the Gibbs expression must still be kelvin.
For T > 0 in a range where the signs remain valid: ΔH° < 0 and ΔS° > 0 gives ΔG° < 0 (E° > 0); ΔH° > 0 and ΔS° < 0 gives ΔG° > 0 (E° < 0).
With both ΔH° and ΔS° negative, the standard-state forward reaction is favored below Tcross. With both positive, it is favored above Tcross. At Tcross, ΔG° = E° = 0. Actual direction also depends on Q.
Setting ΔG° = 0, which is the same as E°cell = 0, gives T = ΔH°/ΔS°. Watch the units: enthalpy is usually tabulated in kJ/mol and entropy in J/(mol K), so one of them must be converted before dividing.
For ΔrH° = +40.0 kJ/mol and ΔrS° = +100 J/(mol·K), Tcross ≈ 400 K. Below it, ΔrG° > 0 and E° < 0; above it the standard forward reaction becomes favored. If both changes are negative, the direction of this temperature trend reverses. If their signs are opposite, the calculated ratio is negative and does not describe a physical positive-temperature crossover. If ΔS° = 0, no finite crossover follows from the ratio formula.
At a standard crossover K = 1, whereas actual-cell equilibrium requires Q = K. An imposed composition can therefore produce nonzero E even when E° = 0. The approximation also fails if phases change or ΔH° and ΔS° vary substantially across the extrapolated interval.
T = ΔH° / ΔS°
Two separate effects appear in the cold. The thermodynamic one, from the equation above, is small. The larger practical effect is kinetic: cold electrolyte is more viscous and ion transport slows, raising internal resistance and cutting the current the battery can deliver. Thermodynamics tells you what is possible, kinetics tells you how fast.
The open-circuit reversible voltage and the voltage while supplying current are different measurements. A battery under load loses terminal voltage through internal resistance, transport limitations and electrode overpotentials. Cooling can worsen those losses even if the standard thermodynamic voltage changes only modestly or in the opposite direction. Conversely heating is not unconditionally beneficial: side reactions and material stability may change. Use ΔS° to predict the reversible standard-voltage slope, and kinetic/transport information to discuss useful current.
Electrolysis of water is endothermic with a positive entropy change, so raising the temperature lowers the required applied voltage. High-temperature electrolyzers exploit exactly this, spending cheap heat to save expensive electricity.
Combining ΔG°=ΔH°-TΔS° with ΔG°=-nFE° gives E°=(TΔS°-ΔH°)/(nF). If ΔH° and ΔS° are approximately constant across a small temperature range, dE°/dT≈ΔS°/(nF). A positive reaction entropy then makes the standard cell voltage rise with temperature.
E°cell = (-ΔH° + T ΔS°)/(nF).
d E°/dT = ΔS°/(nF); positive entropy means heating helps.
T = ΔH°/ΔS°, where E°cell = 0.
ΔH° < 0 and ΔS° > 0 give ΔG° < 0 at positive T within the stated range, not unconditional forward direction at every Q.
ΔH° > 0 and ΔS° < 0 give ΔG° > 0 at positive T within the stated range. Actual mixtures also require the Q term.
ΔG° = ΔH° - T ΔS°
E°cell = (-ΔH° + T ΔS°) / (n F)
T = ΔH° / ΔS°
A cell reaction has ΔH° = −180 kJ/mol and ΔS° = +25 J/(mol·K), with n = 2. Within a range where these values are approximately constant, does heating increase E°cell, and what is the sign of ΔG°?
Try it first: Look at the signs before doing any arithmetic.
0 of 3 steps revealed.
For the same cell, calculate E°cell at 298 K.
Try it first: Put enthalpy and entropy in the same energy unit first.
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A reaction has ΔH° = +55.0 kJ/mol and ΔS° = +150 J/(mol·K). Assuming these are approximately constant near the crossover, above what temperature is ΔG° negative?
Try it first: Identify which of the four cases this is.
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For a cell reaction with n=2 and ΔS°=+100 J/(mol·K), estimate the E° change for warming by 20 K. Take F=96485 C/mol.
Try it first: Use ΔE°≈ΔS°ΔT/(nF).
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| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Mixing kJ and J in the Gibbs equation. | Entropy is tabulated in J/(mol K) while enthalpy is in kJ/mol, so one must be converted. | Convert everything to joules before combining. |
| Using Celsius in the T ΔS° term. | The equation requires absolute temperature. | Add 273.15 to convert to kelvin. |
| Assuming heating always makes a reaction more favorable. | When ΔS° is negative, heating makes ΔG° more positive. | Check the entropy sign before predicting. |
| Looking for a positive crossover when ΔH° and ΔS° have opposite signs. | Opposite-sign ΔH° and ΔS° give no positive crossover in the constant-value model; same-sign values can give Tcross = ΔH°/ΔS°. | Check the signs first; compute a physical crossover only when ΔH°/ΔS° is positive and the model is valid. |
| Blaming cold-weather battery failure entirely on E°. | The thermodynamic shift is only millivolts; slower ion transport is the dominant effect. | Separate thermodynamic limits from kinetic ones. |
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Standard-state relationships ΔG° = ΔH° − TΔS° and ΔG° = −nFE°cell connect temperature, entropy and voltage. Treating ΔH° and ΔS° as approximately constant over a stated range gives an approximately linear E°(T). Same-sign ΔH° and ΔS° permit a positive crossover; opposite-sign cases do not. At ΔG° = 0, E° = 0, but nonstandard mixtures require the Q term. Cold battery performance also depends on kinetics and resistance, not just standard-state voltage.
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