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.
Thermodynamics gives ΔG° = ΔH° - T ΔS°, and electrochemistry gives ΔG° = -nFE°cell. Setting them equal and solving for E° produces a straight line in temperature with intercept -ΔH°/(nF) and slope ΔS°/(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.
Exothermic with positive entropy change: E° positive at every temperature. Endothermic with negative entropy change: E° negative at every temperature. Exothermic with negative entropy change: favorable only below the crossover. Endothermic with positive entropy change: favorable only above the crossover.
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.
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.
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.
E°cell = (-ΔH° + T ΔS°)/(nF).
d E°/dT = ΔS°/(nF); positive entropy means heating helps.
T = ΔH°/ΔS°, where E°cell = 0.
ΔH° negative with ΔS° positive.
ΔH° positive with ΔS° negative.
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. Does heating increase or decrease E°cell, and is the reaction favorable at all temperatures?
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.
0 of 3 steps revealed.
A reaction has ΔH° = +55.0 kJ/mol and ΔS° = +150 J/(mol K). Above what temperature does it become thermodynamically favorable?
Try it first: Identify which of the four cases this is.
0 of 3 steps revealed.
Why it's wrong: Entropy is tabulated in J/(mol K) while enthalpy is in kJ/mol, so one must be converted.
Check instead: Convert everything to joules before combining.
Why it's wrong: The equation requires absolute temperature.
Check instead: Add 273.15 to convert to kelvin.
Why it's wrong: When ΔS° is negative, heating makes ΔG° more positive.
Check instead: Check the entropy sign before predicting.
Why it's wrong: Same-sign cases are favorable or unfavorable at all temperatures, so no crossover exists.
Check instead: Classify the case first, then compute only if it is a mixed case.
Why it's wrong: The thermodynamic shift is only millivolts; slower ion transport is the dominant effect.
Check instead: Separate thermodynamic limits from kinetic ones.
No practice questions are available for this topic yet. You can still practice the whole unit.
Combining ΔG° = ΔH° - T ΔS° with ΔG° = -nFE°cell gives E°cell = (-ΔH° + T ΔS°)/(nF), so a cell potential varies linearly with temperature and its slope is ΔS°/(nF). A reaction with a positive entropy change produces a larger potential as temperature rises, while a negative entropy change means the potential falls with heating. This produces the four familiar thermodynamic cases: exothermic with increasing entropy is favorable at all temperatures, endothermic with decreasing entropy is never favorable, and the two mixed cases switch at the crossover temperature T = ΔH°/ΔS°, where ΔG° = 0 and E°cell = 0. This reasoning explains real behavior: car batteries lose cranking power in the cold, and industrial electrolysis of water is run hot because heating lowers the voltage that must be supplied. Note that these relationships describe standard conditions, and that E° values in reference tables are quoted at 25 °C.
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