Thermodynamics from Cell Potentials

A single electrochemical cell can report the full thermodynamic profile of a reaction. From one measured voltage, and how that voltage drifts with temperature, you can extract $\Delta G$ , $\Delta S$ , $\Delta H$ , and the equilibrium constant $K$ . No calorimeter needed. This matters because cell potentials are far easier to measure precisely than heats of reaction. A potentiometer resolves microvolts. That precision translates directly into thermodynamic data accurate enough to test theories of ion solvation, battery chemistry, and biological electron transport. This article builds the bridge from $E_{\text{cell}}$ to the four central thermodynamic quantities. You will see the master equation linking $\Delta G$ to $E$ . Then you will see how its temperature derivative yields $\Delta S$ . Finally you will combine both to get $\Delta H$ and $K$ , with two fully worked numerical examples.

Key takeaways

$\Delta G^\circ = -nFE^\circ_{\text{cell}}$ converts a measured voltage directly into a free energy change.
The temperature coefficient of the cell potential gives entropy: $\Delta S^\circ = nF(\partial E^\circ/\partial T)_p$ .
Combining $\Delta G^\circ$ and $\Delta S^\circ$ through $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ yields $\Delta H^\circ$ without any calorimetry.
The equilibrium constant follows from $\ln K = nFE^\circ/RT$ , linking electrochemistry directly to chemical equilibrium.
Real cells require care: $n$ must match the balanced equation, and $E^\circ$ must be extrapolated to true standard-state conditions.
Key idea: A cell's voltage and its temperature dependence together encode the complete thermodynamics of the cell reaction.

The core insight

Because electrical work at constant temperature and pressure equals the reversible non-expansion work, $\Delta G$ for a cell reaction is exactly $-nFE_{\text{cell}}$ , and the temperature slope of $E_{\text{cell}}$ is a direct probe of $\Delta S$ , so a voltmeter functions as a thermodynamic instrument.

The physical picture

A galvanic cell converts chemical change into electrical work. Under reversible conditions—meaning the cell is balanced against an external potential so no net current flows—the electrical work extracted equals the maximum non-expansion work available from the reaction. Thermodynamics identifies that maximum non-expansion work with $\Delta G$ . This single identification is the entire bridge between electrochemistry and chemical thermodynamics. The charge transferred when the balanced cell reaction proceeds by $n$ moles of electrons is $nF$ , where $F$ is the Faraday constant (96,485 C/mol). The work done by the cell pushing that charge through potential difference $E$ is $nFE$ . Because the cell does work on the surroundings, this appears as a decrease in Gibbs energy of the reacting system:

\Delta G = -nFE

Now consider what happens as temperature changes. The Gibbs–Helmholtz relation tells us that the temperature derivative of $\Delta G$ at constant pressure is $-\Delta S$ . Since $\Delta G$ is proportional to $-E$ , the temperature derivative of $E$ must be proportional to $\Delta S$ . Measuring how a cell's voltage drifts with a degree or two of temperature change therefore reports the reaction entropy directly—no heat measurement required. Once $\Delta G$ and $\Delta S$ are both known, $\Delta H$ falls out algebraically from the definition of Gibbs energy. And because $\Delta G^\circ$ also fixes the equilibrium constant through $\Delta G^\circ = -RT\ln K$ , the same voltage measurement pins down where the reaction sits at equilibrium.

Figure

From voltage to the four thermodynamic quantities

Figure

Typical temperature coefficient of a cell — nearly linear over small ranges

The core relations

\Delta_r G^\circ = -nFE^\circ_{\text{cell}}

where $n$ is moles of electrons transferred per reaction event, $F = 96{,}485\ \text{C mol}^{-1}$ , $E^\circ_{\text{cell}}$ is the standard cell potential.

\Delta_r S^\circ = nF\left(\frac{\partial E^\circ_{\text{cell}}}{\partial T}\right)_p

where the partial derivative is the measured temperature coefficient of the standard cell potential.

\Delta_r H^\circ = \Delta_r G^\circ + T\Delta_r S^\circ = -nFE^\circ_{\text{cell}} + nFT\left(\frac{\partial E^\circ_{\text{cell}}}{\partial T}\right)_p

where $T$ is the absolute temperature at which $E^\circ_{\text{cell}}$ was measured.

\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}

where $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ and $K$ is the thermodynamic equilibrium constant for the cell reaction as written.

Derivation AFrom electrical work to the four thermodynamic functions

Goal: show that $E^\circ_{\text{cell}}$ and its temperature slope together determine $\Delta G^\circ$ , $\Delta S^\circ$ , $\Delta H^\circ$ , and $K$ .

Step 1 — Identify electrical work with non-expansion work. For a reversible cell operating at constant $T$ and $p$ , the maximum non-expansion (electrical) work obtainable equals $\Delta G$ of the cell reaction, a standard result of combining the first and second laws for a reversible process:

w_{\text{add,max}} = \Delta G

The charge passed when the reaction advances by one mole of reaction (as written, with stoichiometric coefficient set by $n$ electrons) is $nF$ . The reversible electrical work delivered by the cell is $-nFE_{\text{cell}}$ (negative because the system does work on the surroundings). Equating the two:

\Delta G = -nFE_{\text{cell}}

Applying this at standard states (unit activities) gives the first core relation.

Step 2 — Differentiate with respect to temperature. The Gibbs–Helmholtz–type relation from the fundamental equation $dG = -SdT + Vdp$ gives, at constant pressure,

\left(\frac{\partial \Delta G^\circ}{\partial T}\right)_p = -\Delta S^\circ

Substitute $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ , treating $n$ and $F$ as constants:

-nF\left(\frac{\partial E^\circ_{\text{cell}}}{\partial T}\right)_p = -\Delta S^\circ

Rearranging gives the entropy relation:

\Delta S^\circ = nF\left(\frac{\partial E^\circ_{\text{cell}}}{\partial T}\right)_p

Step 3 — Recover the enthalpy from the definition of Gibbs energy. Since $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ at any fixed $T$ , solve for $\Delta H^\circ$ :

\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ

Substitute both expressions found above:

\Delta H^\circ = -nFE^\circ_{\text{cell}} + nFT\left(\frac{\partial E^\circ_{\text{cell}}}{\partial T}\right)_p

This is the enthalpy relation — obtained entirely from electrical measurements, with no calorimetry.

Step 4 — Connect to the equilibrium constant. The general link between standard Gibbs energy and the thermodynamic equilibrium constant is $\Delta G^\circ = -RT\ln K$ . Equate this to $-nFE^\circ_{\text{cell}}$ :

-RT\ln K = -nFE^\circ_{\text{cell}}

Solving for $\ln K$ :

\boxed{\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}, \qquad \Delta S^\circ = nF\left(\frac{\partial E^\circ_{\text{cell}}}{\partial T}\right)_p, \qquad \Delta H^\circ = -nFE^\circ_{\text{cell}} + nFT\left(\frac{\partial E^\circ_{\text{cell}}}{\partial T}\right)_p}

Worked examples

Example 1 — The Weston standard cell (foundational). The Weston cell, Cd(Hg) amalgam | CdSO₄·8/3H₂O(s), Hg₂SO₄(s) | Hg(l), has $E^\circ = 1.01832\ \text{V}$ at $298\ \text{K}$ and a well-known temperature coefficient

\left(\frac{\partial E^\circ}{\partial T}\right)_p = -4.06\times10^{-4}\ \text{V K}^{-1}

The cell reaction transfers $n = 2$ electrons. Find $\Delta G^\circ$ , $\Delta S^\circ$ , and $\Delta H^\circ$ at 298 K. Step 1 — Gibbs energy.

\Delta G^\circ = -nFE^\circ = -(2)(96{,}485\ \text{C mol}^{-1})(1.01832\ \text{V})

\Delta G^\circ = -1.9654\times10^{5}\ \text{J mol}^{-1} = -196.5\ \text{kJ mol}^{-1}

Step 2 — Entropy.

\Delta S^\circ = nF\left(\frac{\partial E^\circ}{\partial T}\right)_p = (2)(96{,}485)( -4.06\times10^{-4}\ \text{V K}^{-1})

\Delta S^\circ = -78.4\ \text{J K}^{-1}\text{mol}^{-1}

The negative sign is physically sensible: the cell reaction converts dissolved and solid reactants into a more ordered solid/liquid product assembly, decreasing entropy. Step 3 — Enthalpy.

\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ = -196{,}540\ \text{J mol}^{-1} + (298\ \text{K})(-78.4\ \text{J K}^{-1}\text{mol}^{-1})

\Delta H^\circ = -196{,}540 - 23{,}363 = -219{,}900\ \text{J mol}^{-1} = -219.9\ \text{kJ mol}^{-1}

Sanity check: $\Delta H^\circ$ is more negative than $\Delta G^\circ$ , consistent with $\Delta S^\circ < 0$ (the $-T\Delta S^\circ$ term is positive, so $\Delta G^\circ$ is "less negative" than $\Delta H^\circ$ — exactly what we found).

Example 2 — The Daniell cell: equilibrium constant and enthalpy (applied, exam-level). Consider the Daniell cell,

\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}

with $n = 2$ . At 298 K, $E^\circ_{\text{cell}} = 1.10\ \text{V}$ (from standard reduction potentials $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$ and $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$ ). Careful measurement gives

\left(\frac{\partial E^\circ}{\partial T}\right)_p = -4.5\times10^{-4}\ \text{V K}^{-1}

Find $K$ at 298 K and $\Delta H^\circ$ . Step 1 — Equilibrium constant.

\ln K = \frac{nFE^\circ}{RT} = \frac{(2)(96{,}485\ \text{C mol}^{-1})(1.10\ \text{V})}{(8.314\ \text{J K}^{-1}\text{mol}^{-1})(298\ \text{K})}

\ln K = \frac{212{,}267}{2477.6} = 85.68

K = e^{85.68} \approx 1.6\times10^{37}

This enormous $K$ confirms the reaction runs essentially to completion — consistent with zinc displacing copper spontaneously and completely in practice. Step 2 — Gibbs energy and entropy.

\Delta G^\circ = -nFE^\circ = -(2)(96{,}485)(1.10) = -212{,}270\ \text{J mol}^{-1} = -212.3\ \text{kJ mol}^{-1}

\Delta S^\circ = nF\left(\frac{\partial E^\circ}{\partial T}\right)_p = (2)(96{,}485)(-4.5\times10^{-4}) = -86.8\ \text{J K}^{-1}\text{mol}^{-1}

Step 3 — Enthalpy.

\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ = -212{,}270 + (298)(-86.8)

\Delta H^\circ = -212{,}270 - 25{,}866 = -238{,}100\ \text{J mol}^{-1} = -238.1\ \text{kJ mol}^{-1}

Sanity check: This compares reasonably with the calorimetric enthalpy of the Daniell reaction (literature values cluster near $-217$ to $-230\ \text{kJ mol}^{-1}$ depending on activity corrections), confirming the electrochemical route gives thermodynamically consistent numbers without a calorimeter.

Limiting cases & common traps

Situation	What to watch for
$\partial E^\circ/\partial T \approx 0$	$\Delta S^\circ \approx 0$ ; then $\Delta H^\circ \approx \Delta G^\circ$
Large negative slope	$\Delta S^\circ$ strongly negative; $\Delta H^\circ$ more negative than $\Delta G^\circ$
$n$ chosen inconsistently with the balanced equation	All four quantities scale incorrectly — always fix $n$ from the balanced half-reactions first
Using measured (non-standard) $E$ instead of $E^\circ$	Must correct with the Nernst equation to standard-state activities before applying these relations
Assuming $\Delta S^\circ$ and $\Delta H^\circ$ are temperature-independent over wide ranges	The coefficient $\partial E^\circ/\partial T$ itself can vary with $T$ ; treat it as locally linear only over narrow intervals

The chemist behind it

Walther Nernst's theoretical framework connecting electrochemical potentials to chemical thermodynamics (culminating in his third law work) made this whole program possible, but the practical exploitation of temperature-coefficient measurements to extract entropies and enthalpies owes much to the American physical chemist Gilbert N. Lewis, whose meticulous electrochemical and thermodynamic tables — compiled with Merle Randall — turned cell-potential thermodynamics into a routine laboratory technique still taught today.

Origins · 1 minGilbert N. Lewis and how this was discovered

Gilbert N. Lewis

1875–1946 · American physical chemist who systematized free energy, activity, and electrochemical thermodynamics into the modern data tables still used to extract ΔH and ΔS from cell voltages

Ask the tutor: How would these relations change if the cell reaction also produced or consumed gas, adding a pressure-dependence to $E^\circ$ ?

Why is the electrical work equal to ΔG in the first place

For a reversible process at constant temperature and pressure, the maximum non-expansion work obtainable is exactly $\Delta G$ — this follows from combining the first and second laws for a reversible path. A cell operated reversibly (balanced against an external potential so current flow is infinitesimal) delivers exactly this maximum work as electrical energy, $-nFE$ . Any real, current-carrying cell delivers less work than this because of internal resistance and concentration polarization, which is why the relations here strictly require the reversible, near-equilibrium limit.

Why does the temperature coefficient give entropy rather than enthalpy directly

The Gibbs-Helmholtz relation states that the temperature derivative of $\Delta G$ at constant pressure equals $-\Delta S$ , not $\Delta H$ . Since $\Delta G$ is directly proportional to $-E^\circ$ , differentiating $E^\circ$ with respect to $T$ isolates $\Delta S^\circ$ cleanly. Enthalpy only appears afterward, as an algebraic combination of $\Delta G^\circ$ (from $E^\circ$ itself) and $\Delta S^\circ$ (from the slope), via $\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ$ .

How precise does the temperature coefficient measurement need to be

Very precise, because $\Delta S^\circ$ depends linearly on a small slope multiplied by the large factor $nF \approx 10^5$ C/mol. A temperature-coefficient error of just $10^{-5}\ \text{V K}^{-1}$ translates into roughly $2\ \text{J K}^{-1}\text{mol}^{-1}$ of error in $\Delta S^\circ$ for $n=2$ . Precision electrochemical thermodynamics typically requires temperature control to millikelvins and voltage resolution to microvolts across several closely spaced temperatures, then a linear or low-order polynomial fit to extract the slope reliably.

Can these relations be applied to a cell that is not at standard conditions

Not directly — you must first use the Nernst equation to correct the measured potential to standard-state activities, or work with the reaction quotient $Q$ explicitly. The relations here use $E^\circ$ and $\Delta G^\circ$ , both defined at unit activity for all reactants and products. If concentrations deviate from standard states, the appropriate substitution is $\Delta G = \Delta G^\circ + RT\ln Q$ , and the temperature-coefficient analysis must track how both $E^\circ$ and $Q$ depend on $T$ .