Result

Free energy and cell potential

T-057Home CU-205Threads energy · equilibrium

Statement

The relation dG = -nFE.

Why it matters

galvanic-cell-emf showed how to assemble a cell's standard potential from tabulated half-reactions, but a potential by itself does not directly say how much useful work a cell can deliver, nor does it connect to the equilibrium constant of the underlying reaction. This result supplies exactly that missing link, tying the purely electrical quantity \(E\) to the thermodynamic quantity \(\Delta G\) that governs spontaneity and equilibrium throughout the rest of chemistry; nernst-equation and faraday-electrolysis both build directly on the \(nF\) bookkeeping introduced here.

Hypotheses

The cell operates reversibly (infinitesimally close to equilibrium, delivering its maximum possible non-expansion work).Any real, finite-current cell dissipates some energy irreversibly (as heat, through internal resistance and overpotential), delivering strictly less electrical work than the reversible maximum; the relation below therefore uses the thermodynamic (open-circuit, essentially zero-current) cell potential, not the potential measured while current is actually flowing under load. All non-expansion work done by the cell is exclusively electrical work.This is what allows the general thermodynamic identity, that reversible non-expansion work equals \(\Delta G\) at constant temperature and pressure, to be applied directly to a cell without needing to separately account for any other form of work being done simultaneously.

Proof

\[ w_{\text{elec}} = -nFE \]

Moving \(n\) moles of electrons through a potential difference \(E\) does electrical work equal to charge times potential; with \(n\) moles of electrons carrying charge \(nF\) (faraday-electrolysis), and adopting the sign convention that work done by the system on the surroundings is negative, the electrical work delivered by the cell is \(-nFE\). A

\[ \Delta G = w_{\text{non-expansion, reversible, max}} \]

A standard result of thermodynamics: at constant temperature and pressure, the maximum non-expansion work obtainable from a process (achieved only in the reversible limit, Hypotheses) equals the free-energy change of that process. A

\[ \Delta G = -nFE \]

Since the cell's only non-expansion work is electrical (Hypotheses), Steps 1 and 2 identify the same quantity, giving the central relation directly; at standard conditions this specialises to \(\Delta G^\circ=-nFE^\circ\). A

\[ \Delta G^\circ = -RT\ln K \quad\Rightarrow\quad \ln K = \frac{nFE^\circ}{RT} \]

Combining Step 3 at standard conditions with the standard relation between \(\Delta G^\circ\) and the equilibrium constant \(K\) gives a second, equally important consequence: a cell's standard potential directly determines the equilibrium constant of its overall reaction, without needing \(\Delta G^\circ\) as an intermediate quantity at all if \(E^\circ\) is already known. A

Result

\[ \Delta G = -nFE, \qquad \Delta G^\circ=-nFE^\circ = -RT\ln K \]

Reading. Cell potential, free-energy change, and equilibrium constant are three equivalent descriptions of the same underlying spontaneity of a redox reaction; any one can be computed directly from either of the other two.

Scope. The reversible-operation assumption (Hypotheses) means \(E\) here is the thermodynamic (zero-current, open-circuit) potential; a working cell under load delivers strictly less usable electrical energy than \(-nFE\) predicts, the difference lost to internal irreversibility.

Corollaries & converses

Since \(F,n>0\) always, \(E^\circ>0\) is exactly equivalent to \(\Delta G^\circ\lt0\): the Result confirms directly what galvanic-cell-emf's Step 4 argued qualitatively, that the spontaneous cell reaction is the one giving a positive standard potential.
nernst-equation extends this relation to non-standard conditions by combining Step 3 with the general \(\Delta G=\Delta G^\circ+RT\ln Q\) relation, giving cell potential as a function of the actual reaction quotient \(Q\) rather than only at standard state.
Converse: measuring a cell's standard potential experimentally, via galvanic-cell-emf's construction, gives an indirect but often very precise route to an equilibrium constant that might otherwise be difficult to measure directly (Worked examples).

Fails without

Use the potential measured under load (current flowing) rather than the reversible, open-circuit potential (Hypotheses): the true reversible work is overestimated in magnitude by the Result if the lower, under-load voltage is mistakenly substituted, or the calculation simply gives a value that does not correspond to any genuine thermodynamic quantity.
Substitute the wrong \(n\), mismatched to the actual balanced overall cell reaction as written: since \(\Delta G\) scales directly with \(n\) while \(E\) does not, using an \(n\) inconsistent with the specific reaction equation gives a \(\Delta G\) that does not correspond to that equation at all.

Common errors

Dropping the negative sign in \(\Delta G=-nFE\), which reverses the predicted spontaneity relative to the sign of \(E\).
Using the cell potential measured under load (with current flowing, subject to internal resistance losses) in place of the reversible, open-circuit potential the Result strictly requires (Hypotheses).
Forgetting that \(n\) in \(\Delta G=-nFE\) must match the actual number of electrons transferred in the balanced overall cell reaction as written, which changes if the reaction is rescaled (unlike \(E\) itself, which as an intensive quantity does not rescale).
Confusing \(\Delta G^\circ\) (standard-state free energy, used with \(E^\circ\)) with the non-standard \(\Delta G\) (used with the actual, non-standard \(E\)) when applying Step 4's link to \(K\), which strictly requires standard-state quantities throughout.

Discussion

The relation \(\Delta G=-nFE\) is one of the cleanest bridges in physical chemistry between an electrical measurement and a chemical thermodynamic quantity, and it is precisely why electrochemical cells are used as a standard laboratory method for measuring equilibrium constants and free energies that would be difficult or impossible to access by direct chemical means — particularly for reactions with very large or very small \(K\), where a modest, easily measured cell potential (through the exponential in Step 4) corresponds to an enormous range of \(K\).

Because \(\ln K\) depends on \(E^\circ\) linearly while \(K\) itself depends on it exponentially, even quite modest experimental uncertainty in a measured \(E^\circ\) (a few millivolts) can translate into a substantial relative uncertainty in the inferred \(K\) for reactions with large \(n\); this sensitivity is a genuine practical consideration in using electrochemical measurements for precise equilibrium-constant determination.

Common misconception: that a cell's rated or measured operating voltage under load is the same quantity that belongs in \(\Delta G=-nFE\). The Result strictly requires the reversible, zero-current thermodynamic potential (Hypotheses); a cell delivering current to an external circuit always operates below this value, the gap representing energy lost to internal irreversibility rather than converted to useful work.

Worked examples

\[ \text{Daniell cell: } E^\circ=+1.10\,\text{V},\quad n=2 \]

Using the standard potential found for the Daniell cell (galvanic-cell-emf's Worked example), \(\Delta G^\circ = -nFE^\circ = -2\times96485\times1.10 = -212{,}267\,\text{J/mol} \approx -212\,\text{kJ/mol}\), confirming the reaction is strongly spontaneous under standard conditions. A

\[ \ln K = \frac{nFE^\circ}{RT} = \frac{2\times96485\times1.10}{8.314\times298} = \frac{212{,}267}{2477.6} \approx 85.7 \]

Substituting into Step 4 with \(R=8.314\,\text{J/mol/K}\) and \(T=298\,\text{K}\) gives an enormous predicted equilibrium constant for the Daniell cell reaction, \(K=e^{85.7}\approx1.6\times10^{37}\), reflecting how strongly the reaction favours products at equilibrium. A

\[ \Delta G^\circ_{\text{Daniell}}\approx-212\,\text{kJ/mol}, \qquad K\approx1.6\times10^{37} \]

Reading. A modest, easily measured cell potential of just over one volt corresponds, through the Result, to an equilibrium constant so large that the reverse reaction is essentially never observed at equilibrium.

Scope. Any measured standard cell potential can be converted this way; the exponential sensitivity in Step 4 is why electrochemical methods are especially valuable for reactions with very large \(K\), inaccessible to direct concentration measurement.

Further example. Consider a galvanic cell based on the reaction between silver ions and copper metal: \(\text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)\). Given the standard reduction potentials \(E^\circ(\text{Ag}^+/\text{Ag}) = +0.80\,\text{V}\) and \(E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\,\text{V}\), calculate the standard cell potential \(E^\circ_{\text{cell}}\), the standard Gibbs free energy change \(\Delta G^\circ\), and the equilibrium constant \(K\) at \(298\,\text{K}\).

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80\,\text{V} - 0.34\,\text{V} = +0.46\,\text{V} \]

The cell potential is the difference between the reduction potential of the cathode (where reduction occurs, \(\text{Ag}^+\)) and the anode (where oxidation occurs, \(\text{Cu}\)). A

\[ n = 2 \]

The balanced equation shows that 2 moles of electrons are transferred per mole of reaction (oxidation of \(\text{Cu}\) to \(\text{Cu}^{2+}\) releases 2e⁻, reduction of 2 \(\text{Ag}^+\) consumes 2e⁻). A

\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} = -2 \times 96485\,\text{C/mol} \times 0.46\,\text{V} = -88{,}760.6\,\text{J/mol} \]

Using the fundamental relation \(\Delta G^\circ = -nFE^\circ\), we convert the electrical work potential into chemical free energy. The negative sign indicates spontaneity. A

\[ \ln K = \frac{nFE^\circ}{RT} = \frac{88{,}760.6}{8.314 \times 298} \]

Relating \(\Delta G^\circ\) to the equilibrium constant via \(\Delta G^\circ = -RT \ln K\) allows us to find the position of equilibrium. A

\[ \ln K = \frac{88{,}760.6}{2477.572} \approx 35.83 \] \[ K = e^{35.83} \approx 3.6 \times 10^{15} \]

Exponentiating the result gives the equilibrium constant. The large value confirms the reaction strongly favors products, consistent with the positive cell potential. A

\[ E^\circ_{\text{cell}} = +0.46\,\text{V}, \quad \Delta G^\circ \approx -88.8\,\text{kJ/mol}, \quad K \approx 3.6 \times 10^{15} \]

Reading. The positive cell potential and large equilibrium constant indicate that the displacement of silver by copper is thermodynamically favorable. The free energy change quantifies the maximum non-expansion work obtainable from this reaction under standard conditions.

Scope. This calculation holds for standard conditions (1 M concentrations, 298 K). For non-standard conditions, the Nernst equation must be used to adjust the potential before calculating \(\Delta G\).

Problems

A cell has \(E^\circ=+0.320\,\text{V}\) and \(n=2\). Compute \(\Delta G^\circ\) in kJ/mol.
Solution
\(\Delta G^\circ=-nFE^\circ = -2\times96485\times0.320 = -61{,}750\,\text{J/mol} = -61.7\,\text{kJ/mol}\).
Using the same cell as Problem 1 at \(T=298\,\text{K}\), find the equilibrium constant \(K\) for its overall reaction.
Solution
\(\ln K = nFE^\circ/RT = 61{,}750/(8.314\times298) = 61{,}750/2477.6 = 24.9\). \(K=e^{24.9}\approx6.6\times10^{10}\), a large but far more modest equilibrium constant than the Daniell cell example, reflecting its smaller standard potential.
Explain why measuring a cell's potential while it is actively powering a device (under load) would give an unreliable estimate of \(\Delta G\) via the Result, and state which potential should be used instead.
Solution
The Result's derivation (Step 2) specifically requires the maximum, reversible non-expansion work, obtained only in the limit of an infinitesimally slow, essentially zero-current process. A cell under load loses energy irreversibly to internal resistance and electrode overpotentials, so its measured operating voltage is lower than the true thermodynamic potential, and using it would underestimate the magnitude of \(\Delta G\). The correct quantity is the open-circuit (zero-current) cell potential, measured with a high-impedance voltmeter that draws negligible current.
A galvanic cell is constructed using the \(\text{Zn}^{2+}/\text{Zn}\) and \(\text{Cu}^{2+}/\text{Cu}\) half-cells. The standard reduction potentials are \(E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\,\text{V}\) and \(E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\,\text{V}\). Calculate the standard cell potential \(E^\circ_{\text{cell}}\) and the standard Gibbs free energy change \(\Delta G^\circ\) for the overall reaction.
Solution
The cathode is the reduction of \(\text{Cu}^{2+}\) and the anode is the oxidation of \(\text{Zn}\). \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10\,\text{V}\). The reaction is \(\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)\), which involves \(n=2\) electrons. \(\Delta G^\circ = -nFE^\circ_{\text{cell}} = -2 \times 96485 \times 1.10 = -212{,}267\,\text{J/mol}\). Converting to kJ/mol: \(-212.3\,\text{kJ/mol}\).
For a certain redox reaction with \(n=1\), the equilibrium constant is measured to be \(K = 10^{10}\) at \(298\,\text{K}\). Calculate the standard cell potential \(E^\circ\) for this reaction.
Solution
Using the relation \(\Delta G^\circ = -RT \ln K = -nFE^\circ\), we have: \(E^\circ = \frac{RT \ln K}{nF}\). Given \(K=10^{10}\), \(\ln K = 10 \ln 10 \approx 10 \times 2.3026 = 23.026\). \(E^\circ = \frac{8.314 \times 298 \times 23.026}{1 \times 96485}\). Numerator: \(8.314 \times 298 = 2477.572\). \(2477.572 \times 23.026 \approx 57{,}049\). \(E^\circ = 57{,}049 / 96485 \approx 0.591\,\text{V}\).