Thermodynamic data from EMF

Gibbs energy from a single EMF reading. The maximum non-expansion (electrical) work obtainable from a spontaneous cell reaction equals the decrease in Gibbs energy of that reaction. This is the electrochemical restatement of $\Delta_r G = \sum \nu_i \mu_i$, and it is the anchor relation for everything that follows:

where: $n$ is the number of electrons transferred per formula unit of reaction (from the balanced half-reactions), $F = 96485\ \text{C mol}^{-1}$ is the Faraday constant, and $E$ is the measured cell EMF in volts. Under standard conditions this becomes $\Delta G^\circ = -nFE^\circ$.

From EMF to equilibrium constant. Because $\Delta G^\circ$ is also related to $K$ through $\Delta G^\circ = -RT\ln K$, equating the two expressions for $\Delta G^\circ$ gives a direct EMF–equilibrium bridge — no separate equilibrium experiment is needed once $E^\circ$ is known:

where: $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$, $T$ is the absolute temperature, and the $0.0592$ form is simply $2.303RT/F$ evaluated at $298\ \text{K}$. A modestly large $E^\circ$ already implies an enormous $K$, because $n$ and $F$ multiply it before it appears in an exponent.

Entropy from the temperature coefficient of EMF. $\Delta G$ is a state function, so the general thermodynamic slope $(\partial \Delta G/\partial T)_P = -\Delta S$ applies to it exactly as it would to any Gibbs energy. Differentiating $\Delta G = -nFE$ with respect to temperature at constant pressure and substituting gives

where: $(\partial E/\partial T)_P$, the temperature coefficient of the cell EMF, is obtained purely experimentally — measure $E$ at a few nearby temperatures with everything else (concentrations, pressure) held fixed, and take the slope of $E$ against $T$. No independent entropy measurement is required.

Enthalpy via Gibbs–Helmholtz. Combine the definition $\Delta G = \Delta H - T\Delta S$ with the two relations above by substituting $\Delta G = -nFE$ and $\Delta S = nF(\partial E/\partial T)_P$:

This is the electrochemical form of the Gibbs–Helmholtz equation. Read its sign structure carefully: if $(\partial E/\partial T)_P > 0$, the EMF rises with temperature, so $\Delta S > 0$ and the reaction absorbs heat reversibly from the surroundings as it runs — making $\Delta H$ less negative (or more positive) than the bare $-nFE$ term alone would suggest. If $(\partial E/\partial T)_P < 0$, the correction pushes $\Delta H$ more negative than $\Delta G$. The figure below traces the whole chain of reasoning in one glance — a single measured $E$ and a measured slope feed four different thermodynamic quantities without ever touching a bomb calorimeter.

Problem. For the Daniell cell reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$, $E^\circ_{cell} = 1.10\ \text{V}$ at $298\ \text{K}$, and $(\partial E/\partial T)_P = -1.4\times10^{-4}\ \text{V K}^{-1}$. Find $\Delta G^\circ$, $\Delta S^\circ$, $\Delta H^\circ$, and $K$.

Step 1. Identify $n$ from the balanced half-reactions: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ and $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$, so two electrons transfer and $n = 2$.

Step 2. Compute $\Delta G^\circ$ from $\Delta G^\circ = -nFE^\circ = -(2)(96485)(1.10) = -212\,267\ \text{J mol}^{-1} \approx -212.3\ \text{kJ mol}^{-1}$.

Step 3. Compute $\Delta S^\circ$ from the measured slope: $\Delta S^\circ = nF(\partial E/\partial T)_P = (2)(96485)(-1.4\times10^{-4}) = -27.0\ \text{J K}^{-1}\text{mol}^{-1}$. The negative sign says the reaction becomes slightly less favoured entropically as temperature rises, consistent with the negative slope.

Step 4. Compute $\Delta H^\circ$ via the Gibbs–Helmholtz result just derived: $\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ = -212\,267 + (298)(-27.0) = -212\,267 - 8046 = -220\,313\ \text{J mol}^{-1} \approx -220.3\ \text{kJ mol}^{-1}$.

Step 5. Compute $K$ from the standard EMF: $\log_{10}K = nE^\circ/0.0592 = (2)(1.10)/0.0592 = 37.16$, so $K = 10^{37.16} \approx 1.4\times10^{37}$. The reaction has $\Delta G^\circ \approx -212.3\ \text{kJ mol}^{-1}$, $\Delta S^\circ \approx -27.0\ \text{J K}^{-1}\text{mol}^{-1}$, $\Delta H^\circ \approx -220.3\ \text{kJ mol}^{-1}$, and $K \approx 1.4\times10^{37}$ — the astronomically large $K$ confirms the reaction runs essentially to completion, matching the large positive $E^\circ$.

Competitive stretch

A related cell shows $E = 0.640\ \text{V}$ at $283\ \text{K}$ and $E = 0.652\ \text{V}$ at $313\ \text{K}$, with $n = 1$. The slope is $(\partial E/\partial T)_P \approx (0.652-0.640)/(313-283) = 4.0\times10^{-4}\ \text{V K}^{-1}$, giving $\Delta S = (1)(96485)(4.0\times10^{-4}) = 38.6\ \text{J K}^{-1}\text{mol}^{-1}$. Interpolating $E$ at $298\ \text{K}$ gives $E_{298} \approx 0.646\ \text{V}$, so $\Delta G \approx -62.3\ \text{kJ mol}^{-1}$, and $\Delta H = \Delta G + T\Delta S \approx -50.8\ \text{kJ mol}^{-1}$ — the positive slope here makes $\Delta H$ less negative than $\Delta G$, opposite to the Daniell cell above.

• Flipping the sign in $\Delta S = nF(\partial E/\partial T)_P$: a positive slope (EMF rising with $T$) gives positive $\Delta S$ directly, with no extra minus sign — the minus sign from $(\partial \Delta G/\partial T)_P = -\Delta S$ has already been absorbed into the derivation.
• Mixing measured (possibly non-standard) $E$ with the equilibrium-constant relation, which strictly needs $E^\circ$: $\ln K = nFE^\circ/RT$ only holds for the standard cell EMF.
• Using two different temperatures in the two terms of $\Delta H = -nFE + nFT(\partial E/\partial T)_P$ — both the $E$ and the $T$ multiplying the slope must refer to the same measurement temperature, in kelvin.
• Assuming $(\partial E/\partial T)_P$ is always small and negative; its sign and size depend entirely on the reaction and must be read from the data, as the two worked examples above show opposite signs.

Ask the tutor: if turning $\Delta G = \Delta H - T\Delta S$ into the final Gibbs–Helmholtz expression for $\Delta H$ feels like copying steps rather than deriving them, ask the AI tutor to rebuild it with you line by line and then test you with a fresh EMF-vs-temperature data set.