10.3 Cell Potential, Gibbs Free Energy & Equilibrium (Nernst Equation)

10.3 Cell Potential, Gibbs Free Energy & Equilibrium (Nernst Equation)

Quick Summary: Electrochemistry directly bridges thermodynamics and chemical equilibrium through fundamental quantitative relationships. The standard cell potential ($E^\circ_{\text{cell}}$) is proportional to standard Gibbs free energy ($\Delta G^\circ = -nFE^\circ_{\text{cell}}$) and dictates the equilibrium constant ($K$). Under non-standard concentrations and pressures, operational cell potential is governed by the Nernst equation ($E_{\text{cell}} = E^\circ_{\text{cell}} - (0.0592\text{ V}/n)\log_{10} Q$). As a cell discharges, reactants convert to products, increasing the reaction quotient $Q$ until $E_{\text{cell}}$ drops to zero at dynamic chemical equilibrium.

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1. Thermodynamic Link: Cell Potential & Gibbs Free Energy

The thermodynamic driving force of a chemical process is measured by the change in Gibbs free energy ($\Delta G$), representing the maximum non-expansion electrical work ($w_{\text{elec, max}}$) a system can deliver to its surroundings: $\Delta G = w_{\text{elec, max}} = -q E_{\text{cell}}$

In an electrochemical cell, the total charge transferred by $n$ moles of electrons is $q = nF$, where:

• $n$ is the number of moles of electrons transferred in the balanced redox equation.
• $F$ is Faraday's constant, the charge carried by one mole of electrons: $F = 96,485\text{ C/mol }e^- = 96,485\text{ J/(V}\cdot\text{mol)}$

Combining these yields the core thermodynamic connection: $\Delta G = -n F E_{\text{cell}}$ Under standard-state conditions ($1.0\text{ M}$ solutes, $1.0\text{ atm}$ gases, $298.15\text{ K}$): $\Delta G^\circ = -n F E^\circ_{\text{cell}}$

Because $n$ and $F$ are positive constants, the algebraic signs of $E_{\text{cell}}$ and $\Delta G$ are inverted:

• Spontaneous: $E^\circ_{\text{cell}} > 0 \iff \Delta G^\circ < 0$
• Nonspontaneous: $E^\circ_{\text{cell}} < 0 \iff \Delta G^\circ > 0$
• Neither side favored under standard conditions: $E^\circ_{\text{cell}} = 0 \iff \Delta G^\circ = 0 \iff K = 1$. (At actual equilibrium it is the non-standard $E_{\text{cell}}$ and $\Delta G$ that equal zero, whatever the value of $K$.)

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2. Standard Cell Potential & The Equilibrium Constant ($K$)

Standard free energy connects to the chemical equilibrium constant ($K$) through: $\Delta G^\circ = -RT \ln K$

Equating the two expressions for $\Delta G^\circ$: $-n F E^\circ_{\text{cell}} = -RT \ln K \implies E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K$

At $298.15\text{ K}$ ($25\ ^\circ\text{C}$), substituting universal constants ($R = 8.314\text{ J/(mol}\cdot\text{K)}$) and converting to base-10 logarithm yields: $E^\circ_{\text{cell}} = \frac{0.0592\text{ V}}{n} \log_{10} K \quad \iff \quad \log_{10} K = \frac{n E^\circ_{\text{cell}}}{0.0592\text{ V}}$

Electrochemistry-Thermodynamics-Equilibrium Triangle

Do not confuse $\Delta G^\circ = 0$ (the special case $K = 1$) with equilibrium itself: every reaction mixture at equilibrium has $\Delta G = 0$ and $E_{\text{cell}} = 0$.

Even small positive cell potentials correspond to enormous equilibrium constants. For example, a reaction with $n = 2$ and $E^\circ_{\text{cell}} = +0.60\text{ V}$ yields: $\log_{10} K = \frac{2 \times 0.60}{0.0592} \approx 20.27 \implies K \approx 1.9 \times 10^{20}$

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3. Non-Standard Conditions & The Nernst Equation

When electrolyte concentrations deviate from standard $1.0\text{ M}$ conditions, free energy follows: $\Delta G = \Delta G^\circ + RT \ln Q$ where $Q$ is the reaction quotient.

Substituting $\Delta G = -nFE_{\text{cell}}$ and $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ produces the Nernst Equation: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q$ At $298.15\text{ K}$, this simplifies to: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592\text{ V}}{n} \log_{10} Q$

Concentration Effects and Le Chatelier's Principle

• Increasing Reactants: Decreases $Q$ ($Q < 1$), making $\log_{10} Q < 0$. The subtracted term becomes positive, raising potential ($E_{\text{cell}} > E^\circ_{\text{cell}}$).
• Increasing Products: Increases $Q$ ($Q > 1$), making $\log_{10} Q > 0$. The potential decreases ($E_{\text{cell}} < E^\circ_{\text{cell}}$).

The Equilibrium State of a "Dead" Battery

As a battery discharges, reactants are consumed while products accumulate. Thus, $Q$ progressively increases toward $K$. When dynamic equilibrium is reached ($Q = K$): $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592\text{ V}}{n} \log_{10} K = 0.00\text{ V}$ At this point, $\Delta G = 0$, and the cell can no longer generate electrical work. Notice that while $E_{\text{cell}}$ drops to zero, $E^\circ_{\text{cell}}$ remains invariant because standard potentials reflect fixed reference states.

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4. Concentration Cells

A concentration cell employs identical electrodes in both half-cells immersed in solutions differing only in solute concentration: $\text{Ag}(s) \mid \text{Ag}^+(aq, \text{dilute}) \parallel \text{Ag}^+(aq, \text{concentrated}) \mid \text{Ag}(s)$

Because both half-cells contain the same chemical couple: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.00\text{ V}$

The cell is driven purely by the entropy of dilution, acting to equalize ionic concentrations:

• Anode (Dilute Compartment): Metal oxidizes to raise cation concentration: $\text{Ag}(s) \to \text{Ag}^+(aq, \text{dil}) + e^-$.
• Cathode (Concentrated Compartment): Cations reduce to deposit metal: $\text{Ag}^+(aq, \text{conc}) + e^- \to \text{Ag}(s)$.
• Potential Formula: $E_{\text{cell}} = -\frac{0.0592\text{ V}}{n} \log_{10} \frac{[\text{Cation}]_{\text{dilute}}}{[\text{Cation}]_{\text{concentrated}}}$

Applications

• pH Meters: Use a glass electrode concentration cell where the potential difference across a thin membrane reflects hydronium ion concentration ($0.0592\text{ V}$ change per pH unit).
• Nerve Impulse Transmission: Neurons maintain transmembrane sodium/potassium concentration gradients, generating resting potentials across cellular membranes.

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5. Worked Problem: Nernst Calculation

Problem: Determine $E_{\text{cell}}$ at $25\ ^\circ\text{C}$ for the cell: $\text{Zn}(s) \mid \text{Zn}^{2+}(0.050\text{ M}) \parallel \text{Cu}^{2+}(2.00\text{ M}) \mid \text{Cu}(s)$ Given: $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}$, $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}$.

1. Overall reaction and $n$: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \to \text{Zn}^{2+}(aq) + \text{Cu}(s) \quad (n = 2)$
2. Standard potential: $E^\circ_{\text{cell}} = (+0.34\text{ V}) - (-0.76\text{ V}) = +1.10\text{ V}$
3. Reaction quotient $Q$: $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.050\text{ M}}{2.00\text{ M}} = 0.025$
4. Nernst Equation calculation: $E_{\text{cell}} = 1.10\text{ V} - \frac{0.0592\text{ V}}{2} \log_{10}(0.025)$ $\log_{10}(0.025) = -1.602$ $E_{\text{cell}} = 1.10\text{ V} - (0.0296\text{ V})(-1.602) = 1.10\text{ V} + 0.047\text{ V} = +1.15\text{ V}$ The elevated reactant concentration increases the cell voltage above standard potential.