5.5 Nernst Equation and Electrochemistry

Deriving the Nernst Equation

Relationship between Gibbs free energy and cell potential

The Nernst equation connects the reduction potential of an electrochemical reaction to the standard electrode potential and the activities of the species involved. Its derivation starts from the Gibbs free energy change for the electrochemical process.

Recall that $\Delta G$ relates to cell potential through:

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

where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant (96,485 C mol$^{-1}$). The sign convention here matters: a positive $E_{\text{cell}}$ gives a negative $\Delta G$, which corresponds to a spontaneous process.

Under arbitrary conditions, the Gibbs energy also depends on the reaction quotient $Q$:

$\Delta G = \Delta G° + RT\ln Q$

Substituting $\Delta G = -nFE_{\text{cell}}$ and $\Delta G° = -nFE°_{\text{cell}}$ into this expression and dividing through by $-nF$ yields the Nernst equation.

Nernst equation expression and components

$E_{\text{cell}} = E°_{\text{cell}} - \frac{RT}{nF}\ln Q$

Each term:

• $E°_{\text{cell}}$: standard cell potential (all species at unit activity)
• $R$: gas constant, 8.314 J mol$^{-1}$ K$^{-1}$
• $T$: absolute temperature in Kelvin
• $n$: moles of electrons transferred in the balanced cell reaction
• $F$: Faraday's constant, 96,485 C mol$^{-1}$
• $Q$: reaction quotient, constructed from activities (approximated by concentrations for solutes and partial pressures for gases). For a reaction $aA + bB \rightleftharpoons cC + dD$:

$Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

Pure solids and pure liquids have activity = 1 and don't appear in $Q$.

At 298.15 K, the prefactor $\frac{RT}{F}$ evaluates to 0.02569 V, so the equation is often written as:

$E_{\text{cell}} = E°_{\text{cell}} - \frac{0.02569 \text{ V}}{n}\ln Q$

or equivalently, converting to base-10 logarithm:

$E_{\text{cell}} = E°_{\text{cell}} - \frac{0.05916 \text{ V}}{n}\log Q$

Relationship between standard cell potential and equilibrium constant

At equilibrium, $\Delta G = 0$ and therefore $E_{\text{cell}} = 0$. The reaction quotient equals the equilibrium constant ($Q = K$), so the Nernst equation becomes:

$0 = E°_{\text{cell}} - \frac{RT}{nF}\ln K$

Rearranging:

$E°_{\text{cell}} = \frac{RT}{nF}\ln K$

This is a direct bridge between electrochemistry and chemical equilibrium. You can also reach it by combining $\Delta G° = -nFE°_{\text{cell}}$ with $\Delta G° = -RT\ln K$, which gives the same result.

A few things to note:

• A large positive $E°_{\text{cell}}$ corresponds to a large $K$, meaning products are heavily favored at equilibrium.
• A negative $E°_{\text{cell}}$ gives $K < 1$, meaning reactants are favored.
• Even a modest $E°_{\text{cell}}$ translates to a very large or very small $K$ because of the exponential relationship. For example, at 298 K with $n = 2$, an $E°_{\text{cell}}$ of just 0.30 V gives $K \approx 10^{10}$.

Cell Potential Under Non-Standard Conditions

Calculating cell potential using the Nernst equation

Under non-standard conditions (concentrations ≠ 1 M, pressures ≠ 1 atm, or temperature ≠ 298 K), the cell potential shifts from $E°_{\text{cell}}$. The Nernst equation accounts for this through the $Q$ term.

To calculate $E_{\text{cell}}$:

1. Write the balanced overall cell reaction.
2. Identify $n$, the number of electrons transferred.
3. Look up or calculate $E°_{\text{cell}}$.
4. Construct $Q$ from the actual concentrations and partial pressures (omitting pure solids and liquids).
5. Plug everything into $E_{\text{cell}} = E°_{\text{cell}} - \frac{RT}{nF}\ln Q$.

Comparing non-standard cell potential to standard cell potential

The sign of $\ln Q$ determines whether the actual cell potential is above or below the standard value:

• If $Q < 1$ (reactants dominate): $\ln Q < 0$, so $E_{\text{cell}} > E°_{\text{cell}}$. The cell has a stronger driving force than under standard conditions.
• If $Q > 1$ (products dominate): $\ln Q > 0$, so $E_{\text{cell}} < E°_{\text{cell}}$. The driving force is weaker.
• If $Q = 1$: $E_{\text{cell}} = E°_{\text{cell}}$ exactly.

As the cell operates and products accumulate, $Q$ increases and $E_{\text{cell}}$ drops. Eventually $Q = K$, $E_{\text{cell}} = 0$, and the cell is "dead" (at equilibrium).

Concentration cell example

A concentration cell uses the same redox couple in both half-cells but at different concentrations. Because the electrode materials are identical, $E°_{\text{cell}} = 0$, and the entire driving force comes from the concentration difference.

For a Cu$^{2+}$/Cu concentration cell with $[\text{Cu}^{2+}]_{\text{cathode}} = 0.10$ M and $[\text{Cu}^{2+}]_{\text{anode}} = 0.010$ M:

$E_{\text{cell}} = 0 - \frac{RT}{nF}\ln\frac{[\text{Cu}^{2+}]_{\text{anode}}}{[\text{Cu}^{2+}]_{\text{cathode}}} = -\frac{RT}{nF}\ln\frac{0.010}{0.10}$

With $n = 2$ and $T = 298$ K:

$E_{\text{cell}} = -\frac{0.02569}{2}\ln(0.10) = -0.01285 \times (-2.303) = 0.0296 \text{ V}$

The cell spontaneously drives Cu$^{2+}$ reduction in the more concentrated compartment and Cu oxidation in the dilute one, working to equalize the concentrations. This is a purely entropic driving force.

Predicting Spontaneous Redox Reactions

Standard reduction potential and spontaneity

Every half-reaction has a standard reduction potential ($E°$), measured relative to the standard hydrogen electrode (SHE, defined as 0.00 V). These values tell you how strongly a species "wants" to gain electrons.

In any redox pair:

• The species with the more positive $E°$ is preferentially reduced (acts as the cathode).
• The species with the more negative $E°$ is preferentially oxidized (acts as the anode).

Calculating standard cell potential

$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$

• If $E°_{\text{cell}} > 0$: the reaction is spontaneous as written under standard conditions ($\Delta G° < 0$).
• If $E°_{\text{cell}} < 0$: the reverse reaction is spontaneous under standard conditions.

An important point: standard reduction potentials are intensive properties. You do not multiply $E°$ by stoichiometric coefficients when combining half-reactions. (You do multiply $n$ when calculating $\Delta G°$.)

Examples of spontaneous redox reactions

Zinc–Copper cell (Daniell cell):

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

• $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$, $E° = +0.34$ V (cathode)
• $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$, $E° = -0.76$ V (anode)
• $E°_{\text{cell}} = 0.34 - (-0.76) = 1.10$ V

A strongly positive value, so this reaction is clearly spontaneous.

Silver–Iron(III) cell:

$\text{Ag}(s) + \text{Fe}^{3+}(aq) \rightarrow \text{Ag}^+(aq) + \text{Fe}^{2+}(aq)$

• $\text{Ag}^+ + e^- \rightarrow \text{Ag}$, $E° = +0.80$ V (anode, since it's being oxidized)
• $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$, $E° = +0.77$ V (cathode)

Wait: here Fe$^{3+}$/Fe$^{2+}$ has the lower reduction potential (+0.77 V vs. +0.80 V for Ag$^+$/Ag). For the reaction as written, Ag is oxidized and Fe$^{3+}$ is reduced, giving:

$E°_{\text{cell}} = 0.77 - 0.80 = -0.03 \text{ V}$

This is slightly negative, meaning the reaction as written is actually non-spontaneous under standard conditions. The reverse reaction (Fe$^{2+}$ reducing Ag$^+$) is the spontaneous direction, though the driving force is small.

Concentration of Electroactive Species

Using the Nernst equation to calculate concentration

If you measure $E_{\text{cell}}$ experimentally and know $E°_{\text{cell}}$ plus the concentrations of all species except one, you can rearrange the Nernst equation to solve for the unknown concentration.

Steps:

1. Write the Nernst equation for the cell reaction.
2. Express $Q$ in terms of the known concentrations and the unknown $[X]$.
3. Substitute the measured $E_{\text{cell}}$ and solve for $\ln Q$.
4. Isolate $[X]$ algebraically.

For example, if the unknown appears only in the numerator of $Q$:

$\ln Q = \frac{nF}{RT}(E°_{\text{cell}} - E_{\text{cell}})$

Then extract $[X]$ from the expression for $Q$.

Applications in analytical chemistry

This approach is the foundation of potentiometry, where you determine an analyte's concentration by measuring a cell voltage rather than performing a chemical analysis.

In a potentiometric titration, you monitor $E_{\text{cell}}$ as titrant is added. The potential changes gradually at first, then shifts sharply near the equivalence point. The inflection point in the $E$ vs. volume curve marks the endpoint. This technique is especially useful for colored or turbid solutions where visual indicators fail.

Ion-selective electrodes

Ion-selective electrodes (ISEs) are sensors built on the Nernst equation. Each ISE contains a membrane that responds selectively to one ion, generating a potential proportional to the logarithm of that ion's activity.

The response follows:

$E = E° + \frac{RT}{nF}\ln a_{\text{ion}}$

where $a_{\text{ion}}$ is the activity of the target ion (often approximated by concentration in dilute solutions) and $n$ is the charge on the ion.

The most familiar example is the glass pH electrode, which is selective for H$^+$. At 298 K, its potential changes by about 59.16 mV per unit change in pH (since $n = 1$ for H$^+$). Other common ISEs target F$^-$, K$^+$, Ca$^{2+}$, and NO$_3^-$.

For accurate measurements, ISEs are calibrated against standard solutions of known concentration, and the measured potential is compared to a stable reference electrode (typically Ag/AgCl or saturated calomel).