Electrochemical cells and EMF

Dip a strip of zinc into zinc sulfate, a strip of copper into copper sulfate, join the solutions with a salt bridge and the metals with a wire, and something remarkable happens: electrons flow, a voltmeter registers a steady reading, and that reading never changes no matter how thick the wires or how large the electrodes are. That reading is the electromotive force, EMF, of the cell — and it is not a vague "voltage" but a precise thermodynamic quantity, the Gibbs energy of the underlying redox reaction expressed in volts. Once you can measure EMF reliably, you can read off Gibbs energies, entropies, equilibrium constants, activity coefficients and even pH from nothing more than a wire and a meter. This article rebuilds cell notation and the Nernst equation from first principles so the rest of the Electrochemical Systems unit can put that machinery to work.

The core insight

The EMF of a galvanic cell is a direct, measurable window into the Gibbs energy of its redox reaction, $\Delta G = -nFE$ , and the Nernst equation extends that link to any concentration away from standard conditions.

The core idea

A galvanic (voltaic) cell is two half-reactions kept physically apart so that the electron transfer between them is forced through an external wire instead of happening by direct collision. The half-cell where oxidation occurs is the anode; the half-cell where reduction occurs is the cathode. Electrons leave the anode, travel through the wire, and enter the cathode, while ions move through a salt bridge (or porous partition) to keep both compartments electrically neutral.

Cell notation. IUPAC convention writes the anode on the left and the cathode on the right, with a single vertical line $|$ for each phase boundary and a double line $\|$ for the salt bridge:

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

where: the leftmost and rightmost symbols are the actual electrode metals; species inside the vertical bars are read left-to-right in the order a phase boundary is crossed.

Standard EMF. Every half-reaction has a tabulated standard reduction potential $E^\circ$ , measured against the standard hydrogen electrode (SHE), which is fixed at $E^\circ = 0$ by convention. The cell EMF is always cathode minus anode, using reduction potentials for both:

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

where: $E^\circ_{cathode}$ and $E^\circ_{anode}$ are the standard reduction potentials of the two half-reactions as tabulated (never flip the sign of the anode value just because it is being oxidised).

Away from standard conditions, concentrations shift the EMF, and the tool that captures this is the Nernst equation, developed formally below.

Figure

Daniell cell showing zinc anode and copper cathode joined by a wire and a salt bridge with electron flow left to right through the wire and ion flow through the bridge

The Nernst equation

Derivation AThe Nernst equation for cell EMF

The goal is to connect the measurable cell EMF $E$ at any concentration to the standard EMF $E^\circ$ and the reaction quotient $Q$ , starting only from the definition of Gibbs energy for a reaction away from standard conditions and the relation between $\Delta G$ and electrical work — this is the single most-used formula in the entire Electrochemical Systems unit.

Step 1. Start from thermodynamics. For a reaction at any composition, the Gibbs energy of reaction relates to its standard value through the reaction quotient $Q$ : $\Delta G = \Delta G^\circ + RT\ln Q$ , where $R$ is the gas constant, $T$ the absolute temperature, and $Q$ is built from the activities (concentrations, for dilute solutions) of products over reactants, each raised to its stoichiometric power.

Step 2. Bring in the electrical link. The maximum electrical work available from a cell equals the Gibbs energy released, and that work is also charge moved times potential: $\Delta G = -nFE$ and, at standard conditions, $\Delta G^\circ = -nFE^\circ$ , where $n$ is the number of electrons transferred per formula unit of the balanced reaction and $F = 96\,485\ \text{C mol}^{-1}$ is the Faraday constant (the charge on one mole of electrons).

Step 3. Substitute both electrical relations into the Step 1 equation. $-nFE = -nFE^\circ + RT\ln Q$ . Divide every term by $-nF$ to isolate $E$ :

E = E^\circ - \frac{RT}{nF}\ln Q

Step 4. Convert to a numerical, base-10 form at room temperature, the version used in nearly every worked problem in this unit. At $T = 298\ \text{K}$ , the constant $RT/F \times \ln(10) = (8.314)(298)(2.303)/96\,485 = 0.0592\ \text{V}$ , so

E = E^\circ - \frac{0.0592}{n}\log_{10} Q \quad (298\ \text{K})

Verified: $RT/F$ at $298\ \text{K}$ is $(8.314)(298)/96\,485 = 0.02569\ \text{V}$ ; multiplying by $\ln(10) = 2.303$ gives $0.05916\ \text{V} \approx 0.0592\ \text{V}$ , matching the standard textbook constant.

The physical reading of the Nernst equation: increasing the concentration of a product-side species raises $Q$ and lowers $E$ ; increasing a reactant-side species lowers $Q$ and raises $E$ — exactly the Le Chatelier intuition translated into a formula you can compute with. When $Q = 1$ (all species at unit activity), the log term vanishes and $E = E^\circ$ , as it must.

Worked example

Problem. For the cell $\text{Zn}(s) \mid \text{Zn}^{2+}(0.024\ \text{M}) \parallel \text{Cu}^{2+}(0.20\ \text{M}) \mid \text{Cu}(s)$ , with $E^\circ_{cell} = 1.10\ \text{V}$ , find $E_{cell}$ at $298\ \text{K}$ and the Gibbs energy change of the reaction at these concentrations.

Step 1. Write the overall reaction and identify $n$ . Cathode reduces $\text{Cu}^{2+}$ , anode oxidises $\text{Zn}$ : $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$ , transferring $n = 2$ electrons.

Step 2. Build the reaction quotient. Pure solids carry activity 1 and drop out: $Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}] = 0.024/0.20 = 0.12$ .

Step 3. Apply the Nernst equation. $E_{cell} = E^\circ_{cell} - \dfrac{0.0592}{n}\log_{10}Q = 1.10 - \dfrac{0.0592}{2}\log_{10}(0.12)$ . Since $\log_{10}(0.12) = -0.921$ : $E_{cell} = 1.10 - (0.0296)(-0.921) = 1.10 + 0.0273 = 1.127\ \text{V}$ .

Step 4. Compute $\Delta G$ at these concentrations using $\Delta G = -nFE$ : $\Delta G = -(2)(96\,485)(1.127) = -2.175\times10^{5}\ \text{J mol}^{-1} = -217.5\ \text{kJ mol}^{-1}$ .

The cell EMF is $E_{cell} = 1.127\ \text{V}$ , and the reaction is more strongly spontaneous ( $\Delta G = -217.5\ \text{kJ mol}^{-1}$ ) than at standard conditions, because diluting the product ion $\text{Zn}^{2+}$ relative to the reactant ion $\text{Cu}^{2+}$ pulled $Q$ below 1 and raised $E$ above $E^\circ$ .

Competitive stretch

At what ratio $[\text{Zn}^{2+}]/[\text{Cu}^{2+}]$ would this cell read exactly $E_{cell} = 1.00\ \text{V}$ ? Rearranging the Nernst equation: $1.00 = 1.10 - 0.0296\log_{10}Q \Rightarrow \log_{10}Q = 0.10/0.0296 = 3.378 \Rightarrow Q = 10^{3.378} \approx 2390$ . The product ion would need to be nearly $2400$ times more concentrated than the reactant ion — a reminder that the logarithmic form makes $E$ fairly insensitive to concentration except at extreme ratios.

Common traps

Writing $Q$ with solid or pure-liquid phases included — only species of variable activity (dissolved ions, gases at non-unit pressure) belong in $Q$ .
Forgetting $E^\circ_{cell}$ is intensive (unchanged by scaling the equation), while $\Delta G^\circ = -nFE^\circ$ is extensive and doubles if you double the balanced equation.
Flipping the sign of the anode potential "because it is oxidation" — $E_{cell} = E_{cathode} - E_{anode}$ always uses tabulated reduction potentials for both electrodes, unmodified.
Mixing the natural-log form $RT/nF$ with numbers meant for the base-10 constant $0.0592/n$ (or vice versa) — confirm which form you are using before substituting $R$ , $T$ and $F$ .

Ask the tutor: if the Step 1-to-Step 3 jump in the derivation — combining $\Delta G = \Delta G^\circ + RT\ln Q$ with $\Delta G = -nFE$ — still feels like two formulas glued together rather than one continuous argument, ask the AI tutor to re-derive it line by line with you filling in the algebra.

Why does the Nernst equation use reduction potentials for both electrodes instead of flipping the anode's sign for oxidation?

Because $E_{cell} = E_{cathode} - E_{anode}$ already accounts for the fact that one electrode is being oxidised — the subtraction itself builds in the reversal, so plugging in tabulated reduction potentials directly (with no sign change) gives the correct answer. If you also flip the anode's sign "for oxidation," you double-count the reversal and get the wrong EMF, often with the wrong overall sign. Trust the formula as written and always look up reduction potentials, never oxidation potentials, for both terminals.

Why does diluting the Zn2+ ion raise the cell EMF above the standard value in the worked example?

The Nernst equation subtracts a term proportional to $\log_{10}Q$ from $E^\circ$ . Diluting $\text{Zn}^{2+}$ , a product of the reaction, shrinks $Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}]$ below 1, which makes $\log_{10}Q$ negative, and subtracting a negative number increases $E$ . Physically this matches Le Chatelier's principle: removing a product (by dilution) pulls the reaction further forward, releasing more Gibbs energy per mole reacted, which shows up as a larger EMF.

Is EMF the same thing as the voltage read on a multimeter connected to the cell?

Only if the meter draws essentially no current, because EMF is defined under conditions where no net current flows and the cell reaction has not been pushed away from equilibrium at each electrode. A real multimeter has extremely high internal resistance, so it draws a negligible current and its reading is an excellent approximation to the true EMF. If you instead connect a low-resistance load, the reading drops below EMF because of internal resistance and polarisation losses inside the cell itself — that terminal voltage is not the same quantity as $E_{cell}$ in the Nernst equation.