Electrochemical Cell and Equilibrium Challenge

Introduction

A standard cell potential describes a cell in which every dissolved species is at 1 mol/dm³ and every gas at 1 bar. Real batteries and laboratory cells almost never meet that condition, and as they discharge their composition keeps changing. This challenge links electrochemistry to equilibrium: the Nernst equation adjusts the potential for the actual composition, and the same algebra tells you the equilibrium constant and why a "flat" battery has reached equilibrium.

Core explanation

Energy and potential. The electrical work a cell can do per mole of reaction is related to its Gibbs energy change by ΔG = −nFE, where n is the number of moles of electrons transferred in the balanced equation and F is the Faraday constant. Under standard conditions, ΔG° = −nFE°. A positive E means a negative ΔG and a spontaneous cell reaction in the direction written.

The Nernst equation. Substituting these into ΔG = ΔG° + RT ln Q and dividing by −nF gives E = E° − (RT/nF) ln Q. At 298 K this is often written E = E° − (0.0592/n) log₁₀ Q. Q is written for the overall cell reaction: products over reactants, with solids omitted. When Q is less than 1, the log term is negative and E exceeds E°; when Q is greater than 1, E is lower than E°.

Link to K. At equilibrium, ΔG = 0, so E = 0 and Q = K. Putting these into the Nernst equation gives ln K = nFE°/RT, or log₁₀ K = nE°/0.0592 at 298 K. Because E° appears in an exponent, even a modest cell potential corresponds to an enormous K. For the Daniell cell, Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with E° = 1.10 V and n = 2, log₁₀ K ≈ 37, so K is about 10³⁷.

Discharge as a journey to equilibrium. As a Daniell cell operates, [Cu²⁺] falls and [Zn²⁺] rises, so Q increases and E decreases. The cell is "dead" when E reaches zero, which is exactly the point at which Q = K. With K so large, the potential stays close to E° for most of the discharge and then drops sharply when Cu²⁺ is nearly exhausted.

Concentration cells. If both half-cells use the same couple, E° = 0 and the potential comes only from the concentration difference. Electrons flow so as to equalise the concentrations.

Formulae

ΔG = −nFE; ΔG° = −nFE°. E = E° − (RT/nF) ln Q = E° − (0.0592/n) log₁₀ Q at 298 K. ln K = nFE°/RT. F = 96 485 C/mol; RT/F = 0.02569 V at 298 K.

Step-by-step reasoning

1. Write the half-equations and the overall cell reaction; find n. 2. Calculate E° = E°(cathode) − E°(anode). 3. Write Q for the overall reaction, omitting solids. 4. Substitute concentrations into the Nernst equation. 5. If needed, find ΔG from −nFE or K from E°.

Visual explanation

Sketch E against log₁₀ Q as a straight line of slope −0.0592/n that passes through E° at log Q = 0 and cuts the axis at log Q = log K. The discharge of a cell is a journey along this line from left to right until the potential reaches zero.

Real-world analogy

A cell is like water held behind a dam. E° is the height difference when both reservoirs are at their standard levels. As water flows, the upper level falls and the lower level rises, the head of water shrinks, and flow stops when the levels match, just as the potential falls to zero at equilibrium.

Real-world example

Glass-electrode pH meters and ion-selective electrodes rely on the Nernst equation: the measured potential changes by about 59 mV for every tenfold change in the activity of a singly charged ion at 25 °C. Instruments are calibrated with buffers so that measured potentials can be converted into concentrations.

Why?

Why does the Nernst equation use ln Q? Gibbs energy depends on concentration through the term RT ln Q, which comes from the entropy of mixing. The cell potential is simply Gibbs energy per unit charge, so it inherits the same logarithmic dependence on composition.

Common misconception

"Doubling the equation doubles E." Multiplying a cell equation by 2 doubles n and doubles ΔG, but E = −ΔG/nF is unchanged, and the Nernst equation gives the same potential because Q is squared while n is doubled.

Worked example

Question: For a Daniell cell at 298 K with [Zn²⁺] = 1.00 mol/dm³ and [Cu²⁺] = 0.0100 mol/dm³, find E and ΔG.

Reasoning: Q = [Zn²⁺]/[Cu²⁺] = 1.00/0.0100 = 100. E = 1.10 − (0.0592/2) × log₁₀ 100 = 1.10 − 0.0296 × 2 = 1.10 − 0.059 = 1.04 V. ΔG = −nFE = −2 × 96 485 × 1.041 ≈ −2.01 × 10⁵ J/mol.

Answer: E ≈ 1.04 V and ΔG ≈ −201 kJ per mole of reaction.

Quick check

1. A copper concentration cell has 0.0010 mol/dm³ Cu²⁺ in one half-cell and 0.10 mol/dm³ in the other. What is its potential at 298 K? Answer: E = (0.0592/2) × log₁₀(0.10/0.0010) = 0.0296 × 2 ≈ 0.059 V, with the dilute side as the anode.

Exam focus

State n clearly and write Q the right way round for the overall reaction. Examiners often ask you to explain the sign of the change from E° in words: a lower product or higher reactant concentration increases E. Remember E° is fixed at a given temperature; only E depends on composition.

Advanced insight

Real cells use activities, not concentrations, and in concentrated electrolytes activity coefficients can differ greatly from 1. Measured potentials also include junction potentials and overpotentials when current flows, so the Nernst value is the zero-current, reversible limit rather than the voltage under load.

Summary

The Nernst equation, E = E° − (RT/nF) ln Q, links the potential of a cell to its actual composition. Combining ΔG = −nFE with ΔG = ΔG° + RT ln Q shows that E falls to zero exactly when Q = K, and that ln K = nFE°/RT, so small potentials correspond to very large equilibrium constants.

Practice questions

1. Calculate log₁₀ K for the Daniell cell at 298 K. Answer: log₁₀ K = nE°/0.0592 = 2 × 1.10 ÷ 0.0592 ≈ 37.2, so K ≈ 10³⁷. 2. For the Daniell cell, what happens to E if [Zn²⁺] is raised to 10.0 mol/dm³ while [Cu²⁺] stays at 1.00 mol/dm³ (ignoring non-ideality)? Answer: Q = 10, so E = 1.10 − 0.0296 × 1 ≈ 1.07 V; the potential falls slightly. 3. A cell has E° = +0.46 V with n = 2. Find ΔG°. Answer: ΔG° = −2 × 96 485 × 0.46 ≈ −88.8 kJ/mol. 4. Why does a battery voltage stay nearly constant for much of its life and then fall quickly? Answer: E depends on log Q, which changes slowly while reactant and product concentrations are comparable, but changes rapidly as a reactant approaches exhaustion.