Electrochemistry · Lesson 3 of 13
The Nernst Equation and Nonstandard Cells
“Derive concentration-dependent cell potentials and use balanced reaction quotients for metal, gas and hydrogen electrodes.”
• Derive the Nernst equation from Gibbs energy and electrical work. • Construct dimensionless reaction quotients with correct exponents and phases. • Calculate cell potentials at nonstandard concentrations. • Relate hydrogen-electrode potential to pH and recognise when a written cell direction is unfavourable.
Why the same cell does not always give the same voltage
A Daniell cell consumes Cu²⁺ and produces Zn²⁺. As discharge proceeds, the relative amounts of products and reactants change. The chemical driving force therefore changes even though the electrode materials are unchanged. The Nernst equation quantifies this link between composition and reversible cell emf. It describes an equilibrium electrode response at specified conditions; a heavily loaded working cell can have additional voltage losses.
Deriving the equation
For a balanced reaction, the reaction Gibbs energy satisfies ΔrG = ΔrG° + RT ln Q. Here R is the gas constant, T the absolute temperature, and Q the dimensionless reaction quotient. In a reversible electrochemical cell, ΔrG = −nFE and ΔrG° = −nFE°. The number n is the moles of electrons transferred per mole of reaction as written and F ≈ 96485 C mol⁻¹ is the charge per mole of electrons.
Divide every term by −nF. The negative divisor changes the sign of the logarithmic term. This gives the general Nernst equation. A high product-to-reactant quotient reduces the forward driving force because the reaction has already moved toward products.
Build Q from the balanced reaction
For aA + bB → cC + dD, Q = a(C)ᶜa(D)ᵈ/[a(A)ᵃa(B)ᵇ], where a denotes activity. Include dissolved species and gases, but set activities of pure solids and pure liquids to 1. Use concentration divided by 1 M and gas pressure divided by 1 bar only under appropriate ideal approximations. Electrons are not included in the overall cell reaction quotient.
For the isolated reduction Mⁿ⁺ + ne⁻ → M(s), the quotient is 1/a(Mⁿ⁺), so E = E° + (RT/nF)ln a(Mⁿ⁺). Increasing the metal-ion activity raises its reduction potential. For a complete Daniell cell, raising Cu²⁺ activity raises the cell emf while raising Zn²⁺ activity lowers it. The sign should agree with the reaction direction before any arithmetic is attempted.
Problem
For Mg + 2Ag⁺ → Mg²⁺ + 2Ag, calculate E when [Mg²⁺] = 0.130 M, [Ag⁺] = 0.0001 M, E° = 3.17 V and T = 298.15 K.
- 1.The balanced reaction transfers two electrons, so n = 2. Q ≈ (0.130)/(0.0001)² = 1.30 × 10⁷; the silver concentration must be squared.
- 2.log₁₀Q = 7.11394. E = 3.17 − (0.05916/2)(7.11394) = 2.9596 V ≈ 2.96 V.
- 3.Q exceeds 1, so the emf is below its standard value. The cell remains strongly favourable in the written direction.
Problem
Calculate E for Ni + 2Ag⁺ → Ni²⁺ + 2Ag with [Ni²⁺] = 0.160 M, [Ag⁺] = 0.002 M and E° = 1.05 V at 298.15 K.
- 1.Q ≈ 0.160/(0.002)² = 40000 and n = 2. The numerator contains product ions; the denominator contains reactant ions.
- 2.E = 1.05 − (0.05916/2)log₁₀(40000) = 0.9139 V, about 0.91 V.
- 3.A smaller silver-ion activity reduces the driving force for its reduction. This physical expectation agrees with the calculation.
Hydrogen pressure and pH
At hydrogen pressure 1 bar, the logarithm reduces to −2 log₁₀a(H⁺). As pH = −log₁₀a(H⁺), the result is E = −0.05916 × pH in volts. Raising pH lowers the hydrogen-ion activity, making proton reduction less favourable relative to the standard hydrogen reference. At pH 10, E = −0.5916 V. This potential is not an absolute energy and does not identify a complete cell without another electrode.
A written cell may have a negative emf
Problem
Consider Pt|Br⁻(0.010 M)|Br₂(l)||H⁺(0.030 M)|H₂(1 bar)|Pt at 298.15 K. Use E°(Br₂(l)/Br⁻) = 1.065 V.
- 1.The written left-to-right reaction oxidises bromide and reduces H⁺: 2Br⁻ + 2H⁺ → Br₂(l) + H₂(g). It transfers two electrons.
- 2.E°cell = 0 − 1.065 = −1.065 V. Q ≈ 1/[(0.010)²(0.030)²] = 1.111 × 10⁷.
- 3.E = −1.065 − (0.05916/2)log₁₀Q ≈ −1.273 V. The negative sign means spontaneous operation would favour the reverse reaction under these conditions.
A tabulated potential belongs to the exact half-reaction and phase convention. Br₂(aq)/Br⁻ is commonly tabulated near 1.087 V, while pure liquid Br₂(l)/Br⁻ is near 1.065 V. A liquid-bromine activity is unity in its pure-liquid reference state; it must not be combined silently with a dissolved-bromine standard potential. This is a small numerical distinction with a real thermodynamic meaning.
Do not include the amount of a pure solid in Q, omit the exponent of Ag⁺, or substitute pH for [H⁺]. If the reaction is reversed, Q becomes its reciprocal and E changes sign. If a reaction is doubled, n doubles and Q is squared, leaving E unchanged.
Quick Quiz
Quiz
For Ni + 2Ag⁺ → Ni²⁺ + 2Ag, what is the ideal concentration form of Q?
Stoichiometric coefficients become exponents; pure solid nickel and silver have unit activity.
At 298.15 K, n = 2 and Q = 100. By how much is E below E°?
The decrease is (0.05916/2)log₁₀(100) = (0.05916/2) × 2.
At 1 bar H₂ and pH 10, the ideal hydrogen-electrode potential is approximately what?
E = −0.05916 pH at 298.15 K and unit hydrogen pressure.
What happens to Daniell-cell emf when Cu²⁺ activity increases while Zn²⁺ is fixed?
Q = a(Zn²⁺)/a(Cu²⁺) falls, so subtracting the logarithmic term yields a larger E.
Which coefficient belongs in the Nernst equation at an arbitrary temperature?
RT/(nF) is the general coefficient for natural logarithms. The familiar decimal coefficient is temperature-specific.
Practice Problems
- Calculate E for Mg|Mg²⁺(0.001 M)||Cu²⁺(0.0001 M)|Cu at 298.15 K using Mg²⁺/Mg = −2.37 V and Cu²⁺/Cu = 0.34 V.
- Calculate E for Fe|Fe²⁺(0.001 M)||H⁺(1 M)|H₂(1 bar)|Pt using Fe²⁺/Fe = −0.44 V at 298.15 K.
- Calculate E for Sn|Sn²⁺(0.050 M)||H⁺(0.020 M)|H₂(1 bar)|Pt using Sn²⁺/Sn = −0.14 V at 298.15 K.
- Derive the hydrogen-electrode potential at pH 4 and H₂ pressure 1 bar at 298.15 K.
- A cell reaction is doubled. Show algebraically why the Nernst emf is unchanged.
1. The reaction is Mg + Cu²⁺ → Mg²⁺ + Cu, so n = 2 and E° = 2.71 V. 2. Q = 0.001/0.0001 = 10. E = 2.71 − (0.05916/2) × 1 = 2.6804 V, about 2.68 V. 3. Using −2.36 V for magnesium instead changes the answer by 0.01 V; use the stated dataset consistently.
Key Takeaways
• Composition changes the reversible driving force of a cell reaction. • The Nernst equation follows from Gibbs energy and electrical work. • Q is dimensionless and must match the balanced reaction. • Pure solid and pure liquid activities are unity in their standard states. • Use the balanced electron count and distinguish ln from log₁₀. • At 298.15 K the decimal-log coefficient is 0.05916 V divided by n. • For H₂ at 1 bar, hydrogen-electrode potential is −0.05916 pH in volts. • A negative calculated emf identifies an unfavourable written direction.