Advanced A level theoretical electrochemistry and redox equilibria

Part 7.7 Questions based on Appendix 1. The Nernst Equation and Appendix 2 Free Energy, Cell Emf and K_c

[Author ©  Dr Phil Brown PhD: Doc Brown's Chemistry exam revision notes suitable for A level chemistry students of advanced pre–university/college advanced level theoretical–physical chemistry courses:  redox equilibria contexts [updated April 29th 2026 *]

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7.7 Exemplar questions

Some links to miscellaneous questions on redox titrations, electrolysis products and the redox chemistry of the rusting of iron and its prevention.

The use of the Nernst equation is described i.e. how does the electrode potential of a half–cell reaction vary with the concentration of the electrolyte.

How to use the cell Emf to calculate the free energy change of a cell (redox) reaction and hence the equilibrium constant K_c for a redox equilibrium.

• How to construct full cell reaction equations is dealt with on the Redox Part 2 page section 6.2.

• See also Electrolysis calculations and Redox volumetric titration calculations

• Detailed notes on corrosion of iron and prevention is on the Transition Metal page on iron

Appendix 1. The Nernst Equation

• The Nernst equation allows the calculation–prediction of what a half–cell potential will be for a different ion concentration (or temperature) from that of the standard conditions i.e. 1.0 mol dm^–3 and 298K.

• The Nernst equation is not needed (as far as I know?) for UK GCE–A2 and IB courses, but it may be needed in coursework projects! You do NOT have to present the thermodynamic derivation of the equation!

  • For a reduction half–cell equilibrium i.e. oxidised state + electrons reduced state, the Nernst equation for the variation of the half–cell potential is

  • E = E^θ + (RT/nF) ln {[ox]/[red]}

    • E = half–cell potential in V which varies with the molar concentrations [mol dm^–3] of the oxidised and reduced species.

    • E^θ is the standard half–cell potential e.g. at 298K when the molarities are 1.0 mol dm^–3.

    • in the case of two aqueous ions like Fe²⁺/Fe³⁺, [red] = [ox] i.e. equal concentrations, so that E varies with the ratio of the two ion concentrations.

    • R = 8.314 J mol^–1 K^–1 (the ideal gas constant).

    • T the absolute temperature in K (Kelvin = ^oC + 273).

    • n = moles of electrons transferred per mole of reactants.

    • F =96500 C mol^–1 (the Faraday constant).

    • ln = natural/Napierian logarithm.

    • [ox/red] = molar concentration of the oxidised or reduced species. University students might well be using activity values as well as molarity values, in which case the activity of the metal is considered to be unity, i.e. a_M(s) = 1.000 (see next paragraph).

    • Note the equivalent equations (lg = log to base 10, log or log₁₀)

      • E = E^θ – (RT/nF) ln {[red]/[ox]}

      • E = E^θ – (2.303RT/nF) lg {[red]/[ox]}

      • E = E^θ + (2.303RT/nF) lg {[ox]/[red]}

• For a metal–metal half–cell equilibrium: Mⁿ⁺_(aq) + ne^– M_(s)

  • the Nernst equation is  E_Mⁿ⁺/M = E^θ_Mⁿ⁺/M + (RT/nF) ln {[Mⁿ⁺_(aq)]/[M_(s)]}

  • BUT the concentration of the metal cannot change so its molarity or 'thermodynamic activity' is considered to be unity, in which case the Nernst equation for the simple metal–metal ion equilibrium is

    • (i) E_Mⁿ⁺/M = E^θ_Mⁿ⁺/M + (RT/nF) ln [Mⁿ⁺_(aq)]

    • (using natural logarithm ln, watch on calculator!)

    • or (ii) E_Mⁿ⁺/M = E^θ_Mⁿ⁺/M + (2.303RT/nF) log₁₀ [Mⁿ⁺_(aq)]

    • (log to base 10, lg, log or log₁₀)

    • at 298K for ln expression: RT/nF = (8.314 x 298)/(n x 96500) = 0.0257/n

    • for lg or log₁₀: RT/nF = (2.303 x 8.314 x 298)/(n x 96500) = 0.0591/n

  • Note: (i) , (ii)

• Five examples of calculations using the Nernst equation are outlined below.

  1. What is the half–cell potential for copper when dipped into a 2.0 mol dm^–3 solution of copper(II) sulphate?

     • Cu²⁺_(aq) + 2e^– Cu_(s) (E^θ = +0.342V)

     • E_Cu2+/Cu = E^θ_Cu2+/Cu + (RT/nF) ln [Cu²⁺_(aq)]

     • E = +0.342 + {(8.314 x 298)/(2 x 96500)} ln [2.0]

     • E = +0.342 + 0.01284 ln(2)

     • E_Cu2+/Cu = +0.342 + 0.009 = 0.351 V (3sf, 0.35V 2sf)

     • Le Chatelier's comment: The increase in Cu²⁺ concentration increases the oxidising potential of the half–cell i.e. a more positive half–cell potential acting from left to right in the equilibrium.

  2. What is the half–cell potential of zinc dipped into a 0.1 molar zinc sulphate solution?

     • Zn²⁺_(aq) + 2e^– Zn_(s) (E^θ = –0.763V)

     • E_Zn2+/Zn = E^θ_Zn2+/Zn + (2.303RT/nF) lg [Zn²⁺_(aq)]

     • E = –0.763 + {(2.303 x 8.314 x 298)/(2 x 96500)} lg [0.1]

     • E_Zn2+/Zn = –0.763 + (–0.029) = –0.792 V (3sf, –0.79 V 2sf)

     • Le Chatelier's comment: The decrease in Zn²⁺ concentration increases the reducing potential of the half–cell i.e. a more negative potential half–cell potential acting from right to left in the equilibrium.

  3. What is the theoretical half–cell potential of aluminium dipped into a 0.001 molar aluminium sulphate solution?

     • Al³⁺_(aq) + 3e^– Al_(s) (E^θ = –1.662V)

     • E_Al3+/Al = E^θ_Al3+/Al + (RT/nF) ln [Al³⁺_(aq)]

     • E = –1.662 + {(8.314 x 298)/(3 x 96500)} ln [0.001]

     • E_Al3+/Al = –1.662 + (–0.059) = –1.721 V (4sf, –1.72V 3sf)

     • Le Chatelier's comment: The decrease in Al³⁺ concentration increases the reducing potential of the half–cell i.e. a more negative potential half–cell potential acting from right to left in the equilibrium.

  4. What is the half–cell potential for a solution of iron(II) and iron(III) ions, containing 0.20 mol dm^–3 of  Fe²⁺ and 0.10 mol dm^–3 of Fe³⁺?

     • Fe³⁺_(aq) + e^– Fe²⁺_(aq) (E^θ = +0.771V)

     • E_Fe3+/Fe2+ = E^θ_Fe3+/Fe2+ + (RT/nF) ln [ox]/[red]

     • E_Fe3+/Fe2+ = E^θ_Fe3+/Fe2+ + (2.303RT/nF) lg [Fe³⁺_(aq)]/[Fe²⁺_(aq)]

     • E = +0.771 + (0.0591/1) log₁₀ (0.1/0.2)

     • E_Fe3+/Fe2+ = +0.771 + (–0.018) = +0.753 V (3sf, 0.75V 2sf)

     • Le Chatelier's Principle comment: The lower Fe³⁺ concentration compared to Fe²⁺, decreases the oxidising potential of the half–cell i.e. a less positive potential half–cell potential acting from left to right in the equilibrium.

  5. Very accurate measurement half–cell electrode potentials can be used to determine the concentration of an ion (e.g. pH with glass electrode).

      When a copper strip (+ve pole) is dipped into a copper(II) ion solution, and combined with a saturated KCl–calomel electrode (–ve pole, E = 0.244V at 298K), the cell voltage measured 0.088 V at 298K.

     From the information deduce the concentration of copper(II) ions in the solution.

     • (i) You first need to calculate the copper  half–cell potential.

     • E_cell = E_+pole – E_–pole

     • E_cell = E_Cu/Cu2+ – E_calomel

     • E_cell = 0.088 = E_Cu/Cu2+ – (+0.244) = E_Cu/Cu2+ – 0.244

     • therefore E_Cu/Cu2+ = E_cell + 0.244 = 0.088 + 0.244 = +0.332 V

     • (ii) You then need to rearrange the Nernst equation to calculate the concentration.

     • E^θ_Cu2+/Cu = +0.342V, and from example 1.

     • E_Cu2+/Cu = E^θ_Cu2+/Cu + (RT/2F) ln [Cu²⁺_(aq)]

     • rearranging (with care!) gives

     • (RT/2F) ln [Cu²⁺_(aq)] = E_Cu^2+/Cu – E^θ_Cu2+^/Cu

     • ln [Cu²⁺_(aq)] = (E_Cu2+/Cu – E^θ_Cu2+/Cu)/(RT/2F)

     • [Cu²⁺_(aq)] = e^{{(E_Cu2+/Cu – Eθ_Cu2+/Cu)/(RT/2F)}}

     • [Cu²⁺_(aq)] = e^{{(0.332 – 0.342)/0.01284}}

     • [Cu²⁺_(aq)] = e^{{(–0.01)/0.01284}} = e^–0.7788 = 0.459 mol dm^–3

     • NOTE that even a small and expected error in the cell Emf can lead to a large concentration calculation error e.g. if the E_Cu2+/Cu of 0.332 is assumed to be absolutely correct, but was actually measured as 0.328 (≈ 1.2% measurement error) the calculation would give e^{{(–0.014)/0.01284}} = e^–1.0903 = 0.336 mol dm^–3 which is only 73% of the 'real' concentration, i.e. a 27% calculation error! Check it out for yourself and note I've used 3sf Emf's in the calculation, so you need a very accurate voltmeter or potentiometer system to have any hope of accurate analytical data.

Appendix 2. Free Energy, cell Emf and equilibrium constant K_equilibrium

• The concentration equilibrium constant K_c

  • for a reaction aA + bB ... dD + eE ...

  • K_c = [A]^a [B]^b .../[D]^d [E]^e ...

• The standard Gibbs free energy change ∆G^θ = ∆H^θ – T∆S^θ_sys

• and ∆G^θ = –RT ln(K_c) = –RT ln([A]^a [B]^b .../[D]^d [E]^e ...)

• The free energy change for a reversible electrode reaction is ∆G^θ = –nE^θF

• n is the number of electrons transferred in the theoretical cell reaction and F is the Faraday constant.

• therefore for a redox equilibria ∆G^θ = –nE^θF = –RT ln(K_c)

• so nE^θF = RT ln(K_c) from which the equilibrium constant K_c can be calculated.

• REMEMBER ...

• ∆G^θ must be negative, and E^θ positive, for the theoretical overall redox reaction, for it to be feasible.

• For more details see Thermodynamics Part 3 section 3.5b

WHAT NEXT?

INDEX 7. Redox equilibria, half–cells, electrode potentials, electrolysis, electrochemical series

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. K_c and K_p equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4 Partition between two phases, solubility product K_sp, common ion effect, ion–exchange systems * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria–vapour pressure, boiling point and intermolecular forces watch out for sub-indexes to multiple sections or pages

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