Electrochemistry · Lesson 4 of 13
Electrical Work, Gibbs Energy and Equilibrium
“Connect potential with maximum electrical work and equilibrium, and distinguish intensive voltage from the amount of reaction.”
• Derive the relationship between reaction Gibbs energy and cell emf. • Calculate equilibrium constants from standard cell potentials. • Distinguish equilibrium emf from a zero current caused by an open circuit or opposing source. • Track how reversing or scaling a reaction changes E, ΔrG and K.
Voltage is work per unit charge
A potential difference of one volt corresponds to one joule per coulomb. If n moles of electrons are transferred per mole of a written cell reaction, the charge magnitude is nF coulombs per mole of reaction. Multiplying that charge by the reversible emf gives the maximum electrical work the cell can deliver at constant temperature and pressure. A practical cell delivers less useful work because current flow produces irreversible losses.
If E is positive, ΔrG is negative and the reaction is thermodynamically favoured in the written direction at the specified composition. If E is negative, external work is required to drive that direction. At reaction equilibrium E = 0 and ΔrG = 0. Standard values refer to standard activities, so E° can stay positive even though the actual E becomes zero after the composition reaches equilibrium.
Problem
Find ΔrG° for Zn + Cu²⁺ → Zn²⁺ + Cu when E° = 1.10 V.
- 1.The oxidation and reduction exchange two electrons, hence n = 2. Use F = 96485 C mol⁻¹.
- 2.ΔrG° = −2 × 96485 × 1.10 = −212267 J mol⁻¹.
- 3.Divide by 1000 J per kJ: ΔrG° ≈ −212.27 kJ mol⁻¹, or about −2.1 × 10² kJ mol⁻¹ at two significant figures. The negative sign agrees with spontaneous discharge.
Why doubling a reaction does not double voltage
Voltage measures energy per charge, not the total energy held by a cell. Doubling Zn + Cu²⁺ → Zn²⁺ + Cu doubles n and ΔrG, but E remains 1.10 V for unchanged conditions. A larger amount of reactants can provide more charge and energy at a similar voltage. Reversing a reaction changes the signs of E and ΔrG. The tabulated reduction potentials must be combined with the reaction actually written.
Equilibrium from the Nernst equation
At equilibrium, Q becomes the equilibrium constant K and the reversible emf is zero. Substituting E = 0 into E = E° − (RT/nF) ln Q and moving the logarithmic term gives nFE° = RT ln K. Combining this with ΔrG° = −nFE° yields the familiar thermodynamic relationship ΔrG° = −RT ln K.
A large positive E° implies K is much greater than 1. It means the equilibrium strongly favours products relative to standard states; it does not mean the reaction is fast or that literally no reactant remains. If E° = 0, K = 1. A negative E° implies K < 1 for the written direction. Standard-state quantities and equilibrium composition describe different conditions.
Problem
For Cu + 2Ag⁺ → Cu²⁺ + 2Ag, E° = 0.46 V. Find K at 298.15 K.
- 1.The copper atom loses two electrons, so n = 2. log₁₀K = 2(0.46)/0.05915935 = 15.5512.
- 2.K = 10¹⁵·⁵⁵¹² ≈ 3.56 × 10¹⁵. An antilogarithm is needed; 15.55 itself is not K.
- 3.Because E° is given to only two decimal places and K is exponential, report the magnitude with appropriate caution. Using the rounded 0.059 coefficient gives about 3.92 × 10¹⁵; the discrepancy is numerical rounding, not different chemistry.
Problem
For 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂(s), E° = 0.236 V. Find ΔrG° and K at 298.15 K.
- 1.The iodide oxidation releases two electrons and the two Fe³⁺ ions consume them, so n = 2.
- 2.ΔrG° = −2(96485.332)(0.236) = −45541.1 J mol⁻¹ = −45.54 kJ mol⁻¹.
- 3.log₁₀K = 2(0.236)/0.05915935 = 7.97845, hence K ≈ 9.52 × 10⁷. The positive potential, negative Gibbs energy and large K are mutually consistent.
A stopped current does not always mean reaction equilibrium
An open switch stops electron transfer even when the cell still has a positive emf. An equal opposing external potential can also stop net current without making the isolated chemical reaction reach its own equilibrium composition. True chemical equilibrium means ΔrG = 0 for the reaction at the actual activities, and therefore zero reversible cell emf. These are distinct explanations for the same observation of no measured current.
Problem
For 2Cr + 3Cd²⁺ → 2Cr³⁺ + 3Cd, use E°(Cr³⁺/Cr) = −0.74 V and E°(Cd²⁺/Cd) = −0.40 V at 298.15 K.
- 1.E°cell = −0.40 − (−0.74) = 0.34 V. Balance two chromium oxidations and three cadmium reductions to obtain n = 6.
- 2.ΔrG° = −6(96485.332)(0.34) = −196.83 kJ mol⁻¹ of reaction.
- 3.log₁₀K = 6(0.34)/0.05915935 = 34.4832, giving K ≈ 3.04 × 10³⁴. Six electrons affect energy and K but do not multiply either electrode potential.
Quick Quiz
Quiz
For a written cell reaction with E > 0 and n > 0, what is the sign of ΔrG?
ΔrG = −nFE, so positive reversible emf corresponds to negative reaction Gibbs energy.
What happens when a reaction equation is doubled at the same composition?
Both charge transferred and energy change double, leaving voltage unchanged.
At chemical equilibrium, which relation is correct?
The actual driving force vanishes at equilibrium. Standard driving force need not vanish.
A large positive E° establishes which conclusion?
ln K = nFE°/(RT). Reaction rate requires kinetic information.
A charged cell has an open switch. Why is zero current insufficient evidence of equilibrium?
Zero current can result from an interrupted conducting path, independently of the chemical driving force.
Practice Problems
- For Fe²⁺ + Ag⁺ → Fe³⁺ + Ag, calculate E°, ΔrG° and K using 0.80 V for Ag⁺/Ag and 0.77 V for Fe³⁺/Fe²⁺ at 298.15 K.
- A button-cell reaction is Zn + Ag₂O + H₂O → Zn²⁺ + 2Ag + 2OH⁻. Given standard reduction potentials Ag₂O + H₂O + 2e⁻ → 2Ag + 2OH⁻, E° = 0.342 V, and Zn²⁺ + 2e⁻ → Zn, E° = −0.76 V, find E° and ΔrG°.
- For the Daniell reaction, estimate K at 298.15 K from E° = 1.10 V.
- If a reaction has equilibrium constant K, determine the values for its reverse and for its doubled equation.
- A cell has n = 2 and E = 0.50 V at its current composition. What maximum reversible electrical work is available per mole of reaction?
1. One electron is transferred. E° = 0.80 − 0.77 = 0.03 V. 2. ΔrG° = −1(96485.332)(0.03) = −2894.56 J mol⁻¹ = −2.895 kJ mol⁻¹. 3. log₁₀K = 0.03/0.05915935 = 0.50711, giving K = 3.215, about 3.2. The small positive potential yields only modest product favouring.
Key Takeaways
• A volt is one joule per coulomb. • The maximum reversible electrical work delivered is nFE per mole of reaction. • ΔrG = −nFE links chemical and electrical driving forces. • At equilibrium E = 0, Q = K and ΔrG = 0. • Standard potential determines K through ln K = nFE°/(RT). • Voltage is intensive while Gibbs energy scales with the written reaction. • Open circuit and electrical opposition can stop current without chemical equilibrium. • Potential predicts thermodynamic direction rather than reaction rate.