Daniell cell away from standard conditions

Worked example

A Daniell cell (Zn | Zn²⁺ ‖ Cu²⁺ | Cu) has E°(Cu²⁺/Cu) = 0.34 V and E°(Zn²⁺/Zn) = −0.76 V. With [Zn²⁺] = 0.17 mol dm⁻³ and [Cu²⁺] = 0.0032 mol dm⁻³ at 298 K (n = 2), find ΔrG for the cell reaction.

1. 1. Cell potential from half-cells: solve for E°cell

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

   = 1.100 V

2. 2. Divide [Zn²⁺] ÷ [Cu²⁺] → Q

   = 53.13

3. 3. Nernst equation: solve for E

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

   = 1.049 V

4. 4. Gibbs energy and cell potential: solve for ΔG°

   $\Delta G^\circ = -nFE^\circ$

   = −202.4 kJ mol⁻¹

Answer: ΔrG = −202 kJ mol⁻¹

Method

E° = 0.34 − (−0.76) = 1.10 V. Q = [Zn²⁺]/[Cu²⁺]; E = E° − (RT/2F) ln Q; then ΔrG = −2FE.

Common mistakes

• Q is upside down. For Zn + Cu²⁺ → Zn²⁺ + Cu, Q = [Zn²⁺]/[Cu²⁺].
• That is ΔrG° (standard conditions). Correct E for the actual concentrations with the Nernst equation first.

Formulas used

• Cell potential from half-cells

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

• Nernst equation

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

• Gibbs energy and cell potential

  $\Delta G^\circ = -nFE^\circ$

More electrochemistry questions

• ΔG° from E°
• Nernst equation
• Mass deposited by electrolysis
• E°cell from reduction potentials
• Equilibrium constant from E°cell
• Electrode potential away from standard conditions
• Latimer diagram: skipping an oxidation state
• Disproportionation constant from a Latimer diagram