Question

For the cell:

Zn | Zn²⁺ (a = 1) || Cu²⁺ (a = 1) | Cu

Given that \[\ce{E^{\circ}_{Zn/Zn^{2+}}}\] = 0.761 volt; \[\ce{E^{\circ}_{Cu^{2+}/Cu}}\] = +0.339 volt

1. Write the cell reaction.
2. Calculate the emf and free energy change at 298 K involved in the cell.

[Faraday’s constant, F = 96500 coulomb mol⁻¹]

Answer 1

Given: Zn | Zn²⁺ (a = 1) || Cu²⁺ (a = 1) | Cu

\[\ce{E^{\circ}_{Zn/Zn^{2+}}}\] = 0.761 volt;

\[\ce{E^{\circ}_{Cu^{2+}/Cu}}\] = +0.339 volt

Temperature (T) = 298 K

Faraday’s constant, F = 96500 C mol⁻¹

a. Since zinc has a more negative standard potential, it gets oxidised:

\[\ce{Zn_{(s)} + Cu^{2+}_{ (aq)} −> Zn^{2+}_{ (aq)} + Cu_{(s)}}\]

This is the spontaneous cell reaction.

b. \[\ce{E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}}\]

= 0.339 V − (−0.761 V)

= 1.100 V

Use the formula

\[\ce{\Delta G^{\circ} = -nFE^{\circ}_{cell}}\]

Where:

n = 2 electrons transferred

F = 96500 C mol⁻¹

\[\ce{E^{\circ}_{cell}}\] = 1.100 V

ΔG° = −2 × 96500 × 1.100

= −212300 J mol⁻¹

= −212.3 kJ mol⁻¹