Question

The following electrochemical cell is set up at 298 K: 

\[\ce{Zn | Zn^{2+}_{ (aq)} (1 M) || Cu^{2+}_{ (aq)} (1 M) | Cu}\]

Given: \[\ce{E^{\circ}_{Zn^{2+}/Zn}}\] = −0.761 V, \[\ce{E^{\circ}_{Cu^{2+}/Cu}}\] = +0.339 V

Calculate the emf and free energy change at 298 K.

Answer 1

Given: \[\ce{Zn | Zn^{2+}_{ (aq)} (1 M) || Cu^{2+}_{ (aq)} (1 M) | Cu}\]

\[\ce{E^{\circ}_{Zn^{2+}/Zn}}\] = −0.761 V,

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

Temperature (T) = 298 K

Faraday’s constant (F) = 96500 C/mol

Formula:

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

= 0.339 − (−0.761)

= 0.339 + 0.761

= 1.10 V

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

Where:

n = 2 (electrons transferred in the redox reaction)

F = 96500 C/mol

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

ΔG° = −2 × 96500 × 1.10

ΔG° = −212.3 kJ