Question

Calculate the cell potential of the following cell at 298K.
Zn(s)|Zn^2+(0.6 M)||Cu^2+(0.3 M)|Cu(s)
E0 (Zn^2+| Zn) = -0.76 volt, E0(Cu^2+ |Cu) =0.34 volt.

Answer 1

Cell notation = anode||cathode (oxidation||reduction)

Zn(s) ------> Zn²⁺ + 2e^- E^o = -0.76 V
Cu²⁺ + 2e^- --------> Cu E^o = 0.34 V

E^o = 0.76 + 0.34 = 1.1 V

E = E^o - 0.0591/2 * log ([Zn²⁺]/[Cu²⁺])

E = 1.1 - 0.0296 log ([0.6] / [0.3])

E = 1.1 - 0.00891 = 1.091 V