Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2 ] = 0.771 M and [Ni2 ] = 0.0200 M.

Answer 1

Eo ( Zn2+/Zn) = -0.76 V
Eo ( Ni2+/Ni) = -0.25 V

since Zn is getting oxidised and Ni is getting reduced hence Zn2+/Zn will be anode and Ni2+/Ni will be cathode

Eo ( cell) = Eo ( cathode) - Eo(anode) = Eo(Ni2+/Ni) - Eo(Zn2+/Zn) = -0.25 - -0.76 = -0.44 + 0.76 = 0.51 V

now if the conc. of Ni2+ was 1 M and conc. of Zn2+ was 1 M ...then E (cell) would have been equal to 0.51V .....but since it is not ...hence we have to use the Nernst equation ...

E (cell) = Eo (cell) - RT/nF ln [Zn2+]/[Ni2+]

where R = 8.314 J/K/mole
T = 25 + 273 = 298 K
n = no.of electrons transffered = 2
F = 96500 Coulombs
[Zn2+] = 0.771 M
[Fe2+] = 0.02 M

E (cell) = 0.51 - ( 8.314 X 298) / ( 2 X 96500) X ln 0.771/0.02

E (cell) = 0.51 - 2477.57/193,000 X ln 38.55

E (cell) = 0.51 - 0.0128 X 3.651

E (cell) = 0.51 - 0.046

E (cell) = 0.464 V