Question

A galvanic cell consists of a zinc electrode in 0.0514 M Zn(NO₃)₂ and a nickel electrode in 0.379 M Ni(NO₃)₂. What is the cell potential (emf, in V) of this cell at 25^oC?

Answer 1

Lets find Eo 1st

from data table:

Eo(Zn2+/Zn(s)) = -0.7618 V

Eo(Ni2+/Ni(s)) = -0.25 V

As per given reaction/cell notation,

cathode is (Ni2+/Ni(s))

anode is (Zn2+/Zn(s))

Eocell = Eocathode - Eoanode

= (-0.25) - (-0.7618)

= 0.5118 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

use:

E = Eo - (2.303*RT/nF) log {[Zn2+]^1/[Ni2+]^1}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Zn2+]^1/[Ni2+]^1}

E = 0.5118 - (0.0591/2) log (0.0514^1/0.379^1)

E = 0.5118-(-2.565*10^-2)

E = 0.5375 V

Answer: 0.537 V