Question

Given the following cell notation for a Galvanic cell, Ni(s)/N2+(0.0010M/cu2+(0.010M/Cu(s) What is the cell potential? 0.59 V 0.56 V 0.62 V 0.69 V

Answer 1

Lets find Eo 1st

from data table:

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

Eo(Cu2+/Cu(s)) = 0.337 V

As per given reaction/cell notation,

cathode is (Cu2+/Cu(s))

anode is (Ni2+/Ni(s))

Eocell = Eocathode - Eoanode

= (0.337) - (-0.25)

= 0.587 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

use:

E = Eo - (2.303*RT/nF) log {[Ni2+]^1/[Cu2+]^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 {[Ni2+]^1/[Cu2+]^1}

E = 0.587 - (0.0591/2) log (0.001^1/0.01^1)

E = 0.587-(-2.956*10^-2)

E = 0.62 V

Answer: 0.62 V