Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.893 M and [Ni2 ] = 0.0130 M. Standard reduction potentials can be found here.

Cr (s) + Ni^2+ (aq) --> <-- Cr^2+ (aq) + Ni (s)

Answer 1

Lets find Eo 1st
from data table:
Eo(Cr2+/Cr(s)) = -0.91 V
Eo(Ni2+/Ni(s)) = -0.25 V

As per given reaction/cell notation,
cathode is (Ni2+/Ni(s))
anode is (Cr2+/Cr(s))

Eocell = Eocathode - Eoanode
= (-0.25) - (-0.91)
= 0.66 V

Number of electron being transferred in balanced reaction is 2
So, n = 2

use:
E = Eo - (2.303*RT/nF) log {[Cr2+]^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 {[Cr2+]^1/[Ni2+]^1}
E = 0.66 - (0.0591/2) log (0.893^1/0.013^1)
E = 0.66-(5.431*10^-2)
E = 0.606 V
Answer: 0.606 V