Question

Calculate the emf for the following reaction. Will the reaction occur spontaneously at 25°C, given that [Fe2+] = 0.600 M and [Cd2+] = 0.00450 M? Cd(s) + Fe2+(aq)→Cd2+(aq) + Fe(s) E o Cd2+/Cd = −0.40 V E o Fe2+/Fe = −0.44 V E = V The reaction as written is spontaneous not spontaneous

Answer 1

Lets find Eo 1st
from data table:
Eo(Cd2+/Cd(s)) = -0.4 V
Eo(Fe2+/Fe(s)) = -0.44 V

As per given reaction/cell notation,
cathode is (Fe2+/Fe(s))
anode is (Cd2+/Cd(s))

Eocell = Eocathode - Eoanode
= (-0.44) - (-0.4)
= -0.04 V

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

use:
E = Eo - (2.303*RT/nF) log {[Cd2+]^1/[Fe2+]^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 {[Cd2+]^1/[Fe2+]^1}
E = -4*10^-2 - (0.0591/2) log (0.0045^1/0.6^1)
E = -4*10^-2-(-6.282*10^-2)
E = 2.282*10^-2 V
Answer: 2.28*10^-2 V

Since E is positive, it is spontaneous