Question

Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s) E° (Fe2+ / Fe) =-0.4400 V, Eº (Cd2+ / Cd) = -0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+] = 0.85 M and [Cd2+] = 0.010 M. E= Lv Will the reaction occur spontaneously at these conditions? o o yes no cannot predict o

Answer 1

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 (0.01^1/0.85^1)
E = -4*10^-2-(-5.704*10^-2)
E = 1.704*10^-2 V
Answer: 1.70*10^-2 V

2)
E is positive.
So, reaction is spontenoues.
Answer: yes