Question

Consider the cell described below at 285 K:

Fe | Fe²⁺ (0.919 M) || Cd²⁺ (0.965 M) | Cd

Given E^o_Cd²⁺→Cd = -0.403 V, E^o_Fe²⁺→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe²⁺ to have changed by 0.391 mol/L.

Answer 1

The reaction taking place is:

Fe + Cd2+ —> Fe2+ + Cd

since Fe2+ is in product, its concentration must be increasing

finally,

[Fe2+] = 0.919 + 0.391 = 1.31 M

[Cd2+] will decrease by same amount

[Cd2+] = 0.965 - 0.391 = 0.574 M

Lets find Eo 1st

Eo(Fe2+/Fe(s)) = -0.441 V

Eo(Cd2+/Cd(s)) = -0.403 V

As per given reaction/cell notation,

cathode is (Cd2+/Cd(s))

anode is (Fe2+/Fe(s))

Eocell = Eocathode - Eoanode

= (-0.403) - (-0.441)

= 0.038 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

we have below equation to be used

E = Eo - (2.303*RT/nF) log {[Fe2+]/[Cd2+]}

Here:

2.303*R*T/n

= 2.303*8.314*298.0/F

= 0.0591

So, above expression becomes:

E = Eo - (2.303*RT/nF) log {[Fe2+]/[Cd2+]}

E = 0.038 - (0.0591/2) log (1.31/0.574)

E = 0.038-(0.01059)

E = 2.741*10^-2 V

Answer: 2.741*10^-2 V