Question

What is the cell potential for the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

at 75 ∘C when [Fe2+]= 3.10 M and [Mg2+]= 0.110 M .

Express your answer to three significant figures and include the appropriate units.

Answer 1

Lets find Eo 1st
from data table:
Eo(Mg2+/Mg(s)) = -2.372 V
Eo(Fe2+/Fe(s)) = -0.44 V

As per given reaction/cell notation,
cathode is (Fe2+/Fe(s))
anode is (Mg2+/Mg(s))

Eocell = Eocathode - Eoanode
= (-0.44) - (-2.372)
= 1.932 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 {[Mg2+]^1/[Fe2+]^1}

T = 75 oC = 348 K
Here:
2.303*R*T/F
= 2.303*8.314*348/96500
= 0.069

So, above expression becomes:
E = Eo - (0.069/n) log {[Mg2+]^1/[Fe2+]^1}
E = 1.932 - (0.069/2) log (0.11^1/3.1^1)
E = 1.932-(-5.006*10^-2)
E = 1.982 V
Answer: 1.98 V