Question

A voltaic cell employs the following redox reaction:
2Fe3+(aq)+3Mg(s)→2Fe(s)+3Mg2+(aq)
Calculate the cell potential at 25 ∘C under each of the following conditions.

A)   [Fe3+]= 3.05 M ; [Mg2+]= 2.5×10⁻³ M

Answer 1

Lets find Eo 1st

from data table:

Eo(Mg2+/Mg(s)) = -2.372 V

Eo(Fe3+/Fe(s)) = -0.04 V

As per given reaction/cell notation,

cathode is (Fe3+/Fe(s))

anode is (Mg2+/Mg(s))

Eocell = Eocathode - Eoanode

= (-0.04) - (-2.372)

= 2.332 V

Number of electron being transferred in balanced reaction is 6

So, n = 6

use:

E = Eo - (2.303*RT/nF) log {[Mg2+]^3/[Fe3+]^2}

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 {[Mg2+]^3/[Fe3+]^2}

E = 2.332 - (0.0591/6) log (0.0025^3/3.05^2)

E = 2.332-(-8.647*10^-2)

E = 2.418 V

Answer: 2.42 V