Question

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

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.

1. [Fe3+]= 2.3×10−3 M ; [Mg2+]= 2.90 M

2. [Fe3+]= 2.90 M ; [Mg2+]= 2.3×10−3 M

1)

Lets find Eo 1st

from data table:

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

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

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Fe3+/Fe(s))

anode is (Mg2+/Mg(s))

The chemical reaction taking place is

Fe3+(aq) + Mg(s) --> Fe(s) + Mg2+(aq)

Eocell = Eocathode - Eoanode

= (-0.04) - (-2.372)

= 2.332 V

Number of electron being transferred in balanced reaction is 6

So, n = 6

we have below equation to be used:

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 (2.9^3/0.0023^2)

E = 2.332-(6.567*10^-2)

E = 2.266 V

2)

E = Eo - (0.0591/n) log {[Mg2+]^3/[Fe3+]^2}

E = 2.332 - (0.0591/6) log (0.0023^3/2.9^2)

E = 2.332-(-8.711*10^-2)

E = 2.419 V