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.

Part A: [Fe3+]= 1.3×10⁻³ M ; [Mg2+]= 2.95 M

Part B: [Fe3+]= 2.95 M ; [Mg2+]= 1.3×10⁻³ 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

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

A)

So, above expression becomes:

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

E = 2.332 - (0.0591/6) log (2.95^3/0.0013^2)

E = 2.332-(7.077*10^-2)

E = 2.261 V

Answer: 2.26 V

B)

So, above expression becomes:

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

E = 2.332 - (0.0591/6) log (0.0013^3/2.95^2)

E = 2.332-(-9.458*10^-2)

E = 2.43 V

Answer: 2.43 V