Question

What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 69 ∘C when [Fe2+]= 3.40 M and [Mg2+]= 0.310 M . Express your answer to three significant figures and include the appropriate units.

What is the cell potential for the reaction Mg(s) + Fe2+ (aq)+Mg2+ (aq) + Fe(s) at 69 C when Fe2+] = 3.40 Mand (Mg2+] =0.310 M. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) ? com Value o D Units E- Submit Provide Feedback

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

use:

E = Eo - (2.303*RT/nF) log {[Mg2+]^1/[Fe2+]^1}

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+]^1/[Fe2+]^1}

E = 1.932 - (0.0591/2) log (0.31^1/3.4^1)

E = 1.932-(-3.075*10^-2)

E = 1.963 V

Answer: 1.96 V