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.

1.) [Fe3+]= 1.1×10⁻³ M ; [Mg2+]= 2.60 M

2.) [Fe3+]= 2.60 M ; [Mg2+]= 1.1×10⁻³ M

Answer 1

When the cell is NOT under standard conditions, i.e. 1M of each reactants at T = 25°C and P = 1 atm; then we must use Nernst Equation.

The equation relates E°cell, number of electrons transferred, charge of 1 mol of electron to Faraday and finally, the Quotient retio between products/reactants

The Nernst Equation:

Ecell = E0cell - (RT/nF) x lnQ

In which:

Ecell = non-standard value

E° or E0cell or E°cell or EMF = Standard EMF: standard cell potential
R is the gas constant (8.3145 J/mol-K)
T is the absolute temperature = 298 K
n is the number of moles of electrons transferred by the cell's reaction
F is Faraday's constant = 96485.337 C/mol or typically 96500 C/mol
Q is the reaction quotient, where

Q = [C]^c * [D]^d / [A]^a*[B]^b

pure solids and pure liquids are not included. Also note that if we use partial pressure (for gases)

Q = P-A^a / (P-B)^b

substitute in Nernst Equation:

Ecell = E° - (RT/nF) x lnQ

Now, from the equation given

First let us get Eº, R, T, n, F, and Q

the reaction quotient

Q = [Mg+2]^3/[Fe+3]^2

n = 6 electrons are being transferred (3x2= 6)

F 96500 C/mol, R = 8.31 4J/molK, T = 298K,

For Eºcell, we need:

Fe3+ + 3 e− ⇌ Fe(s) −0.04

Mg2+ + 2 e− ⇌ Mg(s) −2.372

Eºcell = -0.04 - -2.372 = 2.332 V

now, substitute in equation

a)

Ecell = E0cell - (RT/nF) x lnQ

Ecell = 2.332-8.314*298/(6*96500)*ln( [Mg+2]^3/[Fe+3]^2 )

substitute

Ecell = 2.332-8.314*298/(6*96500)*ln( (2.6^3)/(1.1*10^-3)^2)

Ecell = 2.2614

b)

do similar:

Ecell = E0cell - (RT/nF) x lnQ

Ecell = 2.332-8.314*298/(6*96500)*ln( [Mg+2]^3/[Fe+3]^2 )

substitute

Ecell = 2.332-8.314*298/(6*96500)*ln( (1.1*10^-3)^3 / (2.6^2))

Ecell = 2.427 V