Question

Consider the following cell at 277 K:

Fe | Fe²⁺ (0.567) || Cd²⁺ (1.063) | Cd

which has a standard cell potential of 0.0400 V. What will be the potential of the cell be when [Fe²⁺] changes by 0.305 M?

The answer is 0.0383 V

Answer 1

Nernst Equation

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

then,

Fe2+ + 2 e− ⇌ Fe(s) −0.44

Cd2+ + 2 e− ⇌ Cd(s) −0.40

E°= ERed - Eox = -0.40 - - 0.44 = 0.04 V (as shown)

so

Ecell = E° - (RT/nF) x lnQ

Q = [Fe2+]/[Cd2+]

n = 2 electrons, R = 8.314, T = 277, F = 96500

substitute

Ecell = 0.04 - (8.314*277/(2*96500) * ln ( [Fe2+]/[Cd2+] )

Ecell = 0.04 - 0.011932* ln ( [Fe2+]/[Cd2+] )

initially

[Fe2+] = 0.567+ x

[Cd2+] = 1.063 - x

since Fe2+ is changing by 0.305 , then x = 0.305

[Fe2+] = 0.567+ x = 0.567+0.305 = 0.872

[Cd2+] = 1.063 - x = 1.063-0.305 = 0.758

now;

substitute

Ecell = 0.04 - 0.011932* ln ( [Fe2+]/[Cd2+] )

Ecell = 0.04 - 0.011932* ln ( 0.872/0.758)

Ecell = 0.038328 V, as shown in answer