Question

Consider the following cell at 280 K:

Fe | Fe²⁺ (0.573) || 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 0.341 M?

Answer 1

Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.

Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V

All other samples are based on this reference.

Find the Reduction Potential of each reaction (Tables)

Cd2+ + 2 e− ⇌ Cd(s) −0.40 reduces

Fe2+ + 2 e− ⇌ Fe(s) −0.44 oxidizes

The most positive has more potential to reduce, it will be reduced

The most negative will be oxidized, since it will donate it selectrons

For total E°cell potential:

E°cell = Ered – Eox

Eox = -Ered of the one being oxidized

E°cell = -0.40 - -0.44 = 0.04 V

E°cell = 0.04 V

Now.... this is NOT in standard state:

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

Q = [Fe+2]/[Cd2+]

n = 2 electrons, R = 8.314, T = 298K, F = 96500

Ecell = E° - (RT/nF) x lnQ

Ecell = 0.04 - (8.314*298/2/96500) x ln( [Fe+2]/[Cd2+] )

initially

[Fe+2] = 0.573

[Cd2+] = 1.063

as reaction passes by

[Fe+2] = 0.573 + x

[Cd2+] = 1.063 - x

note that x, is the change of Fe2+, since it is "changeing by"

x = 0.341

so

[Fe+2] = 0.573 + 0.341 = 0.914

[Cd2+] = 1.063 - 0.341 = 0.722

now,

Ecell = 0.04 - (8.314*298/2/96500) x ln( [Fe+2]/[Cd2+] )

Ecell = 0.04 - (8.314*298/2/96500)*ln( 0.914/0.722)

Ecell = 0.036972 V

it is decreasing, since it is reacting, eventually, as time passes by to infinity, the Ecell should have a 0 value