Question

A concentration cell is built based on the following half reactions by using two pieces of gold as electrodes, two Au3+ solutions, 0.101 M and 0.423 M, and all other materials needed for a galvanic cell. What will the potential of this cell be when the cathode concentration of Au3+ has changed by 0.035 M at 307 K?

Au3+ + 3 e- → Au Eo = 1.50 V

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

Eº = Ered - Eox = 0, since it is the SAME value

Q = [Au+3]ox / [Au+3]red

Ecell = E° - (RT/nF) x lnQ

Ecell = 0 - (8.314*307)/(3*96500)* ln ([Au+3]ox / [Au+3]red)

Ecell = 0 - (8.314*307)/(3*96500)* ln (0.101/0.423 )

initially

[Au+3]ox = 0.101

[Au+3]red =0.423

in equilibrium

[Au+3]ox = 0.101 - x

[Au+3]red =0.423 + x

and we know that x = 0.035 so

[Au+3]ox = 0.101 - 0.035 = 0.0126275

[Au+3]red =0.423 + 0.035 = 0.458

substitute:

Ecell = 0 - (8.314*307)/(3*96500)* ln (0.0126275/0.458 )

Ecell = 0.031660 V