A concentration cell is built based on the following half reactions by using two pieces of copper as electrodes, two Cu2+ solutions, 0.145 M and 0.405 M, and all other materials needed for a galvanic cell. What will the potential of this cell be when the cathode concentration of Cu2+ has changed by 0.031 M at 279 K? Cu2+ + 2 e- → Cu Eo = 0.341 V

## Expert Answer

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

so..

Eºcell = 0, since it is the SAME material, copper

so, the Q is driving this effect, n = 2

Ecell = E° – (RT/nF) x lnQ

Ecell = 0 – (8.314*279)O/(2*96500) * ln([Cu+2]ox / [Cu+2]red)

initially

[Cu+2]ox = 0.145

[Cu2+]red = 0.450

in equilibrium

[Cu+2]ox = 0.145 + x

[Cu2+]red = 0.450 – x

and we know thatr x = 0.031 so

[Cu+2]ox = 0.145 + 0.031 = 0.176

[Cu2+]red = 0.450 – 0.031 = 0.419

substitute

Ecell = 0 – (8.314*279)/(2*96500) * ln([Cu+2]ox / [Cu+2]red)

Ecell = 0 – (8.314*279)/(2*96500) * ln(0.176 / 0.419)

Ecell = 0.0104248 V