Question

A Cu/Cu2+ concentration cell has a voltage of 0.22 V at 25 ∘C. The concentration of Cu2+ in one of the half-cells is 1.6×10−3 M . What is the concentration of Cu2+ in the other half-cell? (Assume the concentration in the unknown cell to be the lower of the two concentrations.)

Answer 1

A Cu/Cu2+ concentration cell has a voltage of 0.22 V at 25 ∘C.

The concentration of Cu2+ in one of the half-cells is 1.6×10−3 M .

What is the concentration of Cu2+ in the other half-cell? (Assume the concentration in the unknown cell to be the lower of the two concentrations.)

E°cell = 0, sinc E Cu/Cu+2 is the same for both sides

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

0.22 = 0 - (8.314*298)/(2*96500) * ln(Q)

Q = [Cu+2]ox / [Cu+2]red

0.22 = 0 - (8.314*298)/(2*96500) * ln([Cu+2]ox / [Cu+2]red)

0.22 = -0.01283 * ln([Cu+2]ox / (1.6*10^-3))

0.22 /-0.01283 = ln([Cu+2]ox / (1.6*10^-3))

exp(-17.147) = [Cu+2]ox / (1.6*10^-3))

[Cu+2]ox = (1.6*10^-3)*exp(-17.147) = 5.79*10^-11 --> 5.8 *10^-11 M