Question

A concentration cell based on the following half reaction at 318 K

Cu2+ + 2 e- → Cu SRP = 0.340 V

has initial concentrations of 1.31 M Cu2+, 0.291 M Cu2+, and a potential of 0.02061 V at these conditions. After 8.1 hours, the new potential of the cell is found to be 0.006576 V. What is the concentration of Cu2+ at the cathode at this new potential?

Answer 1

This is driven due to concentration change

so

E° = Ecathode - Eanode = 0 V

We need to use:

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

n = 2 electrons, F = 96500 C/mol, T = 318K, R = 8.314, Q = [Cu2+]ox / [Cu2+]red

substitute

Ecell = 0 - 8.314*318/(2*96500) * ln( [Cu2+]ox / [Cu2+]red)

ECell = -0.01369 ln( [Cu2+]ox / [Cu2+]red)

initially

Ecell = 0.02061

0.02061= - 0.01369 * ln( [Cu2+]ox / [Cu2+]red)

exp(-0.02061/0.01369 ) =  [Cu2+]ox / [Cu2+]red

0.22191  =  [Cu2+]ox / [Cu2+]red

[Cu2+]ox = 0.291

[Cu2+]red = 1.31; this is th eonly way the ratio becomes 0.0222

so

initially

[Cu2+]ox = 0.291 + x

[Cu2+]red = 1.31 - x

substitute in the new value

NEW:

ECell = -0.01369 ln( [Cu2+]ox / [Cu2+]red)

Ecell new = 0.006576

0.006576= -0.01369 ln( [Cu2+]ox / [Cu2+]red)

substitute

[Cu2+]ox = 0.291 + x

[Cu2+]red = 1.31 - x

0.006576= -0.01369 ln( ( 0.291 + x) / (1.31-x))

exp(0.006576/-0.01369) =  (0.291 + x) / (1.31-x)

0.61856*(1.31-x) = 0.291+x

0.8103136 - 0.61856x =  0.291+x

(1+0.61856)x = 0.8103136-0.291

x = ( 0.8103136-0.291)/((1+0.61856)) = 0.3208

substitute

note that reduction = cathode so

[Cu2+]red = 1.31 - 0.3208 = 0.9892 M