Question

A concentration cell based on the following half reaction at 299 K

Zn²⁺ + 2 e^- → Zn       SRP = -0.760 V

has initial concentrations of 1.27 M Zn²⁺, 0.319 M Zn²⁺, and a potential of 0.01780 V at these conditions. After 5.6 hours, the new potential of the cell is found to be 0.002816 V. What is the concentration of Zn²⁺ at the cathode at this new potential?

Answer 1

initially

[Zn+2]ox = 1.27M

[Zn+2]red = 0.319 M

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

Ecell = E° - 0.0592/n*log(Q)

E° = 0, since same material, n = 2 , Q = [Zn+2]ox / [Zn+2] red

Ecell = - 0.0592/2*log(Q)

Ecell = - 0.0296*log( [Zn+2]ox / [Zn+2] red)

Proof:

Ecell = - 0.0296*log( 0.319/1.27) =0.0178 V

[Zn+2]ox / [Zn+2] red) = 0.319 / 1.27

then

initially

[Zn+2]ox = 0.319

[Zn+2] red = 1.27

as reaciton occurs:

[Zn+2]ox = 0.319 + x

[Zn+2] red = 1.27 - x

and we know, th enew E value:

Ecell = - 0.0296*log( [Zn+2]ox / [Zn+2] red)

0.002816 = - 0.0296*log( [Zn+2]ox / [Zn+2] red)

[Zn+2]ox / [Zn+2] red= 10^(0.002816 /-0.0296) = 0.8032761

[Zn+2]ox / [Zn+2] red= 0.8032761

substitute all data:

[Zn+2]ox = 0.319 + x

[Zn+2] red = 1.27 - x

in

[Zn+2]ox / [Zn+2] red= 0.8032761

(0.319 + x ) / (1.27 - x) = 0.8032761

(0.319 + x ) = 0.8032761*(1.27 - x)

(0.319 + x ) = 1.02016 - 0.8032761x

(1+0.8032761)x = 1.02016 -0.319

x = (1.02016 -0.319 ) / ((1+0.8032761)) = 0.38882

[Zn+2]ox = 0.319 + 0.38882 = 0.70782

[Zn+2] red = 1.27 - 0.38882 = 0.88118 M

CATHODE = electrode in which REDUCTION takes place, therefore, the oxidizing agent will get reduced (gain electrons). The electrons are received from the voltmeter ( which come from the anode ). The ions in solution will then form solid over the cathode. Since there is solid formation, there will be an increase in the mass of this electrode.

[Zn+2] cathode = 0.88118 M