Question

The voltage generated by the zinc concentration cell described by,

Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s)

is 26.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.

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

Q = [Zn+2]ox / [Zn2+]red

n = 2electrons, T = 298 Ecell = 26*10^-3 V

substittue

Ecell = E° - (RT/nF) x ln([Zn+2]ox / [Zn2+]red)

26*10^-3 = 0 - (8.314*298)/(2*96500) * ln(0.1/x)

solve for x, concentration of Zn2+ in cathode

- (26*10^-3) / ((8.314*298)) * (2*96500) =  ln(0.1/x)

-2.0253 = ln(0.1/x)

exp(-2.0253 = 0.1/x

x = 0.1/(exp(-2.0253))

x = 0.757838

then

[Cu2+] in cahtode = 0.757838M