Question

Problem #2 A concentration cell based on the following half reaction at 289 K Zn2 + + 2 e-→ Zn SRP =-0.760 V has initial concentrations of 127 M Zn2+, 0.241 M Zn2+, and a potential of 0.02069 v at these conditions. After 8.1 hours, the new potential of the cell is found to be 0.007884 V. what is the concentration of Zn2+ at the cathode at this new potential2 [2n2 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

Then, get

E°cell = Ered - Eox = 0.76- 0.-76 = 0

then, the only driv eis the concentrations, i.e. Q

Ecell = E° - (RT/nF) x lnQ

Ecell = 0 - (8.314*298)/(2*96500) *ln ( [Zn2+]anode/ [Zn+2]cathode )

Ecell = -0.01283 * ln ([Zn2+]anode/ [Zn+2]cathode )

initially

Zn cathode = 1.27

Zn anode = 0.241

Ecell = -0.01283 * ln (0.241/1.27) = 0.0213 V

after reactions

Zn cathode = 1.27 - x

Zn anode = 0.241 + x

new cell potential is 0.007884

0.007884 = -0.01283 * ln ([Zn2+]anode/ [Zn+2]cathode )

[Zn2+]anode/ [Zn+2]cathode = exp(0.007884/-0.01283 ) = 0.54091

[Zn2+]anode = 0.54091 [Zn+2]cathode

and we know.

Zn cathode = 1.27 - x

Zn anode = 0.241 + x

0.241 + x = 0.54091* (1.27 - x)

0.686 - 0.54091x= 0.241 + x

(1+0.54091)x = 0.686 -0.241

x = (0.686 -0.241 ) / (1+0.54091) = 0.28879

Zn cathode = 1.27 - 0.28879 = 0.98121M left

Zn anode = 0.241 + 0.28879 = 0.52979 M formed

[Zn2+]cathode =  0.98121 M left