Question

A concentration cell based on the following half reaction at 309 K Ag+ + e- → Ag SRP = 0.80 V has initial concentrations of 1.37 M Ag+, 0.269 M Ag+, and a potential of 0.04334 V at these conditions. After 9.3 hours, the new potential of the cell is found to be 0.01406 V. What is the concentration of Ag+ at the cathode at this new potential?

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

the equation

Q = [Ag+]anode / [Ag+]cathode

initially

Q = (0.269+x) / (1.37-x)

E° = 0 since the same compound so

Ecell = E° - (RT/nF) x lnQ

0.01406 = 0 - (8.314*309)/(1*96500) * ln(  (0.269+x) / (1.37-x) )

0.01406 /( - (8.314*309)/(1*96500) ) =  ln(  (0.269+x) / (1.37-x) )

exp(-0.5281) =  (0.269+x) / (1.37-x)

0.58972*1.37 - 0.58972x = 0.269+x

0.58972*1.37 - 0.269 = 1+0.58972)x

x = (0.58972*1.37 - 0.269 ) / ( 1+0.58972)

x = 0.3390

Q = (0.269+x) / (1.37-x)

[Ag+] = 0.269+0.3390 = 0.608

[AG+]= 1.37 -0.3390= 1.031