# Answered Essay: A  concentration cell based on the following half reaction at 283 K Ag+ + e- → Ag       SRP = 0.80 V

A  concentration cell based on the following half reaction at 283 K

Ag+ + e → Ag       SRP = 0.80 V

has initial concentrations of 1.35 M Ag+, 0.407 M Ag+, and a potential of 0.02924 V at these conditions. After 3.4 hours, the new potential of the cell is found to be 0.01157 V. What is the concentration of Ag+ at the cathode at this new potential?

If this is only Ag+

then

E° = Ecahtode – Eanode = 0

since it is the same species, at STP this is 0

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 = 0 – (RT/nF) x lnQ

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

Initially

Ecell = 0 – (RT/nF) x ln([Ag+]anode / [Ag+]cathode)

Ecell = 0 – (8.314*283/(1*96500) * ln(0.407/1.35) = 0.02923 V approx..

from the data, we got this right, since E = 0.02924 V from the statement…

now, after reaction happens:

[Ag+]anode = 0.407 + x

[Ag+]cathode = 1.35-x

now, solve

Ecell = 0 – (8.314*283/(1*96500) * ln(0.407+x /1.35-x)

0.01157 = 0 – (8.314*283/(1*96500) * ln(0.407+x /1.35-x)

0.01157 / (- (8.314*283/(1*96500) ) =  ln(0.407+x /1.35-x)

-0.4745  =  ln(0.407+x /1.35-x)

exp(-0.4745) = (0.407+x) / (1.35-x)

0.62219 * 1.35 -0.62219*x = 0.407 + x

(1+0.62219)x = 0.62219*1.35 – 0.407

x =  (0.62219*1.35 – 0.407 ) / ((1+0.62219))

x = 0.26689

[Ag+]anode = 0.407 + 0.26689 = 0.67389M

[Ag+]cathode = 1.35-0.26689 = 1.08311M

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