Question

Consider the cell described below at 267 K:

Sn | Sn2+ (0.999 M) || Pb2+ (0.975 M) | Pb

Given EoPb2+→Pb = -0.131 V, EoSn2+→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.397 mol/L

I've tried almost everything, please help.

Answer 1

Clearly, Sn will oxidize

so

Sn = 0.999 - 0.397= 0.602 M

Pb = 0.975+0.602 = 1.577 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

E° = Ered - Eox = (-0.131) - (-0.143) = 0.012 V

Ecell =0.012 -8.314*298/(2*96500) * ln ( 0.397 / 1.577 )

ECell = 0.02970 V