Question

Consider the following half reactions at 298 K

Ni2+ + 2 e- → Ni Eo = -0.231 V

Pb2+ + 2 e- → Pb Eo = -0.133 V

A galvanice cell based on these half reactions is set up under standard conditions where each solution is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Pb electrode weigh when the nonstandard potential of the cell is 0.09296 V?

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

E°cell

Pb2+ + 2 e- → Pb Eo = -0.133 V

Ni →Ni2+ + 2 e- E0 = 0.231

E°cell = -0.133+0.231 = 0.098 V

mol of Ni(s) = mass/MW = 100/58.6934 = 1.7037

mol of Pb(s) = mass/MW = 100/207.2 = 0.4826254

now..

find Pb2+ when Ecell = 0.09296

Q = [Ni2+]/[Pb+2]

Ecell = E° - (RT/nF) x lnQ

0.09296 = 0.098 - (8.314*298)/(2*96500) * ln ([Ni2+]/[Pb+2])

(0.09296 -0.098 ) /(-(8.314*298)/(2*96500) )= ln ([Ni2+]/[Pb+2])

0.39261 = ln ([Ni2+]/[Pb+2])

exp(0.39261) = [Ni2+]/[Pb+2]

1.4808 = [Ni2+]/[Pb+2]

[Ni2+] is produced, Pb+2 is consumed

1.4808 = (1.7037-x) / (0.4826254+x)

0.4826254*1.4808 + 1.4808 x = 1.7037-x

2.4808 x = 1.7037-0.4826254*1.4808

x = 0.98902 / 2.4808 = 0.3986

mol of Pb = 0.4826254+0.3986 = 0.8812254

mass = mol*MW = 0.8812254*207.2 = 182.58 g of Pb, increase of 82 g