Consider the following half reactions at 298 K Ni2+ + 2 e- Ni Eo = -0.231 V Pb2+

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? 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? 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?

Explanation / Answer

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

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