Part B What is the cell potential when the concentration of Cu2+ has fallen to 0
ID: 693654 • Letter: P
Question
Part B What is the cell potential when the concentration of Cu2+ has fallen to 0.250 M? A voltaic cell consists of a Pb Pb half-cell and a Cu/Cu half-cell at 25 C. The initial concentrations of Pb+ and Cu2+ are 5.30x102 M and 1.50 M respectively Express your answer using two significant figures Submit My Answers Give Up Incorrect: Try Again; 3 attempts remaining Part C What are the concentrations of Pb2 and Cu2+ when the cell potential falls to 0.370 V? Enter your answers numerically separated by a comma. Express your answer using two significant figures. Pb+.(Cu2+11380.34 Submit My Answers Give Up Incorrect; Try Again; 6 attempts remaining Provide FeedExplanation / Answer
initially
[Cu+2] = 1.5
[PB+2] = 5.3*10^-12
after "x" reaciton
[Cu+2] = 1.5 - x
[PB+2] = 5.3*10^-2+ x
is[Cu+2] = 0.25
[Cu+2] = 1.5 - x = 0.25
x = 1.5-0.25 = 1.25
[PB+2] = 5.3*10^-2+ x = 1.303
now,
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
Cu2+ + 2 e Cu(s) +0.337; Pb2+ + 2 e Pb(s) 0.126
E° = 0.337 - -0.126 = 0.463 V
Q = 1.303/0.25 = 5.212
Ecell = E° - (RT/nF) x lnQ
Ecell = 0.463 - (8.314*298)/(2*96500)* ln(5.212) = 0.4418 V
C)
when E = 0.37
Ecell = E° - (RT/nF) x lnQ
0.370 = 0.463 - (8.314*298)/(2*96500)* ln([Pb+2]/[Cu+2])
(0.37-0.463)/(-8.314*298)*2*96500 = ln((0.053 + x )/(1.5-x))
(0.370 -0.463 )/(-0.0592)*2 = ln((0.053 + x )/(1.5-x))
exp(3.1418) = (0.053 + x )/(1.5-x))
23.145*1.5 - 23.145x = 0.053+x
(1+23.145)x = 23.145*1.5-0.053
x = (23.145*1.5-0.053)/(1+23.145)
x = 1.435
[PB+2] = 0.053+1.435 = 1.488 M
[Cu+2] = 1.5-1.435 = 0.065 M