total] For your first determination of lead concentration in your aliquots, you
ID: 694927 • Letter: T
Question
total] For your first determination of lead concentration in your aliquots, you decide you should be able to determine the concentration of the Pb2 in solution by careful measurement of the voltage of an electrochemical cell, or potentiometry a. 3 points] Consider a galvanic cell having one Pb/Pb redox couple and one Zn/Zn2 redox couple. Write and balance the reduction oxidation reaction for this electrochemical cell at standard conditions. b. [4 points] What voltage would be obtained from the above reaction at standard conditions? [5 points] Now that you have thought about the electrochemistry of this cell at known concentrations of Pb2, you are ready to think about using a measured voltage to calculate an unknown concentration. You make an electrochemical cell in which the Pb/Pb2 electrode is a pure solid lead rod and one of your 40.00 mL aliquots. The other electrode is a pure Zinc rod and a 1.000 molar solution of Zn2 solutions in the cell is 25.00 °C. When you attach a voltmeter to your cell you measure a voltage of +0.5538 volts. Set up a Nernst equation and determine the unknown concentration of Pb2 in your cell c. ture of theExplanation / Answer
a)
Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.
Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V
All other samples are based on this reference.
Find the Reduction Potential of each reaction (Tables)
Zn2+ + 2 e Zn(s) 0.7618;
Pb2+ + 2 e Pb(s) 0.126
The most positive has more potential to reduce, it will be reduced
The most negative will be oxidized, since it will donate it selectrons
For total E°cell potential:
E°cell = Ered – Eox
E°cell = EPb - EZn = -0.126 - - 0.7618
E°cell = 0.6358 V
c.
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
0.5538 = 0.6358 -(8.314*298)/(2*96500) * ln([Zn+2]/[Pb+2])
[Zn+2] = 1 M
0.5538 = 0.6358 -(8.314*298)/(2*96500) * ln(1/[Pb+2])
(0.5538 -0.6358 )/((8.314*298)) * -2*96500 = ln(1/[Pb+2])
[Pb+2] = 1/(exp(6.387))
[PB+2] = 0.001683 M
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