A cell potential of 0.427V is measured in an experiment where a solution contain
ID: 968095 • Letter: A
Question
A cell potential of 0.427V is measured in an experiment where a solution containing a 0.010 M_ solution of Pb(NO_1)_2 and a lead electrode is measured relative to a saturated calomel electrode (0.242V relative to a SHE). The cell potential changes to 0.486V when the Pb(NO_3)_2 concentration is changed to 1.0 Times 10^-4 M. Is this data consistent with the Nernst equation? Enough NaF is now added to the 10^-4 M solution of Pb^2+ to completely precipitate the Pb as PbF_2 and establish an equilibrium concentration of excess F equal to 0.190 M. The new cell potential is 0.545V. What is the K_sp of PbF_2? From the data in part b), determine the E^0 value for the 1/2 cell reaction. PbF_2+ 2e^- Pb^0 +2F^-Explanation / Answer
Nernst Equation
Ecell = Eo -RT/nF ln Q
It can be rewritten as
Ecell = Eo - 0.0592/n log Q
Ecell = 0.427V - 0.242 = 0.185
for concentration of 0.01
Ecell for concentration of 0.0001 is
Ecell = 0.486 - 0.242 = 0.244
setting up the nernst equatuion in the 2 cases we have
0.242 = -0.13 - 0.0592/2 logQ
log Q = 12.63
Q = 4.31 x 1012
0.186 = -0.13 - 0.0592/2 logQ
log Q = 10.64
Q = 4.36 x 1010
Since the Q in both cases differ by 100 and the concentration of Pb differ by 100 they are consistent with Nernst equation.
a) New cell potential is 0.545 V which is 0.545-0.242 = 0.303
0.303 = -0.13 - 0.0592/2 logQ
log Q = 14.62
Q = 4.24 x 1014
so Pb concentration is 1 x 10-6 M
Ksp = [Pb2+][F-]2
Ksp = 1 x 10-6 x (0.19)2
Ksp = 3.6 x 10-8
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