Pb^2+ +2e^- ==> Pb Eo red = -0.13v (copy to below)
Ni^2+ +2e^- Ni Eo red =-0.25 (reverse this and change sign of Eo red to Eo ox)
Pb^2+ +2e^- ==> Pb Eo = -0.13 v
Ni ==> Ni^2+ + 2e Eo = +0.25 v (Add these to get Eo cell)
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Pb^2+ + Ni ==> Pb + Ni^2+ Eo cell = 0.25 + (-0.13) = 0.12 v
Ecell = Eo cell - (2.303*RT/n) log Q and Q = Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+)
Ecell = Eo cell - (0.0592/2) log Keq (from step above).
You know Ecell from the problem. You know Eo cell = 0,12 v from above.
For values in K, substitute (Pb) = 1; (Ni) = 1; (Pb^2+) = x M. ; (Ni^2+) = 0.27 M. Then solve for x = (Pb^2+) in M or moles/L
Post your work if you get stuck. I think I have interpreted the problem correctly. You're given the (Ni^2+) and E cell and you want to calculate the (Pb^2+). There is another way to do this but it is much longer and tedious. Having said that my students seem to get confused with the log K when working it this way BUT believe me it is much less math involved and can be done in half the time. Note that since Eo cell = + number that means it is a spontaneous reaction.
Pb(s) Pb^2+ (aq) [pb^2+] =?M
Ni^2+(aq) Ni^2+=.27M Ni(s)
Pb^2+ +2e^- Pb E= -0.13v
Ni^2+ +2e^- Ni E=-0.25
E=-0.020v re wrote the question. want the Molarity of the Pb
4 answers
temp 298
log is base 10
Ecell = Eo cell - (0.0592/2) log Keq (from step above).
.020=.12-(0.0592/2)log Keq log(.27/m)
0.02 = 0.12 - 0.0128551 log(0.27/m)
m=0.000112976
SEEMS LIKE A SMALL AMOUNT. HOW DO I CHECK??
log is base 10
Ecell = Eo cell - (0.0592/2) log Keq (from step above).
.020=.12-(0.0592/2)log Keq log(.27/m)
0.02 = 0.12 - 0.0128551 log(0.27/m)
m=0.000112976
SEEMS LIKE A SMALL AMOUNT. HOW DO I CHECK??
You didn't substitute correctly for Keq. You should calculate
log Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+)
You made log Keq = 0.27. You should follow the instructions I gave.
(Pb) = 1
(Ni^2+) = 0.27
(Ni) = 1
(Pb^2+) = x so the equation is
0.020 = 0.12 - (0.0296)*log[(1)(0.27)/(1)(x)] and solve for x in moles/L.
I don't want to leave without acknowledging an error in what I wrote. It make no difference in the final answer becasue I corrected it later. Here is what I wrote for ONE of the steps.
Ecell = Eo cell - (2.303*RT/n) log Q and Q = Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+) but I omitted Faraday's constant. It should be
Ecell = Eo cell - (2.303*RT/nF) log Q and Q = Keq=(Pb)(Ni^2+)/(Ni)(Pb^2+)
Then if you substitute for R of 8.314, T = 298, F = 96,485 and 2 for n, then
(2.303*8.314*298/2*96,485) = 0.0296. Your post is now down below the line. You may want to repost at the top if you have additional quesitons.
log Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+)
You made log Keq = 0.27. You should follow the instructions I gave.
(Pb) = 1
(Ni^2+) = 0.27
(Ni) = 1
(Pb^2+) = x so the equation is
0.020 = 0.12 - (0.0296)*log[(1)(0.27)/(1)(x)] and solve for x in moles/L.
I don't want to leave without acknowledging an error in what I wrote. It make no difference in the final answer becasue I corrected it later. Here is what I wrote for ONE of the steps.
Ecell = Eo cell - (2.303*RT/n) log Q and Q = Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+) but I omitted Faraday's constant. It should be
Ecell = Eo cell - (2.303*RT/nF) log Q and Q = Keq=(Pb)(Ni^2+)/(Ni)(Pb^2+)
Then if you substitute for R of 8.314, T = 298, F = 96,485 and 2 for n, then
(2.303*8.314*298/2*96,485) = 0.0296. Your post is now down below the line. You may want to repost at the top if you have additional quesitons.
0.020 = 0.12 - (2.303*8.314*298)/(2*96485)*log[((1)(0.27))/((1)(x))]
x= 0.000112046 or 1.12046*10^-4 Molarity
It looks correct
x= 0.000112046 or 1.12046*10^-4 Molarity
It looks correct