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
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.
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??
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.
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
To determine the molarity of Pb^2+ in the solution, we can use the Nernst equation. The Nernst equation relates the standard reduction potential (E°), the actual cell potential (E), the Faraday constant (F), the gas constant (R), and the temperature (T) to the concentration of the species involved in the electrochemical reaction.
The Nernst equation is given as:
E = E° - (RT / (nF)) * ln ([Pb^2+]^x)
In this equation, E is the actual cell potential, E° is the standard reduction potential, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (K), n is the number of moles of electrons transferred in the reaction (in this case, n = 2), F is the Faraday constant (96485 C/mol), [Pb^2+] is the concentration of Pb^2+ ions, and x is the stoichiometric coefficient of Pb^2+ in the balanced equation.
Given that E = -0.020 V and E° = -0.13 V for the reduction of Pb^2+ to Pb, we can substitute these values into the Nernst equation, along with the other constants:
-0.020 V = -0.13 V - ((8.314 J/(mol·K)) * T / (2 * 96485 C/mol)) * ln ([Pb^2+]^x)
To solve for the concentration [Pb^2+], we need to know the temperature (T) and the stoichiometric coefficient (x) from the balanced equation.