How Does Electrode Potential Change in a Mixed Ce4+ and Fe2+ Solution?

AI Thread Summary
The discussion revolves around calculating the concentration of Ce4+ in a mixed solution of Ce4+ and Fe2+. The equilibrium constant (keq) is given as 1.1×10^16, indicating that the reaction favors the formation of products, which suggests a very low concentration of Ce4+ at equilibrium. Participants express confusion over the quadratic equation used to solve for concentrations, with some encountering imaginary numbers in their calculations. The potential of the platinum electrode in the solution is calculated to be 0.767 V, but questions arise regarding the accuracy of the concentration values obtained. Ultimately, the calculations indicate that [Ce4+] is expected to be very small, around 10^-8 M, due to the high equilibrium constant.
jkh4
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1. Homework Statement

Given the following half-reactions:
Ce4+ + e− → Ce3+ E° = 1.72 V
Fe3+ + e− → Fe2+ E° = 0.771 V

A solution is prepared by mixing 7.0 mL of 0.30 M Fe2+ with 8.0 mL of 0.12 M Ce4+.

Calculate [Ce4+] in the solution.

I've got the potential of a platinum electrode dipped into the resulting, equilibrated, solution (relative to SHE) to be 0.767 V, and keq of 1.1×10^16 from this equation: Ce4+ + Fe2+ ⇌ Ce3+ + Fe3+.

How do you get [Ce4+]? Why isn't that the same as the concentration posted in the question?

Thanks!
 
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jkh4 said:
1. Homework Statement

Given the following half-reactions:
Ce4+ + e− → Ce3+ E° = 1.72 V
Fe3+ + e− → Fe2+ E° = 0.771 V

A solution is prepared by mixing 7.0 mL of 0.30 M Fe2+ with 8.0 mL of 0.12 M Ce4+.

Calculate [Ce4+] in the solution.

I've got the potential of a platinum electrode dipped into the resulting, equilibrated, solution (relative to SHE) to be 0.767 V, and keq of 1.1×10^16 from this equation: Ce4+ + Fe2+ ⇌ Ce3+ + Fe3+.

How do you get [Ce4+]? Why isn't that the same as the concentration posted in the question?

Thanks!

R)____Ce4+___ +_________ Fe2+___⇌___Ce3+___+_Fe3+
I) .008*.12/.015 ______.007*.3/.015______0_________0
C)_____-x _______________-x ___________x ________x
E) .008*.12/.015 -x ____.007*.3/.015 _____x ________x

k=\frac{x*x}{(\frac{0.008*0.12}{0.015}-x)(\frac{0.007*0.3}{0.015}-x)}
 
but since the k is such large number, [Ce4+] is going to be large too right?
 
jkh4 said:
but since the k is such large number, [Ce4+] is going to be large too right?
no it will be near 0. X represents Ce3+ concentration and that number minus x represent the Ce4+ Solve x using quadratic. Subtract from that number. Then Ce4+ ends being a pretty small number
 
so in this case, k is 1.1×10^16? cause when i try the quadratic on the internet quadratic solver, it says the answer is an imagary number...
 
i got the same problem too... i just cannot solve the quadratic equation
 
NotMrX said:
R)____Ce4+___ +_________ Fe2+___⇌___Ce3+___+_Fe3+
I) .008*.12/.015 ______.007*.3/.015______0_________0
C)_____-x _______________-x ___________x ________x
E) .008*.12/.015 -x ____.007*.3/.015 _____x ________x

k=\frac{x*x}{(\frac{0.008*0.12}{0.015}-x)(\frac{0.007*0.3}{0.015}-x)}

I used this method too, but i did not get the correct answer

after i found X, it is the [Ce3+], i subtract it from 0.008*0.12/0.015 which is the [Ce4+] at equilibrium
my answer was very close to 0...around 10^-8
why am i not getting the right answer
 
Can anyone explain how did this person get potential of a platinum electrode?
 

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