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

[jkh4]

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!

 

[NotMrX]

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

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

 

[jkh4]

but since the k is such large number, [Ce4+] is going to be large too right?

 

[NotMrX]

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

 

[jkh4]

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...

 

[higherme]

i got the same problem too... i just cannot solve the quadratic equation

 

[higherme]

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

 

[garbagefish]

Can anyone explain how did this person get potential of a platinum electrode?