[Chemistry] Another electrochemistry question

1. Apr 1, 2007

jkh4

1. The problem statement, all variables and given/known data

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!

2. Apr 1, 2007

NotMrX

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

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

3. Apr 1, 2007

jkh4

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

4. Apr 3, 2007

NotMrX

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

5. Apr 4, 2007

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....

6. Nov 15, 2007

higherme

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

7. Nov 15, 2007

higherme

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

8. Mar 25, 2008

garbagefish

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