[jkulier]

PROBLEM

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 6.0 mL of 0.30 M Fe2+ with 4.0 mL of 0.12 M Ce4+. What is the potential of a platinum electrode dipped into the resulting, equilibrated, solution (relative to SHE)?

ATTEMPTED SOLUTION

- Calculated the concentration of Ce4+ and Fe2+ after the mixing of the individual solutions via C1V1 = C2V2
- Found E+ and E- (i.e. for the half reaction involving Ce4+: E+ = 1.72 - 0.05916*log(1/[Ce4+]))
- Took E- and subtracted it from E+

Please tell me where I went wrong? Am I heading in the wrong direction entirely?
Thank you in advance.

[Borek]

Have you taken into account fact, that Ce⁴⁺ and Fe²⁺ react? The one in excess will force final potential.

[jkulier]

Alright, now that I have taken that into consideration, my equilibrium concentrations look like this: Fe2+ = 0.00132 + x, Ce4+ = x, Fe3+ = 0.00048 - x, Ce3+ = 0.00048 - x

I am rather confused about how to go about using this information in finding the potential. Which equation would pertain to this?

Thank you again.

[Borek]

Start assuming reaction went to completion. see what concentrations of ions will be present.

[jkulier]

Start assuming reaction went to completion. see what concentrations of ions will be present.

In that case, would Fe2+ = 0.00132, Ce4+ = 0, Ce3+ = 0.00048, Fe3+ = 0.00048??

[Borek]

You are on the right track. Potential of the system is defined by the Fe³⁺/Fe²⁺ pair. Once you will get the potential calculated you can use this value to calculate concentration of Ce⁴⁺ - it is obviously not zero, just very small. But there is no need for that, as that's not what the question asks.