[FOrbitals]

Hi Guys,

Firstly, thanks for having a look at my post.

I have been given an assignment to do on reactions in general, I've gotten a bit stuck with this one:

Fe(OH)₃ + HOCl  H₂FeO₄ + HCl.

I have to balance the equation, using the half reaction method.
So, what I've done is :
- Removed the hydrogen from HOCl, H₂FeO₄ and HCl as I think that they are spectator ions, seeing as their oxidation state doesn't change throughout the reaction.

- This then left me with Fe(OH)₃ + OCl^-  FeO₄^2- + Cl^-

- From having the above, I took Fe as the reducing agent( as it lost H and gained O) and Cl and the oxidizing agent(as it lost O)

-Presuming those were correct, I went forward with the half reactions;
O: Fe(OH)₃ + 2e^- FeO₄ + H₂O
R:OCl^- :rarrow:Cl^-+2e^-

-From having these half reactions, I moved forward with balancing it, (hoping that it is in a basic solution) I got to this:

2Fe(OH)₃ + 3OCl^- + 4OH^- 2FeO₄^2- + 3Cl^-+5H₂O

Now, the reason I'm stuck, is that I'm not entirely sure if I was correct to take out the H as a spectator and if doing so caused my whole thing to go wrong?
Please let me know if what I did was logical and correct?
The method that I have used, is that the correct method since I have been asked to use the half equation method?

Thanks so much for reading this. All help is greatly appreciated!

[magician4]

starting to check your work from the end, first I was quite at ease:

2Fe(OH)₃ + 3OCl^- + 4OH^- 2FeO₄^2- + 3Cl^-+5H₂O

well, some protons left/right still were missing(equal numbers  no prob) , some final water balancing still needed to be done, but, asides from that , everything looks allright.

then I started to check your half equations

not so good

... and after finishing that, the big question that remained for me was: how the heck did he get from that - to his final equation, which is right?  Did you look it up somewhere and just introduce it as your final result?

well, ok , step by step, let's start with checking the oxidation states (formal analysis):

iron in Fe(OH)₃ corresponds to Fe^(+III) + 3 OH^- , in H₂FeO₄ to 2 H⁺ + Fe^(+VI) + 4 O^2-
  going from (+III) to (+VI) , this is an oxidation, and it will release 3 electrons per iron involved

chlorine in HOCl corresponds to H⁺ + O^2- + Cl^(+I) , in HCl to H⁺ + Cl^(-I)
  going from (+I) to (-I) , this is a reduction, and it requires 2 electrons per chlorine involved

now then, based on this analysis , let's frame some raw half equations ( i.e. not balanced with respect to the total electron exchange) , using water as temporary balance

(ox.) Fe(OH)₃ + H₂O  H₂FeO₄ + 3 H⁺ + 3 e^-

(red.) HOCl + 2 e-  Cl^- + OH^-

(brief checking ... yeah, looks good)

now we need equal numbers of electrons to be exchanged in both half processes, and with 2 and three in each sub-process, respectively,  the answer of course is 6
  we'll need the oxidation twice, the reduction three times to get where we want

therefore:

( 2* ox.)  2 Fe(OH)₃ + 2 H₂O  2 H₂FeO₄ + 6 H⁺ + 6 e^-
(3* red.)  3 HOCl + 6 e-  3 Cl^- + 3 OH^-

=================

(redox) 2 Fe(OH)₃ + 3 HOCl + 2 H₂O  2 H₂FeO₄ + 3 Cl ^- + 6 H⁺ + 3 OH^-

=

(redox) 2 Fe(OH)₃ + 3 HOCl    2 H₂FeO₄ + 3 HCl  +  H₂O


well, here we are

any questions?


regards

Ingo

[FOrbitals]

Hi Ingo. 
Sorry for taking so long to reply.
Only last thing,  is that reaction taking place in an acidic or basic medium?

Thanks so much for your help.

[Hunter2]

I would say alcaline. H₂FeO₄ is only known as salt for example BaFeO₄.

[FOrbitals]

But what about the HCl and HOCl then?

[magician4]

I would agree that the outer pH should be on the mild alkaline side  (in fact , I would recommend to use a buffer to keep it there, as the HCl produced else would spoil the party,as Hunter2 pointed out correctly), somewhere near 7.5 or so.

in this situation , Fe(OH)₃ still would be Fe(OH)₃ ( the limit is somewhere around pH > 3.5 , if memory serves), H₂FeO₄ (pKa1= 3.5 , pKa2 = 7.3 ) would be sufficiently deprotonated to protect it, and HOCl ( pKa 7.54) still would be present ( as part of its own buffer) in concentrations high enough to serve the purpose.

Of course, taking this into account , the redox-equation "in reality" would look somewhat different:
(redox) 2 Fe(OH)₃ + 3 HOCl  + 7 OH^-  2 FeO₄^2- + 3 Cl^- +  8 H₂O

so yes, there is a difference between the what you were asked to show ( with respect to educts, products) and reality's requirements



regards

Ingo