starting to check your work from the end, first I was quite at ease:
2Fe(OH)3 + 3OCl- + 4OH- 2FeO42- + 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 H
2FeO
4 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 + H
2O
H
2FeO
4 + 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 H
2O
2 H
2FeO
4 + 6 H
+ + 6 e
-(3* red.) 3 HOCl + 6 e-
3 Cl
- + 3 OH
-=================
(redox) 2 Fe(OH)
3 + 3 HOCl + 2 H
2O
2 H
2FeO
4 + 3 Cl
- + 6 H
+ + 3 OH
- =
(redox) 2 Fe(OH)
3 + 3 HOCl
2 H
2FeO
4 + 3 HCl + H
2O
well, here we are
any questions?
regards
Ingo