[Mustafa28]

Guys I need some help.

Every explanation on the internet about balancing redox reaction's uses super easy reactions involving A + B --> C + D

I need to learn how to split redox reaction into half reactions and balance when things are more complicated.

I know how to assign oxidation numbers (even in metal-ligand complexes) , and the various steps to balancing after splitting, but what I have difficulty with is splitting a reaction into two half reaction's when the reaction is complicated.

Here's a question from my last mid-term

5. (4 points) Balance the following reaction in basic solution
 
Fe(CN)6 ^4-  + Ce⁴⁺  ----> Ce(OH)₃ + Fe(OH)₃  + CO₃^2- + NO₃-

The Fe goes from oxidation state of +2 to +3 (so it get's oxidised by losing an electron)
The Ce goes from oxidation state 4+ to 3+, (so it gets reduced by gaining an electron)

How would you split the following overall reaction into two half reactions? Please explain your reasoning. What do you do with the CO3 and the NO3 in particular?

[Yakimikku]

What about the C and the N? Are they being oxidized or reduced?

[Mustafa28]

What about the C and the N? Are they being oxidized or reduced?

Can that happen? Multiple oxidation-reductions in same reaction? If so wouldn't I need to write 3 or 4 "half reactions" ?

Also the way the prof set up that question is perplexing. There is no oxygen on the left hand side. How come there are magically oxygens on the right hand of the equation. It makes zero sense.

[Borek]

Can that happen? Multiple oxidation-reductions in same reaction? If so wouldn't I need to write 3 or 4 "half reactions" ?

You don't have multiple reactions - you have CN^- that is getting oxidized. This is one ion, so there is one reaction. If you plan to balance the reaction in terms of oxidation numbers, you have a problem, but if you use half reaction method, it won't be more difficult than any other reaction.

Also the way the prof set up that question is perplexing. There is no oxygen on the left hand side. How come there are magically oxygens on the right hand of the equation. It makes zero sense.

It makes perfect sense, when you read the whole question. Reaction takes place in the basic solution, which means you have access to water and plenty of OH^- ions. They are not listed in the reaction, but they are there and you can freely use them for balancing oxygen and hydrogen.

[Yakimikku]

Can that happen? Multiple oxidation-reductions in same reaction? If so wouldn't I need to write 3 or 4 "half reactions" ?

You don't have multiple reactions - you have CN^- that is getting oxidized. This is one ion, so there is one reaction. If you plan to balance the reaction in terms of oxidation numbers, you have a problem, but if you use half reaction method, it won't be more difficult than any other reaction.

To clarify, the Ce^IV is oxidizing both the CN^- ligands and the Fe^II. There is only one overall reaction that you need to worry about and two half-reactions (one of ox. and the other for red.). You are right however, that there are multiple oxidation-reduction reactions happening. That doesn't complicate the problem that much, so I wouldn't worry about it.

By checking the oxidation states of the C and N on both sides of the equation, you will see without doubt that there is C and N oxidation (Ce^IV is a strong oxidant!). Although at this point we have revealed that this is the case, you should check by yourself for practice and it'll become important with balancing electrons as you solve the problem.

[Mustafa28]

Thanks guys,

Can someone explain to me how I got about splitting the reactions into half reactions? I'm having trouble particularly with the concept of spectator vs "needs to be included"

The problem is that every website/book i have consulted, it is as if the author subconsciously fears the page and half he is about to spend typing up "add h2o" "balance this" "balance that" etc in the multi-step explanation for half reactions ... that they all seem to gloss-over the most important part

Namely how do you split the reaction into half reactions in the first place? Like I said it is super obvious with the easy examples they use, but in questions like this theres so many reagents that I need to make a choice about what gets included and what get's excluded in the half reactions.

Without balancing it, can someone write out what the starting half reactions for this equation are. I'll try to balance it from there and check my answer against the correct answer listed in the marking scheme.

[Borek]

Reaction you listed in your first post doesn't contain water and OH^-, but it contains everything else. Cerium part should be obvious - everything else is the second half reaction.

In this particular case it is pretty obvious - ferrocyanide is getting oxidized, so you need products containing iron, carbon and nitrogen. All three are listed.

[Mustafa28]

Reaction you listed in your first post doesn't contain water and OH^-, but it contains everything else. Cerium part should be obvious - everything else is the second half reaction.

In this particular case it is pretty obvious - ferrocyanide is getting oxidized, so you need products containing iron, carbon and nitrogen. All three are listed.

hmmm seems theres some fundamental problems with my understanding of this whole topic.

For example, I was reviewing an assignment and one of the questions was the one attatched here (the solution given by the prof is in blue)

I would get this question wrong right off the bat because the oxidation states given for Ca(OH)2 make no sense at all

According to hierarchal rules she gave.

Group 2 metals are +2 oxidation state.
Oxygen is very electronegative and will normally be in a -2 oxidation state.
Hydrogen is in a +1 oxidation state unless its bonded to things that are less electronegative than it (ie things like Grp 1/2 metals)

The Ca(OH)2 is neutral, therefore , Ca2+  and  2(oh)-   makes perfect sense.

But how the hell can the hydrogen be in a -1 oxidation state? I would think it needs to be +1. .

Arggg evil electrochemistry

[AWK]

5. (4 points) Balance the following reaction in basic solution
 
Fe(CN)6 ^4-  + Ce⁴⁺  ----> Ce(OH)₃ + Fe(OH)₃  + CO₃^2- + NO₃-

The Fe goes from oxidation state of +2 to +3 (so it get's oxidised by losing an electron)
The Ce goes from oxidation state 4+ to 3+, (so it gets reduced by gaining an electron)

Fe(CN)₆^4- + xe- = Fe³⁺ + 6C⁴⁺ + 6N⁵⁺
You shoud do electron balance for the whole complex ion

[Borek]

I would get this question wrong right off the bat because the oxidation states given for Ca(OH)2 make no sense at all

You pay no attention to what you write, so I guess you also have problems paying attention to what you read. You stated oxidation numbers for atoms in Ca(OH)₂ don't make sense, then you explained - correctly - how they do make sense.

Hydrogen oxidation number is -1 in CaH₂, not in Ca(OH)₂. And hydrides are an exception to the general rule that H is usually +1.

Edit: you even listed this rule!

Hydrogen is in a +1 oxidation state unless its bonded to things that are less electronegative than it (ie things like Grp 1/2 metals)

That's exactly the situation in CaH₂.

Please compare the list of rules here: http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method

[Mustafa28]

I would get this question wrong right off the bat because the oxidation states given for Ca(OH)2 make no sense at all

You pay no attention to what you write, so I guess you also have problems paying attention to what you read. You stated oxidation numbers for atoms in Ca(OH)₂ don't make sense, then you explained - correctly - how they do make sense.

Hydrogen oxidation number is -1 in CaH₂, not in Ca(OH)₂. And hydrides are an exclusion to the general rule that H is usually +1.

Please compare the list of rules here: http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method

I was referring to Professor's solution. If you look at the labels below the equation, she gives the oxidation states for every element. When it comes to Ca(OH)2, the oxidation state given for the H is (-1). My contention is that it should be +1. And I was asking whether I was wrong, and If so why.

[Borek]

I was referring to Professor's solution. If you look at the labels below the equation, she gives the oxidation states for every element. When it comes to Ca(OH)2, the oxidation state given for the H is (-1). My contention is that it should be +1. And I was asking whether I was wrong, and If so why.

Sorry, I misread your post. Its a typo on her side, should be +2, -2, +1.

[Mustafa28]

Oh I see, thanks