Redox Reactions: Identifying Agents & Writing Half-Eqns

[DarylMBCP]

Hey guys, I'm really new to Redox reactions so any help is greatly appreciated. How do I identify the oxidizing and reducing agents from the chemical equation;

3N₂H₄ --> 4NH₃ + N₂

Why is it that the agents are products of the equation, shouldn't they be the reactants? I'm also not really sure how to write the half-equations since there are two compounds with Nitrogen. How do I write out the equation for the half-equation on Hydrogen since it is always supposed to be +1 so it is a spectator ion?

Sry but I'm really unsure of these redox reaction questions. Thanks for the help.

 

[danago]

To find what is being oxidised (and hence the reducing agent) and what is being reduced (and hence the oxidising agent), you could assign oxidation numbers to each species involved. Then note that when something is oxidised, its oxidation number increases, and when it is reduced, its oxidation number decreases.

 

[DarylMBCP]

you could assign oxidation numbers to each species involved.

When you say species, do u mean compounds? All the compounds presented in the chemical equation have neutral charges so how do I find the oxidation numbers of the compounds?

when something is oxidised, its oxidation number increases, and when it is reduced, its oxidation number decreases.

However, for the oxidizing agent to be reduced by the reducing agent and the reducing agent be oxidized by the oxidizing agent, shldn't there be two compounds for the reactants since there are two agents?

Thanks for the help.

 

[Borek]

Oxidation numbers are assigned to indivdual atoms.

In this case the same substance acts as reducing and oxidizing agent, it is called disproportionation.

There is no H⁺ here.

 

[DarylMBCP]

Oh, so N₂H₄ is both agents? Then how am I to write the half-equation for Nitrogen, since there are two compounds that contain Nitrogen? Is there one for NH₃ and one for N₂?

 

[DarylMBCP]

Oh ok, I think I got it, are the half-equations;

N₂H₄ + e^- --> NH₃

&

N₂H₄ --> N₂ + 4e^- ?

Sry but I'm quite new in writing half-equations. Btw, for half-equations, must I balance them to make sure that the number of atoms is the same on both sides (the first half-equation)? Can I also leave the spectator ions in half-equations (like the H₄ in the second half-equation)?

 

[Borek]

Half equations are just ike any other equations - they should be balanced in terms of atoms first, then you finish balancing adding electrons to balance charge.

Here you will need some artficial tricks to balance half-equations, as there is no source of hydrogen. I don't think it makes sense to try, this is realitvely easy to balance by inspection.

 

[DarylMBCP]

Hmm, ok, does this mean tht I omit the H₄ from N₂H₄ in the second equation? Sry, pls be clearer in your explanation.

 

[Borek]

I didn't want to show it, as I don't think it makes sense, but if you insist... Just don't tell anyone

OK, let's assume it all happens in water. This is a stupid assumption, but helpfull. Besides, it won't change the final result.

Let's start with N₂H₄ -> NH₃ part. Obviously we need some hydrogen to balance. In water that means add water on the left and OH^- on the right. That gives first half reaction:

N₂H₄ + H₂O -> NH₃ + OH^-

(this has to be balanced in terms of atoms, then you have to add electrons to balance charge).

Second half reaction is N₂H₄ -> N₂. We have excess hydrogen here, so let's try it as

N₂H₄ -> N₂ + H⁺

(again, balance atoms, balance charge).

Now, combine both half reactions so that electrons balance. Do you see what should be the next logical step?

Edit: it can be also balanced without water/oxygen, just by adding H⁺ wherever you need more hydrogen. It is about as artficial as the water approach, but can be easier to swallow.

 

[DarylMBCP]

K, thnks for the help. It's really much appreciated.