Redox Reactions: Identifying Agents & Writing Half-Eqns

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In summary, the first equation shows that 3N2H4 is reduced to 4NH3 and N2. The second equation shows that N2H4 is oxidized to N2 and H+
  • #1
DarylMBCP
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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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
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)?
 
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  • #7
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.
 
  • #8
Hmm, ok, does this mean tht I omit the H4 from N2H4 in the second equation? Sry, pls be clearer in your explanation.
 
  • #9
I didn't want to show it, as I don't think it makes sense, but if you insist... Just don't tell anyone :wink:

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.
 
  • #10
K, thnks for the help. It's really much appreciated.
 

Related to Redox Reactions: Identifying Agents & Writing Half-Eqns

What are redox reactions?

Redox reactions, short for reduction-oxidation reactions, are chemical reactions that involve the transfer of electrons between molecules. In these reactions, one molecule gains electrons (reduction) while another molecule loses electrons (oxidation).

How do you identify the oxidizing and reducing agents in a redox reaction?

The oxidizing agent is the molecule that causes another molecule to lose electrons and undergo oxidation. It is typically written on the left side of the equation in a redox reaction. The reducing agent is the molecule that causes another molecule to gain electrons and undergo reduction. It is typically written on the right side of the equation.

What is the purpose of writing half-equations in redox reactions?

Half-equations, also known as half-reactions, are used to show the transfer of electrons in a redox reaction. By separating the reaction into two parts (oxidation and reduction), it allows for a clearer understanding of the electron transfer process and helps identify the oxidizing and reducing agents.

How do you balance redox reactions?

To balance a redox reaction, you need to make sure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. This can be achieved by adjusting the coefficients of the reactants and products in the equation.

What are some real-life applications of redox reactions?

Redox reactions are involved in many important processes in our daily lives, such as combustion (burning of fuels), corrosion (rusting of metals), and photosynthesis (conversion of sunlight into energy by plants). They are also used in industrial processes, such as the production of batteries, and in wastewater treatment to remove pollutants.

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