Balancing Gen Chem

[DoCt0rPeTe]

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Having a hard time balancing this quickly... anyone have an efficient method?

KMnO4 + NH3 ----> KNO3 + MnO2 + KOH + H2O

 

[FSUpredent]

These are a pain, but here is the half reaction method.

http://chemistry.about.com/od/generalchemistry/ss/redoxbal.htm

It takes forever to do these, and I didn't get one on the real DAT. In fact, I haven't heard of anyone having to fully balance one of these, but who knows? A more likely question would be to name the oxidizing agent or the reductant. Something along those lines.

 

[DQLEUCD]

Having a hard time balancing this quickly... anyone have an efficient method?

KMnO4 + NH3 ----> KNO3 + MnO2 + KOH + H2O

1. Split into half reaction
2. balance everything except for O and H
3. Balance the O's by adding water
4. Balance the H's by adding H+ ions
5. Balance the charge by adding electrons
6. Multiply the 2 half reactions with a coefficient that will cancel out the electrons
7. add them up and your done. This is if it's in an acidic solution

If in a basic solution, you would add OH- to balance out the H+ which you'll get water.

 

[DoCt0rPeTe]

but what are the two half rxns?

 

[DQLEUCD]

but what are the two half rxns?

Whats the answer? if you multiply the reactants by & and products by 4 it'll balance out. I believe the half reactions are their forms as ions. Therefore you would need to know which state they are in in order to cancel the spectator ions out.

 

[dentalml]

The answer is 8KMnO4 + 3NH3 ----> 3KNO3 + 8MnO2 + 5KOH + 2H2O

I don't think you can use the half reaction method....how would you balance the potassium on the left for NH3 --> KNO3 ?
Is there a faster way to balance this other than trying numbers and checking?

 

[ZanP]

What I did is the following:

Half Reaction 1 :
KMnO4 + NH3 ----> KNO3 + MnO2

Add H2O on the left side and then 5 H+ on the right, and then 5e on the right :

H2O + KMnO4 + NH3 ----> KNO3 + MnO2 + 5H+ + 5e-

Half Reaction 2 :
KMnO4 ----> MnO2 + KOH

Add H2O on the right side, then 3H+ on the left, and then 3e- on the left :

3e- + 3H+ + KMnO4 -----> MnO2 + KOH + H2O

Half Reaction 1 x 3
Half Reaction 2 x 5

3H2O + 3KMnO4 + 3NH3 ----> 3KNO3 + 3MnO2 + 15H+ + 15e-
15e- + 15H+ + 5KMnO4 -----> 5MnO2 + 5KOH + 5H2O

Then of course, add the equations:

3H2O + 8KMnO4 + 3NH3 + 15e- + 15H+ -----> 3 KNO3 + 8MnO2 + 5KOH + 5H2O + 15e- + 15H+

Net :
8KMnO4 + 3NH3 ----> 3KNO3 + 8MnO2 + 5KOH + 2H2O

I don't even know if you are allowed to do that reaction split up like i did, but if this was the test, that is exactly what I would have done.

..

 

[dentalml]

thanks for the explanation
makes perfect sense!