Balancing Gen Chem
Started by DoCt0rPeTe
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.
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.
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.
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.
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?
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?
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.
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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.
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