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Topic: Balancing Redox Reaction  (Read 31264 times)

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Offline dipndots

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Balancing Redox Reaction
« on: October 02, 2007, 11:57:56 PM »
I learned how to balance simple redox reactions by doing two half equations. But this one is very confusing.


The following reaction occurs in basic solution. Balance it by adding only OH- or H2O. Enter your answer as the sum of the coefficients.


MnO4-(aq) + NO2-(aq) --> MnO2(s) + NO3-(aq)

Offline grignardgirl

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Re: Balancing Redox Reaction
« Reply #1 on: October 17, 2007, 12:36:49 AM »
Okay, so the half reactions start out as:

MnO4 + 3e- ---->  MnO2
NO2-  -----> NO3- + 2e-

The next thing you need to do is balance the Oxygens

MnO4 + 3e- ---->  MnO2 + 2H2O
H2O + NO2-  -----> NO3- + 2e-

Now you need to balance the Hs by adding H plusses to the other side (we will get rid of these later with the base)

4H+ + MnO4 + 3e- ---->  MnO2 + 2H2O
H2O + NO2-  -----> NO3- + 2e- + 2H+

Since the Electrons don't balance, we need to multiply to get a common denominator

2(4H+ + MnO4 + 3e- ---->  MnO2 + 2H2O)
3(H2O + NO2-  -----> NO3- + 2e- + 2H+)

Now we can put the equations together and cancel the electrons out

8H+ + 2MnO4 + 6e- + 3H2O + 3NO2-  ----->  2MnO2 + 4H2O + 3NO3- + 6e- + 6 H+


Now we need to cancel out the H+ with -OH.  We should add 8 because that will cancel out all of the -OH on both sides (2 are left over on the right side).  You need to make sure that you add it to both sides :)  When the 2 combine, it makes H2O

8H2O + 2MnO4 + 3H2O +3NO2-   ----> 2MnO2 + 4H2O + 3NO3- + 6H2O + 2 OH-

Now the last thing you need to do is cancel out the extra H2Os and you're done!

H2O + 2MnO4 + 3NO2-  ----> 2MnO2 + 4H2O + 3NO3- + 2 OH-
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Offline Borek

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Re: Balancing Redox Reaction
« Reply #2 on: October 17, 2007, 04:17:01 AM »
Okay, so the half reactions start out as:

MnO4 + 3e- ---->  MnO2
NO2-  -----> NO3- + 2e-

Wrong - you don't know number of electrons at this stage. You use electrons to balance charge AFTER oxygen and hydrogen is balanced. Also you have missed some charges (MnO4-) in the equations you have wrote.

See half reaction method for detailed method explanation.

Quote
H2O + 2MnO4 + 3NO2-  ----> 2MnO2 + 4H2O + 3NO3- + 2 OH-

That's hardly balanced - you have water on both sides. And canceling it won't help, it will be still not balanced.
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Offline Sev

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Re: Balancing Redox Reaction
« Reply #3 on: October 17, 2007, 04:53:51 AM »
Quote
8H2O + 2MnO4 + 3H2O +3NO2-   ----> 2MnO2 + 4H2O + 3NO3- + 6H2O + 2 OH-

...
H2O + 2MnO4 + 3NO2-  ----> 2MnO2 + 4H2O + 3NO3- + 2 OH-

You have forgotten to cancel 4H2O.

Offline grignardgirl

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Re: Balancing Redox Reaction
« Reply #4 on: October 18, 2007, 01:34:03 AM »
Oh, I'm sorry.  I meant to cancel out all but 1 water on the left side.

And the way I learned was adding the electrons first...
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Offline Sev

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Re: Balancing Redox Reaction
« Reply #5 on: October 18, 2007, 01:40:24 AM »
Quote
And the way I learned was adding the electrons first...

How do you know how many e-s to add?

Offline AWK

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Re: Balancing Redox Reaction
« Reply #6 on: October 18, 2007, 02:32:57 AM »
You should balance charges on both side of equation
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Offline Borek

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Re: Balancing Redox Reaction
« Reply #7 on: October 18, 2007, 02:54:07 AM »
You should balance charges on both side of equation

Sev asked how you can know how many electrons to add BEFORE balancing hydrogen and oxygen.

grignardgirl most likely uses oxidation numbers for that, so in reality she uses some mix of both methods, mix that won't work when she will not know what oxidation numbers to assign. Properly applied half reaction method will still work in such case.
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Offline Sev

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Re: Balancing Redox Reaction
« Reply #8 on: October 18, 2007, 05:00:18 AM »
Quote
grignardgirl most likely uses oxidation numbers for that, so in reality she uses some mix of both methods, mix that won't work when she will not know what oxidation numbers to assign. Properly applied half reaction method will still work in such case.

Oh, OK.  I understand now.  Thank you.

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