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Topic: Balancing redox rxn  (Read 4620 times)

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ygao85

  • Guest
Balancing redox rxn
« on: June 13, 2005, 10:15:39 PM »
The question is to balance the following eqn:

Br-(aq)+MnO4--->Br2(l)+Mn2+(aq)

My answer is:
2Br-(aq)+MnO4-+8H+-->Br2(l)+Mn2++4H2O

However the answer is:
16H++2MnO4-+10Br--->5Br2+2Mn2++8H2O

I tried to the this anwser but i couldn't. Can someone explain how to do it plz. Thank you.

arnyk

  • Guest
Re:Balancing redox rxn
« Reply #1 on: June 13, 2005, 10:25:34 PM »
Balance your charges.

yl88

  • Guest
Re:Balancing redox rxn
« Reply #2 on: June 15, 2005, 12:26:33 AM »
General procedure:
1. write oxidation and reduction as separate equations
2. balance everything OTHER THAN O and H
3. balance O and H by adding H2O and H+
4. balance charge (this is the final step for rxn in acidic sln.)
5. multiply each reaction with least common factor of # of electrons to cancel e- on overall rxn.
6. if rxn happens in basic sln, add OH- to the side that has H+, and combine OH- with H+ to form H2O
This question is in acidic condition according to your answer, therefore:
oxidation: (2Br- --> Br2 + 2e-) * 5
reduction: (MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O) * 2
overall rxn (add everthing up from the top two equations): 16H++2MnO4-+10Br--->5Br2+2Mn2++8H2O

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