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### Balancing Redox Reactions

Posted: Sun Feb 07, 2016 8:50 pm
i am very lost trying to balance these equations. Can anyone give a simple method of trying to balance equations like these.

### Re: Balancing Redox Reactions

Posted: Sun Feb 07, 2016 9:15 pm
Hey, don't fear! :)

This is the method I was taught (I'll use an example, question 14.3b):

MnO4- + H2SO3 ---> Mn2+ + HSO4-

In MnO4- (oxidizing agent)
Mn = +7
O = -2

In H2SO3 (reducing agent)
H = +1
S = +4
O = -2

In Mn2+
Mn = 2+

In HSO4-
H = +1
S = +6
O = -2

When balancing half equations, use the following steps:
1. Balance the elements that undergo reduction/oxidation.
2. Balance the oxygen by adding H2O to the other side of the equation.
3. Balance the hydrogen by adding H+ to the side with less hydrogens.
4. Balance the charges by adding electrons to the more positive side of the equation.
5. Multiply each equation appropriately so that the electrons cancel out.

The half equations become:

MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2O [must be multiplied by 2]

H2SO3 + H2O ---> HSO4- + 3H+ + 2e- [must be multiplied by 5]

Full redox equation becomes:

2MnO4- + H+ + 5H2SO3 ---> 2Mn2+ + 3H2O + 5HSO4-