Balancing Redox reactions

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Balancing Redox reactions

Postby Michelle_Ahn_2N » Sun Feb 12, 2017 10:19 pm

When matching the numbers of oxygen and hydrogen, on which side do we put each of these,
H2O, OH, H ?
Is it correct that we put OH- on reactant side and H2O for product side for basic reactions
while we put H2O for reactant side and H+ on product side for acidic reactions?

Thanks in advance!

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Joined: Mon Jul 11, 2016 3:00 am

Re: Balancing Redox reactions

Postby Aishwarya_Natarajan_2F » Sun Feb 12, 2017 10:37 pm

It isn't so much about putting one on the reactant v. product side as it is making sure there are equal numbers of each on both sides.

When balancing redox reactions in basic solutions:
1) split the reaction up into its half reactions

For each half reaction:
2)balance all elements that do not include H and O first
3) to balance out the oxygen, add the amount of H2O necessary to the other side of the equation

example: Br- --> BrO3- can be balanced first by adding 3H2O to the left hand side
Br- + 3H2O --> BrO3

4) Now notice there is an imbalance in H with 6 on the reactant side and 0 on the right. To balance this, we will place twice as much H2O as needed on the right, and because this would unbalance the equation we are also adding the same # of moles of OH- to the left. Then we can cancel like species on both sides.

Br- + 3H2O + 6OH- ---> BrO3- + 6H2O
Br- + 6OH- ---> BrO3- + 3H2O

5) Notice nhow the number of O and H is the same on both sides of the reaction now. The last step is to balance out the charges by counting the total charges on both sides and adding electrons to balaance, and in this case we would add 6 electrons to the right to balance the charge.

6) repeat these steps for the other half reaction and sum the two.

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