How do I balance an equation using the oxidation number method?
When balancing redox reactions we go through several steps.
1) Find half reactions
2) Balance half reactions in both mass and charge.
3) Adjust have reactions so that electrons will cancel.
4) Add half reaction together
1) To find the half reactions we look at the reactants and products. Sn is obviously forming SnO2, and HNO3 is forming the ...
It shows how to balance an equation using the oxidation number method. The solution is detailed and well presented.
Balancing redox equations by the oxidation number method
2KMnO4 + 5H2O2 + 6HCl --> 5O2 + 2MnCl2 + 8H2O + 2KCl
I need to calculate the balanced NET IONIC equation for the above reaction.
I've been using the oxidation number method and determined H2O2 is the reducing agent and KMnO4 is the oxidizing agent.
Mn gained 5 electrons, but I'm not sure how many electrons Oxygen gained, I'm thinking one electron but the two different oxygen oxidation numbers in the products are throwing me off. I know I need to make the electrons lost equal the electrons gained. My problem is that I'm not sure about the number of oxygen electrons gained and then I'm not sure on which compounds containing oxygen I should apply the coefficient to.
Am I even using the right approach?View Full Posting Details