Oxidation numbers- What they are and how they are calculated.
 
An oxidation number is a way to keep track of electrons in redox reactions.  In a redox, the oxidizing agent is reduced and the reducing agent is oxidized. An easier way to remember this is by using the pneumonic OILRIG. This stands for oxidation is loss; reduction is gain.
 
How do I calculate oxidation numbers for Na2SO4 and MnO4-?
In order to follow along with my explanation, please refer to you chemistry book's four oxidation rules. The sulfate, SO4, is a compound. This makes the oxidation state neutral, so it's oxidation number is zero. The oxidation number for sodium is  +1 and it is -2 for oxygen. Since we do not know the oxidation number for S, this is what we will solve for in this problem.
There are two sodium atoms, so the number two is multiplied times sodium's oxidation state. As previously stated, we do know the oxidation number for S. For this reason, we will just call it's oxidation number S for now. The problem contains four oxygen atoms. We should multiply four times oxygen's oxidation state, which is -2. The entire problem will be set equal to zero, since S is part of a neutral species.
0=2(+1)+S+4(-2)
Solving this, we get -6.
-6+ ?= 0
0+-6= +6
The final answer is +6.
   
For the second problem, we already know what the oxidation number for oxygen. There are four oxygen atoms. We will multiply four times oxygen's oxidation state, -2. The oxidation state for Mn is what we want to find. The sum of the oxidations numbers in this problem is -1 because MnO4- has a - charge.
-1=Mn+4(-2)
Solving this, we get -8.
-8+?= -1
-8+1= +7
The final answer is +7.
  

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