00:08
Let's assign oxidation numbers to each of the following ions for a, a c -l -o -3 minus.
00:15
The oxidation number of chlorine will be our unknown.
00:21
Oxidation number of oxygen is minus two.
00:25
Set up an equation here.
00:27
X plus three minus two is equal to minus one.
00:32
Solving for x here is going to yield five.
00:35
Therefore, the oxidation number of chlorine will work out.
00:39
To plus 5.
00:41
For b, s .o3 -2 minus sulfur is unknown.
00:50
Oxygen has an oxidation number of minus 2.
00:53
X plus 3 minus 2 is equal to minus 2.
00:57
So helping for x here will yield 3 and, sorry, will yield 4.
01:06
Therefore the oxidation number of sulfur here will work out to plus 4.
01:10
For c, c2 .042 minus.
01:17
Oxidation number of carbon is unknown...