00:01
So we have multiple redox reactions to balance for this question.
00:04
The first thing that we need to do is we need to figure out the charges on each atom in each ion or molecule.
00:11
So we start with the negative atom.
00:14
So in this case, in the so3, we'll start with the oxygen.
00:17
Each o has a charge of negative two.
00:20
There are three of them, so that's a total of negative six.
00:24
This whole ion has a charge of negative two.
00:26
So the sulfur in this case has to have a charge of plus four.
00:30
So plus four minus six equals minus two.
00:33
Same deal over here for permanganate, negative two.
00:36
There are four of them, so that's negative eight.
00:38
It equals negative one.
00:39
That means the manganese is a plus seven.
00:43
For the sulfate, same deal.
00:46
This equals negative two, so our sulfur is positive six.
00:50
And here we're told that the charge on this is plus two.
00:53
I'm just going to write it above.
00:54
Okay, so we'll start with what is being reduced.
00:58
So the thing that is reduced is also our oxidizing agent.
01:04
So it's important not to be confused by that because one would think, oh, if it's being reduced, it's the reducing agent.
01:14
Okay, think of it like this.
01:16
When you're buying real estate, the real estate agent is not the person buying it.
01:21
They are helping the transaction to occur.
01:24
Same deal with this.
01:25
The thing being reduced is not the reducing agent.
01:30
It has a reducing agent that helps it to be reduced.
01:34
So the thing that is being reduced has its charge going down, and that is the manganese in the permanganate.
01:41
So our oxidizing agent is actually the mno4 minus.
01:46
Okay, so we're answering that part of it.
01:49
So we will start with our mno4 minus goes to mn plus two.
01:57
So we have – i'm just going to put the charge above it.
02:02
The first thing we need to do is make sure that we have the same number of manganese on each side.
02:07
We do, so now we have to balance the charge.
02:09
So we will add five electrons to the left, and we are told to balance this in acidic solution so we can add hydrogen ions and water to help balance it.
02:20
So i need to add four waters because i need four oxygens, and that gives me eight hydrogen ions that i need to add.
02:28
I need to double check that my charge is equal on each side.
02:32
So i have positive eight from the hydrogen ions minus five electrons is positive three, minus one on the permanganate is positive two.
02:40
So total on the left side is positive two.
02:42
Total on the right side is positive two.
02:44
The thing that is being oxidized is also our reducing agent, and that is the so3 2 minus.
02:53
So it's the other reactant.
02:55
So here we have the so3 2 minus going to so4 2 minus.
03:03
The charges are four and six.
03:08
So we have the same number of sulfurs on each side.
03:11
Now we have to fix their charge.
03:13
So we'll add two electrons to the right.
03:15
Then we need to add one water to the left side to balance our oxygens, and then two hydrogen ions to the right to balance the hydrogens.
03:25
The next thing we need to do is cancel out our electrons.
03:29
The smallest number that two and five go into evenly is ten.
03:34
So i need to multiply the top by two and the bottom by five.
03:39
I'm going to drag everything.
03:41
Well, no, i'm going to cancel out the electrons.
03:43
They are gone.
03:44
I'm going to drag everything down for this first one, and then we can reduce later after that.
03:50
So we have 16h plus, plus 2mno4 minus, plus 5h2o, plus 5so3 2 minus, goes to 2mn2 plus, plus 4h2o, plus 5so4 2 minus, plus 10h plus.
04:19
So you'll notice here now that we have some things on each side that will cancel out.
04:24
We have h pluses and we have h2os.
04:28
So there are 16 h pluses on the left and 10 on the right.
04:32
So all of these are going to be gone and we'll be left with just six of them.
04:38
And then we have five waters on the left and four waters on the right.
04:43
So we will get rid of all of those on the right and have just one left on the left.
04:50
And so then this would be your balanced reaction.
04:54
Okay, the next problem.
04:56
Oops, got to move this up.
04:57
We have to first figure out what our charges are on each atom.
05:08
So o is negative 2.
05:10
There are three of them.
05:11
That's negative 6 equals negative 2.
05:14
That means that this s has a total of positive 4.
05:18
So each s is a positive 2 because there are two of them.
05:21
When something's by itself, its overall charge is zero.
05:24
Here, total charge on o is negative 8.
05:27
This equals negative 2.
05:29
So sulfur is a positive 6.
05:31
And then the chloride, we are told, is a negative 1.
05:38
So the sulfur goes from 2 to 6.
05:41
The chlorine goes from 0 to negative 1.
05:44
So for our reduction half reaction, it is the cl2 going to cl minus...