00:01
So in this video we're going to go over question 92 from chapter 4, which says balance the following oxidation reduction reactions that occur in acidic solution using the half reaction method.
00:09
So what's the half reaction method and why doesn't matter that we're an acidic solution? so let's go over this method.
00:17
So first we take our reaction and we split it into oxidation and reduction half reactions.
00:22
So the first thing we need to do is identify where oxidation is happening and where reduction is happening.
00:27
Once i have my two half reactions, i balance all of the elements other than oxygen and hydrogen.
00:34
And then once those are balanced, i balance oxygen by adding water molecules.
00:38
And then i balance hydrogen by adding protons because i'm in an acidic solution.
00:43
And then once i've done that, i probably have unbalanced charges, meaning the net charge in my reactants is not the same as the net charge on my products.
00:52
So i balance that out by adding electrons wherever my charge is uneven.
00:57
And then once i've done that, i can just take my two half reactions and add them together, making sure that the number of electrons that are in the products of my oxidation half reaction matches the number of electrons in my reactants in my reduction half reaction because they need to cancel out.
01:15
So now that we've gone over that, let's just dive right into part a.
01:19
So here's our reaction in part a.
01:21
Let's assign oxidation states to each of our reactants in products so that we can identify our oxidation.
01:27
And reduction half reactions.
01:30
So cus, the oxidation state on this is just zero because we have the elemental form, just the pure element copper.
01:38
And then in our nitrate ion, the charge is minus one.
01:41
All of our oxidation states need to add to equal minus one.
01:45
So for oxygen, since we're bonded to the less electronegative nitrogen atom, oxygen is going to take all of the electrons in the bond and have a minus two oxidation state.
01:55
Since there's three of them, that's minus six overall.
01:58
What plus minus six would give us minus one, the oxidation state on nitrogen must be plus five.
02:06
Then on the product side, any monotomic ion, the oxidation state is just the charge on that ion.
02:12
So we have plus two.
02:15
And then here, again, oxygen is bonded to less elector -negative nitrogen.
02:19
It takes all of the electrons minus two oxidation state, which means the oxidation state on nitrogen here is plus two.
02:26
So where am my oxidation states changing? well, one place they're changing is from copper to copper plus two.
02:33
So i have an increase in my oxidation state that corresponds to a loss of electrons, which is oxidation.
02:47
Where else are my oxidation state changing? well, it's the same on oxygen, but it's different on nitrogen.
02:53
So from nitrogen, i go from a plus five to a plus two oxidation state.
02:57
So that's a decrease in oxidation state that corresponds to a gain in electrons or reduction.
03:09
So now i have my two half reactions.
03:12
Let's write them out.
03:31
Okay.
03:33
So now i have my two half reactions.
03:35
So first we want to balance elements other than oxygen and hydrogen, except it seems like they're already balanced.
03:41
So that's done.
03:42
And then we want to balance oxygen atoms by adding water molecules.
03:46
So in my second half reaction, my reduction half reaction, i have three oxygen atoms on the left and only one on the right.
03:54
So i'm going to add two water molecules to my products.
03:59
And now i want to balance hydrogen atoms by adding protons.
04:02
So now i have four hydrogen atoms in my products and none in my reactants.
04:06
So i'm going to have plus four protons.
04:12
And now i need to balance charge.
04:14
So i have plus four minus one.
04:16
So that's plus three in my reactants.
04:19
And both of my products are neutral.
04:21
So i need three electrons over here.
04:29
And then for my other half reaction, i have a totally neutral reactant and a plus two sign on my product.
04:35
So i'm going to need two electrons in my products.
04:41
So now i have my balanced individual half reactions, but i have two electrons in my oxidation half reaction and three electrons in my reduction half reaction.
04:51
So i need the same number of electrons if they're going to cancel.
04:54
So the least common multiple of two and three is six.
04:57
I'm going to multiply my first equation by three to get six electrons.
05:01
And my second equation by two to get six electrons.
05:08
So that gives me three cus, forming three c .u plus two, plus six electrons.
05:21
And then my other half reaction is six electrons.
05:25
Let's see, i was multiplying by two.
05:26
So now i have eight protons plus two in 03 minus, yields two in no gas, plus four waters.
05:50
Is that correct? yes, that is correct.
05:53
Okay, so now i have the same number of electrons in both sides, so they cancel out, and i can go ahead and add my two half reactions.
06:01
And what i get is three solid copper plus eight protons, plus two nitrate ions, yields three, cu, plus two, plus four h2o.
06:40
So that's part a.
06:44
This is our balanced reaction for part a.
06:46
Let's move on to part b.
06:49
So in b we have cr 207, which has a minus 2 charge, plus cl minus, yield cr plus 3 plus cl2 gas.
07:12
So let's go ahead and assign oxidation states.
07:14
So in cr 207, i know that my oxidation states are going to have to add to equal that minus 2 charge.
07:20
So each of my oxygen atoms bonded to the less electronegative chromium atom has a minus 2 chart, has a minus 2 oxidation state.
07:28
Which means since i have seven of them, that's contributing minus 14 overall.
07:33
Minus 14 plus what equals minus 2, that would be plus 12.
07:37
And since there's two chromium atoms, they must each have plus 6.
07:43
And then in cl minus, i have a monotomic ion, which means my oxidation state is the same as the charge on the ion.
07:50
Same for my chromium plus 3.
07:52
And then co2 gas, my oxidation state is just zero because i have just chlorine bonded to itself, just the pure element.
08:00
So where are my oxidation states changing? well, one place they're changing is from chromium with a plus six oxidation state to chromium with a plus 3 oxidation state.
08:10
So plus 6 to plus 3, that's a decrease in oxidation state.
08:13
That would correspond to a gain in electrons or reduction.
08:17
Then i have my oxidation state changing from from cl minus to cl2 from minus 1 to 0.
08:31
So that's a gain of electrons, i mean that's a loss of electrons or oxidation.
08:44
So now we've identified our two half reactions.
08:47
Let's write them out.
08:52
So first we have our oxidation half reaction, which is cl minus, cl2.
09:03
And then we have the reduction half reaction.
09:05
Which is cr2 -07 to cr plus 3.
09:19
And that's the whole reaction.
09:21
Okay.
09:22
So first we want to balance atoms other than oxygen or hydrogen.
09:26
So in my first half reaction, the oxidation half reaction, i have two chlorine atoms over here and just one over here.
09:31
So i'm going to need a coefficient of two.
09:33
And then in my second half reaction, i have two chromium atoms here and just one over here.
09:38
So i need another coefficient of two.
09:41
Now we want to balance oxygen atoms by adding water molecules.
09:45
So here i have seven oxygen atoms, which means i need seven over here.
09:48
So i'm going to add 7 h2o.
09:55
But now i have 14 hydrogen atoms over here and no hydrogen atoms over here.
10:01
So i'm going to have to add 14 protons.
10:09
But now my charges are uneven.
10:11
So i have plus 3 on this side, and i have 14 minus 2 over here.
10:16
So that's plus 12 on this side.
10:18
So that's a difference of plus 9 towards my reactants.
10:22
So i'm going to need 9 electrons.
10:28
As for my other half reaction, i have a charge of minus 2 over here and no charge over here.
10:33
So i'm going to need 2 electrons in my products.
10:38
So i have 2 electrons in 1 half reaction and 9 in the other half reaction.
10:42
So i'm going to take my first half reaction and multiply it by 9 and my second half reaction and multiply it by 2.
10:50
So what do i get when i do this? well, i get 18 cl minus yields 9 cl2 plus 18 electrons.
11:07
And then for my other half reaction, i now have 18 electrons plus 28 protons plus 2cr207, and that's going to form 4cr3 plus plus 14 waters.
11:43
So now if i add these together, my electron.
11:46
Cancel from both sides of the reaction and what i'm left with is 18 cl minus plus 28 protons plus 2 cr 207 and that's going to form 9 cl2 plus 14 h2 and that's our reaction for part b so now moving on to part c so in c i have solid lead plus solid lead oxide plus h2s .o4 forms pbso4.
13:05
Okay, so let's go ahead and assign oxidation state.
13:08
So for solid lead, that's an oxidation state of zero.
13:11
That's just the pure element.
13:13
Then in pbo2, i have an oxidation state of minus two on each of my oxygen atoms.
13:19
There's two of them, so that's minus four.
13:20
So the oxidation state on lead must be plus four.
13:24
In h2s .4, the oxidation state on oxygen, is minus 2.
13:30
There's 4 of them, so that's minus 8 overall...