This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.
4.07 Oxidation-reduction reactions
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Del curso dictado por University of Kentucky
Chemistry
598 calificaciones
De la lección
Week 4
We will explore how compounds react with one another to form new substances and then write balanced chemical equations to represent what is happening in a reaction. We will explore several different types of reactions including precipitation, acid-base, oxidation-reduction, and combustion reaction.
Conoce a los instructores
Dr. Allison SoultLecturer
Chemistry
Dr. Kim WoodrumSr. Lecturer
Chemistry
In this module,
we're going to look at oxidation-reduction reactions often called redox.
By the end of this module, you should be able to identify both the oxidized and
reduced species in a chemical reaction.
So when we look at acid-base reactions,
we're frequently looking at proton transfer, or
H+ transfer between two species, either a proton donor or a proton acceptor.
When we look at redox reactions, or oxidation-reduction reactions, we're
looking at the transfer of electrons, and when we change the number of electrons
on a particular substance, we see a change in what we call the oxidation number.
And so, we want to talk about how we assign oxidation numbers to species, and
what that tells us about those substances in the reaction.
One important thing to note is that oxidation number is merely a bookkeeping
tool for us to use to be able to determine what is oxidized and what is reduced.
So this is an example of a redox reaction.
When we have a redux reaction we have to have two halves.
We have to have one oxidation, and we have to have at least one reduction.
And so, we can't have two oxidations or two reductions, or just an oxidation or
just a reduction.
They have to happen in that pair.
When we have an oxidation, what we see is that we have something like magnesium,
going to Magnesium 2+, it has lost electrons.
And if we have a reduction we have something that is gaining electrons.
So we have something going from O2 to say O2 minus,
where it has gained those electrons.
And what we can do is then take these two half reactions, and
figure out what the total reaction is, which in
this case is magnesium plus oxygen, yielding magnesium ions and oxide ions.
And so this is one of the reactions that was used in old flashbulbs,
to provide a source of light when taking photos.
Now that we see our electrons ar ethe same on both sides,
we can actually cancel them out.
So, our overall reaction won't actually show those electrons, and
we can rewrite it to look like magnesium plus oxygen yielding magnesium oxide, but
we still need to be able to divide it up to determine what those number of
electrons are to make sure those are balanced as well as all the elements.
We won't get into balancing redox reactions in this course, but
we will look at how we identify what's being oxidized and what's being reduced.
So when we're looking at redox reactions,
we have to be able to show the two half reactions.
Each half reaction explicitly shows the electrons involved in
the overall reaction.
When we get to the overall reaction,
the number of electrons on the left should be equal to the number on the right, so
the electrons don't appear in the overall reaction.
And every time we look at a redox reaction,
we have to have one oxidation and one reduction.
When we're looking at redox reactions,
sometimes it's a challenge to keep track of which one is which.
And so here are a couple of pneumonics that can help you remember remember what's
going on in this processes.
The first one is called OILRIG, so
oxidation is loss of electrons, reduction is gain of electrons.
Another option is LEO the lion says GER, so
lose electrons oxidation, gain electrons reduction.
Mean the same thing,
it's just a little mnemonic to help us remember what we're looking at.
Before we can determine that what species is being oxidized and reduced, and
write our oxidation and
reduction half reactions, we first have to assign oxidation numbers.
And remember I said that oxidation numbers are merely a means of
bookkeeping of keeping track of our electrons.
And what we see is that the oxidation number is
the number of charges the atom would have in a molecule or
ionic compound if the electrons were transferred completely.
And when we're looking at an ionic compound,
the electrons are transferred completely.
The charge on the ion is the same as the oxidation number.
However, for molecular compounds that are covalent bonds.
What we see is the electrons are being shared.
And so we basically have to pretend for
a moment that instead of being shared, the electronics are transferred completely.
They will be able to assign oxidation numbers to every substance, and
as a result determine our reduction and oxidation half reaction.
So let's go through the rules that we use to determine the oxidation numbers of
a species.
The first one is this,
probably the simplest, free elements with no charge have an oxidation number of 0.
This includes both monatomic and diatomic elements, but
any element that doesn't explicitly show a charge will have an oxidation number of 0.
When I see something that already has a charge, some monatomic ion,
what I see is that the ionic charge is equal to the oxidation number.
So for lithium plus 1, we have an oxidation number of plus 1.
For fluoride, we have oxidation number of minus 1.
For titanium our oxidation number is 4.
So, those are also fairly easy to identify.
The thing we have to remember there is are we shown the charge on the substance.
So for lithium, if it doesn't show a charge,
we know that that is a neutral atom and will have an oxidation number of 0.
When we get to oxygen, we see that it usually has an oxidation number of
minus 2, which is the same as the charge it would have in an ionic compound.
Even if we have an oxygen in a molecular compound,
it's oxidation number is generally minus 2.
There are occasions where we have the oxidation of minus one for oxygen.
So something like H2O2 is a peroxide, or
a compound involving our peroxide ion, O2, 2 minus.
In that case, the oxidation number will be minus 1.
For hydrogen, the oxidation number is typically plus 1.
Things like H2O, HCL, CH4, all of these has an oxidation number of plus 1.
If it is paired with a metal compound, lithium, calcium, sodium,
magnesium, then the hydrogen will actually have an oxidation number of minus 1.
We don't see these binary metal compounds or metal hydrides very often.
For fluorine, it will always have an oxidation number of minus 1.
And for other halides, they can have a value of minus 1 or
they can also have positive values.
So if I'm looking at something like HCl,
than chlorine has an oxidation number of minus one.
However, if I'm looking at something like KClO3 or HIO4, basically are acids formed
from our polyatomic ions or the derivative salts from those polyatomic ions,
then my halogen can have a different oxidation number.
And what I'm going to have to do is determine the oxidation number of
the halogen, based on the oxidation number of the other elements.
So a few of the rules that apply when we're assigning oxidation numbers to
elements was the free element or an ion or in a compound.
For a neutral molecule the sum of the oxidation states must be equal to 0.
So if we're looking at something like sodium chloride,
sodium has an oxidation number of plus 1,
chloride has an oxidation number of minus 1, plus 1, minus 1 equals 0.
If we're looking at an ion, such as a polyatomic ion, then what we know is
the sum of the oxidation states must be equal to the net charge of the ion.
So in this case, I'm looking at permanganate ion.
I see I have a minus 1 charge, so I know that tells me that I'm
the sum of my oxidation numbers will be equal to minus 1.
I know that each oxygen has an oxidation state of minus 2.
So, here I have 4 times minus 2 to represent those four oxygens.
We still report the oxidation state as belonging to a single atom of
that element, so we still say the oxidation number is minus 2.
But it contributes a minus 8 to the overall charge of that polyatomic ion.
And then we look at what's left, and we see that manganese must be a plus 7, so
that the rest of the statement is equal to minus 1.
Now what the, most of the time,
our oxidation numbers are going to be equal to integers or whole numbers.
And this is actually an example we did look at in our chart of rules
because typically oxygen is a minus 2 oxidation number, or a minus 1.
On very rare occasions, we can have a fractional oxidation number,
such as this one for O2 minus, which have an oxidation number of minus one half.
Let's look at some examples determining oxidation numbers for
atoms in these compounds and ions.
The first thing I'm going to notice about the substance is,
what is the overall charge, is it neutral or does it have a charge?
Because that's going to tell me some information about the sum of
the oxidation numbers.
For SO2, I see that the su,
the charge on the molecule is neutral, so I know that it's going to be equal to 0.
When I look at my oxygen know that oxygen is going to have an oxidation number of
minus 2, and that means I'm going to have x for my oxidation number of sulfur.
So I have x plus 2,
because there are two oxygen, times the oxidation number of minus 2.
So I have x minus 4 equals 0, and x equals 4.
So I was able to solve for
the value of x, and solve for our oxidation of sulfur as plus 4.
When I look at my next example,
the dichromate ion, I notice that I have a charge of minus 2, so
I know that the sum of my oxidation numbers is going to be equal to minus 2.
Here, we have chromium,
which we didn't have any rules that apply to the oxidation of chromium.
So I'm going to call it x.
When I look at oxygen, it's a minus 2 oxidation number for each of them.
So I can say I have 2, from our subscript dichromium, times x
because that's the oxidation number of chromium, plus 7 times minus 2.
Because, I have seven oxygens, each with an oxidation number minus 2,
and that's going to be equal to a total of -2.
So I have 2x minus 14 equals minus2.
I add 14 to both sides, I get 2x equals 12, and x equals 6.
So now, I add the oxidation number of chromium is equal to plus 6.
I can go back and check my work.
I have 2 times plus 6, gives me 12.
Minus 2 times 7 gives me minus 14.
12 minus 14 equals minus 2, so I know I've set my oxidation numbers up correctly.
One more example where for our oxidation is positive.
We have out polyatomic cation here.
We know that we have a plus charge there, so
I know that the sum of my oxidation numbers is going to be equal to plus 1.
When I look at my nitrogen,
I may not know what the oxidation number's going to be there, but
I do know that the oxidation number on hydrogen is going to be plus one.
Now, typically when hydrogen has a positive oxidation number,
it is written first.
Amino compounds,
these are NHx compounds, are an example of where they are not written first.
And so, remember, the only time hydrogen is going to be a minus 1 is when
it's paired with one of these metals, so
it's written second, but it's still a positive 1 value.
And I'm going to assign x as my oxidation number for nitrogen.
So we have x plus 4 times plus 1.
So we have x plus 4 equals 1 or x equals minus 3.
So now we have an oxidation number of minus three for nitrogen plus four times
plus one of our hydrogen, which gives us a total of plus one, which is the overall
oxidation number of the molecule, or the overall charge on the molecule.
Now, let's set an example to let you assign the oxidation number for
chlorine in perchlroric acid, HCLO4, our correct answer is 7.
Remember, when we're looking at our compound we want to go through and
assign the oxidation numbers that we know.
Hydrogen would be plus 1, chlorine we don't know, so we're going to call it x.
And Oxygen's going to have an oxidation number of minus 2.
Because this is a neutral compound,
the sum of our oxidation number must be equal to 0.
So we have 1 from our hydrogen, plus x and
we only have one chlorine so we don't need a coefficient in front of the x.
Plus 4 times minus 2 equal to 0.
So we have 1 plus x minus 8 equals 0.
X minus 7 equals 0, or x equals 7.
So we have a plus 7 for the oxidation state of chlorine.
When I go back and look at my problem, I see I have plus 1
plus 7 minus 4 times 2, and that in fact does equal 0.
So, I know I've done the calculation correctly.
Now let's look at an example and define which species, in being oxidized and
which species is being reduced.
So here we have iron plus copper ion, yielding iron ion plus copper solid.
Now we can go through and assign our oxidation states.
We have a free element, so our oxidation state will be 0 for iron.
For our ions, it's going to be the same as the charge.
So we have plus 2 for our oxidation state.
For iron, we also have a plus 2.
And for copper over here, we have a 0.
Now what we want to look at is how these substances change.
How does iron change as it goes from left to right, and how does the copper change?
So let's look first at our iron, and what we see is that
as we go from iron to iron 2 plus, that we've actually lost electrons, and
so that tells us we're dealing with an oxidation process.
When I go from copper to plus to copper, we're gaining electrons because we're
gaining negative charge, and the oxidation number goes from plus 2 to 0.
So our charge is going down, our oxidation number is going down,
we're gaining electrons, and that's considered to be a reduction.
Now we know the process, it's undergoing oxidation.
We know that iron is being oxidized, and copper is being reduced.
There's another type of terminology that we also use,
when we're looking at redox reaction.
And that is reducing agent, okay, which is the substance that's being oxidized.
Because it's the agent that's causing someone else to be reduced.
Then we have the oxidizing agent, because it's its species being reduced, and
it's causing someone else to be oxidized.
So be careful in assigning those terms to a particular substance in an equation.
So when we look at redox reactions,
there are actually kind of some subtypes that are all redox reactions.
And it doesn't mean that every time we see these reactions they're always redox, but
just to cover a few of them so you're familiar with them.
One is a combination, two or
more substances combine to form a single product.
So if we have something such as Na plus Cl2 going to NaCl,
that would be a combination reaction.
We can have decomposition reactions,
where a substance is breaking down into its components.
We can have a displacement reaction where an ion and
a compound is replaced by an ion or an atom of another element.
So we frequently see this when we look at the reaction of a metal and an acid for
example, and what we get on the other side is zinc chloride plus hydrogen gas.
All right, so that would be a displacement reaction, and
then we have a disproportionation reaction, it's an element in
one oxidation state, which is simultaneously oxidized and reduced.
In this reaction, we're going to look at what happens when we combine silver
nitrate and solid copper.
This is an oxidation reduction reaction,
which is also known as a redox reaction, because we have the transfer of electrons.
So in our test tube we have a piece of copper wire, and
we're going to pour in some silver nitrate to cover that wire.
And what we're going to see,
is that the copper that's in the wire is actually going to go into solution.
And the silver nitrate that's in solution is actually going to form solid silver and
plate onto the copper wire.
As we can see, the solution is turning blue, and that is from the formation of
the copper two ions that have been oxidized from the copper 0 state.
And we also see the formation of solid silver which happens from the reduction
from the silver plus 1 ion to solid silver which has a 0 oxidation state.
In the next module, we're going to look at Combustion Reactions,
which is one type of a redox reaction.
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