00:00:00
in this video we'll look at everything
00:00:02
you need for experiment 6.
00:00:06
the first topic that we'll look at is a
00:00:09
self-oxidation reduction reaction this
00:00:11
is also called disproportionation and
00:00:15
overall what it means is that one
00:00:18
molecule will act as both the oxidizing
00:00:21
agent and the reducing agent so that you
00:00:24
get one product that goes one way and
00:00:26
one product that goes the other way
00:00:29
so remember that a redox reaction is a
00:00:32
tug of war for electrons
00:00:35
between the reducing agent and the
00:00:38
oxidizing agent so the reducing agent is
00:00:41
the thing that gets oxidized so it's
00:00:45
going to lose electrons
00:00:47
it starts out with the electron and then
00:00:51
whichever one wants that electron more
00:00:54
we'll end up with the electron
00:00:57
so our reducing agent gets oxidized by
00:01:00
losing electrons and its oxidation
00:01:04
number increases
00:01:06
our oxidizing agent gets reduced it
00:01:09
gains electrons and its oxidation number
00:01:12
decreases
00:01:15
when an atom is at its lowest oxidation
00:01:18
number it can only be a reducing agent
00:01:21
so what this means is that there is a
00:01:24
range of possible oxidation numbers for
00:01:27
every atom and if it has already given
00:01:32
up all of its electrons it doesn't have
00:01:35
any more to give up there's no way that
00:01:38
it could be oxidized
00:01:40
and when an atom is at its highest
00:01:43
oxidation number it can only be an
00:01:46
oxidizing agent so if an atom
00:01:51
has
00:01:52
no electrons it will want to get the
00:01:55
electrons so if something is at its
00:01:58
lowest oxidation number it cannot get
00:02:01
any more electrons all right
00:02:04
so some substances are in the middle
00:02:07
because there's a range of possible
00:02:09
oxidation numbers that an atom can have
00:02:12
and when it's in the middle it can act
00:02:15
as a reducing agent or an oxidizing
00:02:18
agent
00:02:19
so for example in hydrogen peroxide
00:02:23
which is H2O2 it has an oxidation number
00:02:28
of negative one
00:02:30
now the range of oxidation numbers for
00:02:34
oxygen the most negative that it can go
00:02:38
is negative two
00:02:40
and the most positive it can go is zero
00:02:45
so if you start with a negative one
00:02:48
oxidation number it could gain one
00:02:50
electron and be reduced
00:02:54
and a molecule in which oxygen has
00:02:58
negative two as an oxidation number
00:03:00
would be water
00:03:02
and if it loses an electron
00:03:05
the oxygen could have an oxidation
00:03:07
number of zero which it does when it is
00:03:10
molecular oxygen so in O2
00:03:14
so hydrogen peroxide is a molecule that
00:03:18
can disproportionate and it does so by
00:03:22
having one of the oxygens become
00:03:25
oxidized to O2 and one of the oxygens
00:03:29
become reduced to H2O
00:03:34
so this is a self-oxidation reduction
00:03:37
reaction
00:03:39
and the way that it works out once
00:03:42
everything is balanced is that two
00:03:44
molecules of hydrogen peroxide will turn
00:03:47
into two molecules of water and one
00:03:50
molecule of o2 and the hydrogen peroxide
00:03:53
that you buy at any drugstore
00:03:57
um that's why it's in that dark bottle
00:04:00
so that it does this a little bit less
00:04:03
readily because sunlight can sometimes
00:04:04
help with that but this is also why
00:04:07
hydrogen peroxide is great for treating
00:04:09
wounds the O2 the oxygen is uh pretty
00:04:15
good at killing bacteria
00:04:17
so again the oxygen going to in peroxide
00:04:21
going to O2 is oxidation the oxygen
00:04:25
going to negative 2 as it is in water is
00:04:29
reduction
00:04:31
okay so predicting products of redox
00:04:34
reactions it is as cool as it sounds and
00:04:38
it is not as difficult as it sounds
00:04:41
so we will give you all of the
00:04:43
information that you need for this to
00:04:45
predict the products in a redox reaction
00:04:47
it will be based on the chemical
00:04:50
properties of the reactants
00:04:52
and your observations in lab
00:04:56
so for example zinc metal zinc will turn
00:05:00
into zinc two plus plus two electrons
00:05:05
all right some things that you know
00:05:07
about need to know about zinc this will
00:05:10
happen in an acidic or a neutral medium
00:05:13
so
00:05:15
one of the things that you might look at
00:05:18
in your experimental conditions is was
00:05:23
there any acid present
00:05:25
or was their base present was it just in
00:05:28
water
00:05:29
well other things were in the solution
00:05:32
that would give you a clue about the
00:05:35
medium that it's in
00:05:37
so what about if it's in a basic medium
00:05:41
well we're going to tell you first of
00:05:43
all so zinc will turn into a zinc
00:05:46
hydroxide plus two electrons well this
00:05:50
is
00:05:51
this is still zinc two plus it's in a
00:05:54
basic medium
00:05:56
it's not
00:05:58
exactly the same the way that it's
00:06:01
listed in this reaction but in both of
00:06:04
these cases zinc is oxidized to zinc two
00:06:07
plus
00:06:08
the way that you know that zinc would be
00:06:12
oxidized to zinc 2 plus is from the
00:06:15
stuff that we did in experiment one
00:06:18
so in those naming tables you will see
00:06:21
the common charges which are also listed
00:06:24
as oxidation States and that will give
00:06:27
you a clue about whether or not a metal
00:06:30
can have more than one oxidation state
00:06:33
and what those oxidation states are for
00:06:37
zinc there's only one option of zinc two
00:06:39
plus so if zinc gets oxidized the only
00:06:43
thing it could turn into is a zinc two
00:06:45
plus now what about reduction
00:06:49
could zinc gain an electron to turn into
00:06:53
a zinc one minus
00:06:56
or a zinc two minus could have gained
00:06:58
two electrons
00:07:00
what do you think
00:07:04
definitely not so you will not see
00:07:06
Metals with negative oxidation States so
00:07:12
zinc cannot be an oxidizing agent
00:07:15
because it cannot be reduced
00:07:19
okay let's look at bromine so br2 is how
00:07:24
the element bromine exists
00:07:27
we could give it some electrons so we
00:07:31
could reduce bromine and turn it into
00:07:34
two BR minuses right the br2 has a zero
00:07:37
and then we get two bromines that each
00:07:40
have a negative one
00:07:41
this can happen
00:07:43
what about oxidation
00:07:46
could you take br2
00:07:48
remove some electrons and get a product
00:07:52
that has bromine in a higher oxidation
00:07:54
number
00:07:56
here again you you might wonder how
00:08:00
could I figure this out
00:08:02
you could look at things that have
00:08:04
bromine in it and check their oxidation
00:08:07
number and in this case I'm going to
00:08:09
tell you what those things are
00:08:11
which is all of these polyatomic ions
00:08:14
now you may not have seen the bromine
00:08:17
version written out we usually deal with
00:08:20
the chlorine version but this is
00:08:23
hypobromite bromite bromate and per
00:08:27
bromate it parallels perfectly what you
00:08:31
would see with
00:08:33
hypochlorite chloride chlorate and
00:08:36
perchlorate so check the oxidation
00:08:39
numbers for Bromine in each of these
00:08:42
molecules and you should see that this
00:08:46
could happen if br2 is oxidized you
00:08:49
should end up with a positive oxidation
00:08:51
number for those
00:08:54
now
00:08:56
which is it kind of has that similar
00:08:59
pattern to what you just saw in this
00:09:01
other polyatomic ion permanganate it's
00:09:04
kind of special we don't really see the
00:09:07
pattern that you see for the halogens we
00:09:10
really just see mno for minus as the
00:09:14
permanganate polyatomic ion now
00:09:17
permanganate has three very specific
00:09:21
things that it'll turn into
00:09:23
if you are in an acidic medium
00:09:26
medium just means
00:09:29
solution mixture whatever you have this
00:09:33
ion in for us it's going to be a
00:09:35
solution
00:09:36
if you're in an acidic solution
00:09:38
permanganate will turn into mn2 Plus
00:09:42
so check the oxidation number of
00:09:46
manganese in permanganate and then in
00:09:49
all of these products to see whether
00:09:50
it's been oxidized or reduced
00:09:53
if you are in a basic medium
00:09:56
permanganate will turn into mno2
00:09:59
which is also a solid so you in addition
00:10:03
to
00:10:04
knowing that the solution is basic
00:10:07
um some other things you're going to see
00:10:09
mno2 is a solid that would precipitate
00:10:13
and if you're in a very basic medium
00:10:16
will give you in the reading some
00:10:18
criteria for this off the top of my head
00:10:21
I think the pH has to be something like
00:10:24
14. so we mean extremely basic
00:10:27
and you will get mno4 2 minus
00:10:31
so watch that really closely because the
00:10:33
only thing we've changed is the charge
00:10:37
and the experimental observations will
00:10:40
help you determine the product so this
00:10:42
will be a case where you'll be able to
00:10:45
identify based on what you put into a
00:10:48
reaction whether the solution is acidic
00:10:51
basic or very basic
00:10:53
and
00:10:55
all of these things appear different
00:10:59
from each other so it should be obvious
00:11:02
and if you have any issues seeing colors
00:11:05
someone should tell you the color so
00:11:07
that it can be obvious that a reaction
00:11:10
has or has not happened
00:11:13
so permanganate is this magenta e
00:11:18
purpley
00:11:20
extremely vibrant color
00:11:24
mn2 plus is colorless
00:11:27
so The Disappearance of the purple color
00:11:31
would be your clue that you have formed
00:11:33
mn2 plus as long as you're in an acidic
00:11:36
solution
00:11:38
mno2 I mentioned that it's a solid it's
00:11:41
also this like brownish black color so
00:11:45
you will get a brown black precipitate
00:11:47
if you form mno2
00:11:50
and mno42 minus is green so our options
00:11:55
are bright magenta e purple colorless
00:11:59
black brown precipitate or green so
00:12:04
between the conditions is it acidic is
00:12:08
it basic and the color and appearance of
00:12:11
what is in your reaction you will be
00:12:13
able to predict the product of this
00:12:16
redox reaction
00:12:19
okay so balancing redox reactions we are
00:12:23
going to need it for this if you have
00:12:26
already done this in class this part
00:12:28
will be a review it was also in the
00:12:30
previous video but it's okay if you did
00:12:32
not watch it in the previous video
00:12:34
because it was not required for the last
00:12:36
experiment
00:12:39
for balancing a redox reaction I'm going
00:12:42
to give you these rules they're in order
00:12:45
they're in priority order so follow them
00:12:49
in order every time and you'll be good
00:12:51
to go so the first thing you need to do
00:12:54
is identify what is oxidized and what is
00:12:57
reduced you'll have to do this by
00:13:00
assigning oxidation numbers and looking
00:13:02
at what changed
00:13:05
then the oxidation part and the
00:13:08
reduction part get separated we call
00:13:11
this a half reaction because it only
00:13:13
shows half of the reaction and we're
00:13:15
going to balance those half reactions
00:13:18
separately from each other before we put
00:13:21
them back together
00:13:23
within a half reaction the first thing
00:13:25
you do is balance all of the elements
00:13:27
but not oxygen and hydrogen
00:13:31
of course there's always going to be one
00:13:33
of those caveats now if we have the
00:13:37
hydrogen peroxide that we talked about
00:13:39
earlier in the video as one of the
00:13:41
things in our reaction we're probably
00:13:44
going to have to deal with oxygen
00:13:46
toward the beginning of this but in all
00:13:49
other cases leave oxygen and hydrogen
00:13:51
for later otherwise you are immediately
00:13:54
setting yourself up for failure so
00:13:57
balance everything else
00:13:59
then balance the number of electrons
00:14:02
gained or lost
00:14:07
then you'll balance charges so you will
00:14:11
need to know if the solution is acidic
00:14:14
or basic if you are just being given a
00:14:17
reaction out of context you will be told
00:14:20
whether it's acidic or basic
00:14:22
if it's related to an experiment you're
00:14:24
doing you might have to look through the
00:14:26
steps to figure it out
00:14:29
and then the very last step is to
00:14:31
balance oxygen and hydrogen by adding
00:14:34
water
00:14:35
when you get to this step the number of
00:14:38
water molecules that you add should be
00:14:42
one
00:14:44
obvious or very easy to calculate
00:14:48
and two should balance perfectly
00:14:52
you should just be able to add a whole
00:14:55
number of water molecules to one of the
00:14:57
sides of the reaction to make it balance
00:15:00
and if you can't that means that
00:15:03
something went wrong in an earlier step
00:15:07
with the two half reactions now balanced
00:15:10
you will set it up so that the number of
00:15:12
electrons gained will cancel with the
00:15:15
number of electrons lost when you add
00:15:17
them together
00:15:19
and then you will add them together once
00:15:21
you add them together you may have water
00:15:25
on both sides and you'll need to do some
00:15:27
canceling you may have
00:15:30
h plus or o h minus on both sides and
00:15:33
need to do some canceling but you should
00:15:36
not see any of the electrons so this is
00:15:39
one of the things that's challenging
00:15:41
about balancing a redox reaction we
00:15:43
never show the electrons because they
00:15:46
never Exist by themselves in the
00:15:48
solution they get transferred from one
00:15:51
thing to another
00:15:54
all right let's do an example so in an
00:15:57
acidic solution we're going to balance
00:15:59
dichromate reacting with iron two plus
00:16:02
and the products are chromium 3 plus and
00:16:05
iron three plus
00:16:07
we call this the skeleton of the
00:16:10
reaction
00:16:11
or the skeleton equation it has the
00:16:15
major reactants in products but it is
00:16:17
not balanced it doesn't have all of the
00:16:19
other things that are going to make it
00:16:21
balance
00:16:23
so assign oxidation numbers chromium and
00:16:26
dichromate is a plus six and then
00:16:29
everything else was a monoatomic ion so
00:16:32
the charge in the oxidation number are
00:16:34
the same
00:16:36
my chromium goes from a plus six to a
00:16:40
plus three this will happen if it gains
00:16:43
electrons so my chromium is reduced
00:16:46
and iron lost electrons it's oxidized
00:16:52
so when we're talking about half
00:16:53
reactions
00:16:55
the oxidation half reaction just shows
00:16:59
what happens to iron it's the thing
00:17:02
that's oxidized
00:17:04
and my reduction half reaction
00:17:08
is dichromate turning into chromium 3
00:17:11
plus and the first part of balancing the
00:17:15
half reactions is also done on here I
00:17:18
have two chromiums on the left
00:17:21
I need to balance the chromium on the
00:17:23
right
00:17:24
forget the oxygen for now we'll come
00:17:26
back to it it'll get fixed up but I need
00:17:29
to know that my elements other than
00:17:32
oxygen and hydrogen are balanced so I
00:17:34
needed to in front of that chromium
00:17:37
feeling
00:17:39
that step the next step is to balance
00:17:41
electrons so iron goes from a plus two
00:17:44
to a plus three so it lost one electron
00:17:49
and when we set up chemical equations we
00:17:54
never subtract anything right I can't
00:17:56
write iron two plus minus one electron
00:17:59
gives me iron three plus
00:18:01
but I can
00:18:04
can imagine that and then write it in a
00:18:07
way that is positive so I will write
00:18:09
plus one electron on the reactant side
00:18:12
to show that my or on the product side
00:18:15
to show that my reactant kicked out one
00:18:18
electron
00:18:19
for chromium
00:18:21
I am going from my plus six to a plus
00:18:24
three
00:18:25
that's three electrons
00:18:28
but that is three electrons
00:18:31
per chromium
00:18:34
I have two chromiums that need to go
00:18:36
from a plus six to a plus three
00:18:38
so I need six electrons to be added to
00:18:43
my reactant
00:18:45
after this it is balancing not charges
00:18:48
so if we look at the oxidation reaction
00:18:53
on the left side the only thing I have
00:18:55
is iron two plus so the overall charge
00:18:59
on that side is plus two
00:19:01
on the product side I have an iron three
00:19:03
plus and an electron
00:19:06
if I have something with a three plus
00:19:07
charge and I add a negative one to it
00:19:11
that is an overall charge of plus two
00:19:16
so I have plus two on both sides that
00:19:19
matches I don't need to do anything to
00:19:21
balance the overall charges
00:19:23
for the reduction half reaction
00:19:26
dichromate is a negative two I've added
00:19:29
six electrons to it so this is a
00:19:32
negative eight total on the left
00:19:36
and on the right I have two things with
00:19:39
a plus three so I have a plus six charge
00:19:42
on the right and a negative eight on the
00:19:44
left not balanced
00:19:47
so I am in an acidic solution which
00:19:51
means that h plus is my only option for
00:19:55
balancing the charges in this half
00:19:58
reaction
00:19:59
which I can balance them if I add 14 h
00:20:03
plus it seems like a lot and it will
00:20:07
seem like a lot sometimes but
00:20:09
we only had h plus the only way to get
00:20:13
the same charge on both sides was to add
00:20:16
14 of them so I now have a plus sex
00:20:19
charge on both sides
00:20:22
and my last step is to balance oxygen
00:20:25
and hydrogen with water well I I have 14
00:20:30
H pluses that's only going to balance if
00:20:33
I can add seven water
00:20:35
and very conveniently
00:20:38
I have seven oxygens that need to be
00:20:41
balanced so if I had seven water
00:20:43
molecules to the right side it's perfect
00:20:47
my half reaction is now perfectly
00:20:49
balanced the number of waters I needed
00:20:51
to add just worked because the other
00:20:54
steps were done correctly
00:20:56
in order to add these together and have
00:20:59
my electrons cancel
00:21:03
I need to multiply the top reaction by
00:21:05
six this is what it means to have the
00:21:08
number gained equal the number lost
00:21:11
the electrons that are being given to
00:21:14
chromium have to come from Iron so I
00:21:18
need enough irons to give the right
00:21:20
number of electrons for the chromiums
00:21:23
that I have and vice versa so if I
00:21:27
multiply by 6 on the top reaction I'll
00:21:30
have six electrons in both of them and
00:21:33
they'll cancel each other out
00:21:35
so add these together and my final
00:21:39
result My overall is 14h plus plus one
00:21:43
dichromate plus six iron two plus will
00:21:47
turn into iron six iron three plus two
00:21:51
chromium three plus and seven water
00:21:54
molecules
00:21:57
let's do another example and this time
00:22:00
let's do something that's a
00:22:01
disproportionation so we have chlorine
00:22:05
turning into CL minus and clo3 minus it
00:22:10
is in an acidic solution
00:22:14
so putting oxidation numbers on here
00:22:17
chlorine and cl2 is a zero it's the
00:22:20
element Chlorine
00:22:22
CL minus is negative 1 and the chlorine
00:22:25
in clo3 minus is plus five
00:22:29
so chlorine is changing oxidation
00:22:32
numbers here
00:22:34
and the oxidation half reaction is that
00:22:39
cl2 is turning into clo3 minus right
00:22:43
going from a zero to a plus five that's
00:22:46
going to be losing electrons so let's
00:22:50
balance this half reaction first and
00:22:52
then look at the reduction side so
00:22:55
balancing this half reaction I need to
00:22:57
balance elements other than oxygen and
00:22:59
hydrogen which means my chlorines need
00:23:02
to be balanced and I need a 2 in front
00:23:05
of the clo3 minus
00:23:08
now each chlorine is going from a zero
00:23:12
to a plus five so that's five electrons
00:23:17
for each chlorine so it's kicking out a
00:23:21
total of 10 electrons
00:23:24
it's it's a lot of electrons but if our
00:23:27
count is right our count is right so
00:23:29
it's kicking out 10 electrons
00:23:32
the charge balancing in The Next Step
00:23:36
I'm in an acidic solution so I only have
00:23:39
h plus is available
00:23:41
on the left side My overall charge is
00:23:44
zero because all I have is that neutral
00:23:46
molecule
00:23:48
on the right side I have two things with
00:23:51
a negative one and then ten more things
00:23:53
that have a negative one I have a
00:23:55
negative 12 charge on the right side so
00:23:59
I need 12 h plus to balance this out
00:24:05
and now I have a zero charge on both
00:24:09
sides
00:24:10
and I have to balance this with water on
00:24:14
the other side well I have 12 H pluses
00:24:17
on the right if I make six water
00:24:19
molecules I know my H pluses will
00:24:21
balance
00:24:22
uh the question is do my oxygens balance
00:24:26
I have six oxygens on the left now
00:24:28
because of my water molecule
00:24:30
uh the clo3 minus has three oxygens
00:24:35
but I have two of them so I do have six
00:24:38
oxygens on the right so this is done
00:24:41
for the reduction half reaction it's
00:24:46
something turning into CL minus right
00:24:49
it's the other product
00:24:51
the reactant is also cl2 right so my cl2
00:24:56
has a zero oxidation number it is
00:25:00
disproportionating so this one molecule
00:25:03
I mean we'll right it will end up
00:25:05
reacting with another molecule but
00:25:07
there's always more than one molecule in
00:25:09
the solution
00:25:10
the cl2 is both being oxidized and being
00:25:16
reduced in this
00:25:19
reaction overall that we have
00:25:22
so the cl2 turning into CL minus
00:25:26
if you go through the steps to balance
00:25:28
it this is where you should end up cl2
00:25:31
plus two electrons turning into two CL
00:25:34
minus
00:25:35
so give this a try and make sure that
00:25:38
you end up in the same place remember
00:25:39
you're going to balance the chlorines
00:25:41
because it's not oxygen or hydrogen
00:25:44
then you're going to balance the
00:25:46
electrons and
00:25:48
once you do that I mean based on how
00:25:49
this turned out the charges will be the
00:25:52
same on both sides so you won't need to
00:25:54
add h plus
00:25:55
and there are no oxygens in here so we
00:25:57
don't need to balance anything out with
00:25:59
water molecules
00:26:01
before I add them together I do need to
00:26:04
make the number of electrons equal in
00:26:07
both which means that my reduction
00:26:09
reaction gets multiplied by 5.
00:26:12
this means my electrons will cancel when
00:26:14
I add it together
00:26:16
and my overall reaction is 6 H2O plus 6
00:26:21
cl2 turning into two clo3 minus
00:26:25
plus 12h plus plus 10 CL minus
00:26:30
now I don't have
00:26:32
h plus on both sides I don't have water
00:26:35
on both sides I don't have to worry
00:26:36
about that
00:26:38
but
00:26:40
all of my coefficients are divisible by
00:26:42
2. so you do still have to reduce the
00:26:48
coefficients in your reaction to get to
00:26:51
whole numbers
00:26:54
there is another example of
00:26:56
disproportionation in this experiment
00:26:59
and thiosulfate is the molecule that can
00:27:04
disproportionate
00:27:05
if you are in a highly acidic solution
00:27:09
it can turn into sulfur so Elemental
00:27:14
sulfur
00:27:15
just solid sulfur I know that many of
00:27:18
you know that there are different
00:27:20
combinations of sulfur that exist if you
00:27:23
need to write Elemental sulfur you can
00:27:25
just write the letter and then call it a
00:27:27
solid
00:27:28
and the other product that it turns into
00:27:32
is sulfurous acid
00:27:35
so I would take a minute here assign the
00:27:39
oxidation number to sulfur and
00:27:42
thiosulfate which is what this s203 two
00:27:45
minus is
00:27:46
the h plus is just telling you it's an
00:27:48
acidic solution obviously this is not
00:27:51
balanced but assign your oxidation
00:27:53
numbers in solid sulfur
00:27:56
and in the sulfurous acid to convince
00:27:59
yourself that this is disproportionation
00:28:03
at this point in the semester we will
00:28:05
start using the spectrophotometers so
00:28:09
the spectrophotometers are instruments
00:28:11
that will give you information about a
00:28:14
solution related to the color of light
00:28:18
it absorbs so let's get into it
00:28:22
so absorbance of light white light is a
00:28:26
combination of many colors of light
00:28:29
the visible region of the spectrum goes
00:28:33
from around 400 nanometers where the
00:28:35
light appears Violet to 700 nanometers
00:28:38
where the light appears red
00:28:40
the regions just outside the visible
00:28:42
region have names related to the colors
00:28:45
they are adjacent to so ultraviolet is
00:28:49
next to Violet and infrared is next to
00:28:52
Red
00:28:53
the energy of a photon of light is also
00:28:57
related to the wavelength and then in
00:29:00
turn to the frequency of that light
00:29:03
when a substance absorbs energy from the
00:29:06
photons of light the energy causes it to
00:29:09
go from a ground state to an excited
00:29:11
state where the difference in energy
00:29:14
between those two two states is exactly
00:29:17
equal to the energy of that photon
00:29:22
if we have a container of a solution and
00:29:25
white light hits the solution
00:29:28
the solution absorbs photons of a
00:29:32
specific energy so when the light comes
00:29:35
out the other side the spectrum is
00:29:37
missing the portion that was absorbed
00:29:40
and in our example the solution looks
00:29:43
blue and the reason it looks blue is
00:29:47
because it absorbed orange light so the
00:29:51
color absorbed and the color scene are
00:29:54
not the same
00:29:56
the color that we see is the
00:29:58
complementary color of the absorbed
00:30:01
light
00:30:02
you might remember using a color wheel
00:30:04
to match up complementary colors so blue
00:30:08
and orange are complementary which is
00:30:10
what we saw in this example
00:30:11
red and green are complementary and it's
00:30:15
not marked here but yellow and violet
00:30:18
are complementary
00:30:21
we are going to use an instrument called
00:30:24
a spectrophotometer and this is what we
00:30:26
will use to quantitatively measure the
00:30:30
amount of light absorbed so this this is
00:30:33
a quantitative
00:30:35
measurement quantitative process we
00:30:39
won't have to use it quantitatively just
00:30:42
yet but we will in the future and in
00:30:45
next semester too
00:30:47
um so to do this we actually take two
00:30:49
measurements
00:30:51
for both of the measurements we shine
00:30:54
white light through a container
00:30:56
any light that is not absorbed is passed
00:30:59
through a prism which separates it into
00:31:02
its component colors this is exactly
00:31:04
like the kind of prism that you might
00:31:06
put up to
00:31:08
um like in the window to make sure that
00:31:11
you get cute rainbows on your walls when
00:31:13
the light shines through so it's the
00:31:14
same type of prism probably a little bit
00:31:17
fancier
00:31:19
um one of the containers that we're
00:31:22
gonna pass this white light through
00:31:24
contains what we call a blank solution
00:31:29
um this is shown in the top sequence on
00:31:30
the slide right now
00:31:32
the measurement from the blank sample is
00:31:34
the Baseline it's like a background
00:31:36
measurement it makes sure that anything
00:31:40
we measure later
00:31:42
is only from the thing we are interested
00:31:45
in
00:31:47
so the light
00:31:48
that is passed through the sample which
00:31:52
is the container that has our colored
00:31:54
complex is also passed through a prism
00:31:57
and split into its component colors and
00:32:00
what the instrument does is it measures
00:32:02
how much of a specific color is missing
00:32:06
because it was absorbed
00:32:09
the way that these are related the
00:32:12
absorbance is the amount of light
00:32:14
absorbed by the sample
00:32:16
where this becomes useful quantitatively
00:32:19
is that the absorbance is proportional
00:32:22
to the concentration of the substance
00:32:24
that is absorbing the light
00:32:26
and if we plot that absorbance as a
00:32:30
function of wavelength we call the
00:32:33
result and absorption Spectrum so this
00:32:36
is all we're going to do right now is
00:32:38
look at absorbance versus wavelength
00:32:41
and pull some information out of it
00:32:45
so here's one example of what an
00:32:48
absorbent Spectrum might look like
00:32:51
from the shape we can tell there's one
00:32:54
specific wavelength at which the
00:32:56
solution absorbed the most light right
00:32:59
the tallest if absorbance is increasing
00:33:02
on the y-axis the tallest point of our
00:33:05
Peak
00:33:06
the wavelength that corresponds to this
00:33:10
highest level of absorbance is what we
00:33:13
call the Lambda Max right Lambda is what
00:33:16
we use for wavelength and then Max
00:33:18
because it is the wavelength of Maximum
00:33:21
absorption
00:33:24
the Lambda Max for this peak is in the
00:33:28
blue region of the visible spectrum
00:33:30
which we can tell from how our range
00:33:34
goes from 400 being Violet 700 being red
00:33:38
and if this absorption is in the blue
00:33:42
region of the spectrum then the solution
00:33:44
would appear Orange
00:33:47
an absorbent Spectrum with a Lambda Max
00:33:49
in the yellow green region would appear
00:33:53
violet
00:33:55
and some solutions have a Lambda Max
00:33:57
outside the visible region
00:34:00
so for example a solution of dichromate
00:34:04
has a Lambda Max around 330 nanometers
00:34:07
which is outside of the visible region
00:34:11
however because the tail of the
00:34:14
absorbent Spectrum
00:34:16
shows that dichromate absorbs some light
00:34:20
in the Violet region it does appear
00:34:23
yellow
00:34:24
and you should carefully read the
00:34:27
operation instructions in the appendix
00:34:30
of your lab manual which these are also
00:34:33
posted on lab archives for the sparta
00:34:36
photometer before you go to lab and or
00:34:40
before you watch The Experiment video
00:34:43
um probably both and if you watch The
00:34:46
Experiment video twice it's maybe okay
00:34:49
to read it the first time not read it
00:34:51
the first time but make sure you have
00:34:52
looked at the instructions
00:34:55
and also looked at the video
00:34:57
in conjunction in some way before going
00:35:00
into lab
00:35:02
when you get data for an absorbent
00:35:05
Spectrum you're going to get two values
00:35:08
so make sure you know what these values
00:35:12
mean
00:35:13
our spectrophotometer will read in the
00:35:18
visible and near visible region so
00:35:23
a number between 400 and 700 is going to
00:35:29
be a wavelength so the the computer so
00:35:33
well it's a little computer but the
00:35:35
computer connected to the sparkle
00:35:36
photometer will find this wavelength for
00:35:39
you but you need to make sure that
00:35:41
you're looking for the bigger number as
00:35:44
your Lambda Max
00:35:46
and the smaller number as your
00:35:49
absorbance
00:35:52
okay
00:35:53
let's take a look at what's happening in
00:35:56
this experiment
00:35:58
so there are a couple of parts to this
00:36:00
experiment the one that I want to really
00:36:02
spend time on is part A where you will
00:36:06
have the progressive reduction of
00:36:08
dioxovanadium five
00:36:11
so this is
00:36:14
um this is really cool the colors are
00:36:16
really pretty which extra reminder don't
00:36:19
forget to have someone tell you what
00:36:21
those colors are if you're having any
00:36:23
trouble distinguishing them
00:36:25
and you will see a color change for
00:36:28
every oxidation state of vanadium that
00:36:31
it's possible for us to achieve
00:36:34
so starting with metavanidate which is
00:36:37
vo3 minus
00:36:40
in which Vanadium has a plus five
00:36:43
oxidation state
00:36:45
if you put this into
00:36:47
sulfuric acid you get VO2 plus which is
00:36:51
dioxovanidium
00:36:54
and we're going to put a 5 after it
00:36:57
because
00:36:58
Vanadium can obviously have multiple
00:37:00
oxidation numbers so we need to label
00:37:03
that when we're doing naming so VO2 plus
00:37:06
is dioxovanadium in which Vanadium has a
00:37:10
plus five oxidation state so just taking
00:37:13
metavanidate and dissolving it in
00:37:15
sulfuric acid does not change the
00:37:19
oxidation number
00:37:20
it just ends up in a different form that
00:37:24
gives us this really pretty yellow color
00:37:28
oh and I guess I do I have a reminder
00:37:30
here of of what's Happening which is
00:37:34
really just some h3o Plus
00:37:38
and or h plus however you like to write
00:37:40
it reacting with the vo3 minus
00:37:43
and turning into H2O so we don't have
00:37:47
any oxidation number changes here
00:37:50
from the dioxovanadium five if you put
00:37:54
zinc metal in the solution you can turn
00:37:58
into
00:37:59
oxovanadium 4 which is v o
00:38:04
with a two plus charge which right these
00:38:07
are really difficult to say because
00:38:09
there's VO2 with a plush charge and then
00:38:12
there's vo with a two plus
00:38:14
in the vo with a two plus the oxidation
00:38:19
number of vanadium is four and it's blue
00:38:25
if this continues to react with zinc
00:38:28
metal you'll go to Vanadium three plus
00:38:31
just the Vanadium Ion with a plus three
00:38:35
this is kind of an aqui green color
00:38:39
and if it reacts with zinc again
00:38:43
you'll get Vanadium two and Vanadium two
00:38:46
is purple
00:38:47
so you will see from yellow to Blue to
00:38:51
Aqua to purple the oxidation number of
00:38:54
vanadium change from five to four to
00:38:56
three to two
00:38:58
and as long as there is zinc metal in
00:39:01
here you can maintain this reaction so
00:39:07
a couple of things to think about and
00:39:10
make sure you understand
00:39:12
what does the thing turn into
00:39:14
and does that mean the zinc is oxidized
00:39:17
or reduced
00:39:19
is the Vanadium oxidized or reduced
00:39:24
what gas could be forming
00:39:30
the Vanadium 2 this is the
00:39:35
um lowest number that you can get for a
00:39:37
Vanadium oxidation number you cannot go
00:39:40
Vanadium one and
00:39:44
when mixed with zinc we're not going to
00:39:46
go to Vanadium zero
00:39:48
so you will not produce Elemental
00:39:51
Vanadium so Vanadium two plus is the
00:39:54
limit
00:39:54
so that means Vanadium 2 plus can only
00:39:58
act as a reducing agent
00:40:02
so if you mixed Vanadium 2 plus with
00:40:04
something
00:40:06
the only thing Vanadium could do is
00:40:09
cause the reduction of that thing
00:40:12
and Vanadium 5 can only act as an
00:40:15
oxidizing agent so our yellow solution
00:40:19
Vanadium with a plus five oxidation
00:40:22
state you cannot go any higher
00:40:27
so the only thing that Vanadium 5 could
00:40:31
do is cause the oxidation of something
00:40:34
else
00:40:37
and the color of the solution here tells
00:40:41
us the oxidation number of vanadium in
00:40:43
the solution so if you had any one of
00:40:47
these Vanadium solutions to start with
00:40:50
from the color you would know the
00:40:52
beginning oxidation number and after
00:40:55
reacting it with something
00:40:56
from the color you would know the ending
00:40:59
oxidation number
