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This content explains how to balance redox reactions using the half-reaction method, emphasizing the need to balance both atoms and charge (electrons) across the reaction.
hi everyone this is our last video on
redox reactions we're going to talk
about balancing redox reactions using a
method called half reactions
so if you remember in our discussion
about redox reactions we know that there
is a transfer of electrons or a change
in the oxidation number with one of the
species involved in our reaction in
order for redox reactions to be balanced
we see here that the electrons lost in
oxidation must equal the number of
electrons gained during reduction
and so when we think about balancing
redox reactions not only do we need to
have the same number of atoms for each
element on both sides of the equation
but we also need to think about having
the same number of electrons on each
side of the equation so the method we're
going to use to balance these reactions
is called the half reaction method
these are the five steps that we're
going to walk through for balancing
simple redox reactions first we're going
to write the two individual half
reactions we'll write a half reaction
for oxidation and another one for
reduction then we're going to balance
all the elements
next we'll balance the charge by adding
electrons and then if necessary we're
going to multiply each half reaction's
coefficients by the smallest possible
integers in order to yield equal number
of electrons in each
and then finally We'll add the two
balanced half reactions together and
cancel any species that appears on both sides
sides
so to start with we're going to balance
just some half reactions on their own
and we'll only need to use the first
three steps in our half reaction method
so here we have two half reactions in
the first we see we have 10 going from
the four plus state to the two plus
state so our oxidation number is
changing so in order to balance this
reaction we first want to make sure that
we're balancing all of our elements we
can see that the the element is balanced
on both sides but we need to balance by
adding electrons so we can balance
charge so on the left hand side we have
the four plus State and on the right
hand side we have the two plus state
so in order for our charge to be
balanced it looks like we need to
decrease the charge on the left hand
side on the reactant side so that means
we'd have to add electrons right in
order to reduce that charge since we go
from 4 down to two it looks like I need
two electrons in order to balance the
charge so this is my balanced half
reaction I can see that this is a
reduction reaction because my oxidation
state is being reduced or decreased
in my next example we have chromium
going from the two plus state
and this is already balanced in terms of
the elements but we do need to balance
in terms of charge so on the left hand
side I have a two plus charge and on the
right hand side I have a three plus so
in order to have the charges balance on
both sides of the equation I need to add
electrons and I can see that I'll need
to add them to the right hand side right
so if I have a 3 plus on the right and I
reduce it by adding electrons I'll need
to add only one electron to get it to
equal to two
so this is a balanced half reaction and
I can see that this is an oxidation
reaction because my oxidation state is increasing
all right so now let's graduate to
balancing a full redox reaction
we're going to start by writing the the
expected products so the question says
complete and balance the following redox
reaction which gives the highest
possible oxidation state for the
oxidized atoms so I have aluminum and I
have fluorine as my reactants now I know
that fluorine usually likes to have a
negative charge associated with it so
that is not the species that's being
oxidized I expect to generate fluoride
ions in this reaction
the aluminum is a metal and so this
likes to be oxidized to a positive state
right so the highest oxidation state of
aluminum that we expect would be a three
plus state
so we have the skeleton for our redox
reaction but in order to balance it we
need to write our half reactions so I'm
going to write a half reaction involving
the aluminum species so I have aluminum
as my reactant and I have the aluminum
ions as my product the other species
involved in my my skeleton for the redox
reaction is the fluorine F2 and the
other species involved with that is f minus
minus foreign
foreign
so I wrote my half reactions and now I
need to go and balance in terms of all
the elements so for my aluminum reaction
I only have one aluminum on both sides
so that's balanced for my fluorine
reaction I have F2 on the left which
means that on the right I'll also need
to have two fluorines
all right so we're balanced in terms of
the elements next we want to balance
charge by adding electrons
so if I look at my reaction involving
aluminum in order to balance the charge
I have a charge of zero on the left for
my Elemental aluminum and on the right I
have a plus three charge so I need to
reduce the charge on the right by adding
three electrons and so now that would
sum to give a charge of zero
for my fluorine reaction I have a charge
of zero on the left but I have a charge
of 2 minus on the right so I will need
to add electrons to the left hand side
in order to balance the charge for this
reaction so now looking at my half
reactions I can see that the reaction
involving the aluminum is my oxidation
right my oxidation state has increased
as I went from 0 to plus 3 for my
aluminum and the reaction involving
the oxidation state is going from zero
in the fluorine Elemental State and to a
negative one for the fluoride
so continuing to balance this reaction
we move to step four step four says
if necessary multiply each half
reactions coefficients by the smallest
possible integers to yield equal numbers
of electrons in each so I can see I have
three electrons in my oxidation half
reaction and I have two in my reduction
half reaction I need to have them be
equal so I want to find the least common
multiple of 3 and 2. so we know that
would be six so in order to get six
electrons in this half reaction I'll
have to multiply by two in order to get
six electrons in this half reaction I'll
have to multiply by three so I'm going
to do that and multiply through all my
coefficients by these integers so this
gives me two aluminum gives two aluminum
ions plus two times three gives me 6 electrons
electrons
and for my reduction reaction I have
three fluorines F2 plus six electrons
gives 3 times 2 6 fluoride ions
all right so now the number of electrons
is the same in both of my half reactions
now step five says to add the balanced
half reactions together and cancel any
species that appears on both sides so
I'm going to write my reactions again so
and so now I want to add them together I
notice that my electrons can cancel
because I have the same number of
electrons on both sides of the reaction
and then I simply add together what I
see on the left hand sides of my two
half reactions and add together what I
so we have
fluorine plus aluminum giving aluminum
ions and fluoride ions we can notice in
this that the products here right could
form an ionic compound if we take the
charge on the aluminum and make that our
subscript on the the fluoride right we
would get three there if we take the
charge on the fluoride and make that our
subscript on the aluminum that would
give us the ionic compound aluminum
fluoride alf3
and it looks like we would have two of
them based upon the balanced chemical reaction
reaction
so let's do a few more examples to get
some practice so here we have another
unbalanced redox reaction and if we look
at this it and just look at the numbers
of atoms we have on the left and the
right it almost appears to be balanced
right I have one tin atom on the left
and I have one on the right I have one
copper Ion on the left I have one copper
Ion on the right and so this appears to
be balanced but when we consider the
charge it is not so if I look at the
charge on the left I have two plus two
plus so I have plus 4 on the left and on
the right I have plus 4 and plus one so
that would be plus five on the right so
obviously our charges are not balanced
and so we need to go through the steps
of balancing this so I'm going to first
write my half reactions
I have the tin on the left and the two
plus State and
this is going to form 10 in the 4 plus state
state
and then I'll write my half reaction for
the copper as well as it goes from the
two plus state to
to
2 plus 1 state
all right so there's my two half
reactions that I'm going to work on I'll
notice that they are already balanced in
terms of the elements so now let's
balance by chart for charge by adding in electrons
electrons
so for my reaction involving 10 I'm
going from the two plus state to the 4
plus state so in order to have the
charge balance on both sides it looks
like I'll need to add in electrons onto
the right as a product
right and in order to have the charges
equal I need to add in two electrons
right so 4 minus 2 would give me the two
plus charge I have on the left
when I go to my copper reaction I have a
plus two charge on the left which needs
to be reduced by 1 in order to equal the
charge on the right so I'll have to add
in one electron on the left for this reaction
reaction
now thinking through my change in
oxidation number I can identify that
this is my balanced half reaction for
the oxidation right because I can see
that the oxidation number of the tin has
increased and increase means oxidation
and when I look at the half reaction for
the copper my oxidation state has been
reduced or decreased so this is my
reduction half reaction
so now we've done step three so next is
step four which says to multiply each
half reactions coefficients by the
smallest possible integer so that we
have the same number of electrons in
each so I have two electrons in this
half reaction and I have only one
electron in my copper half reaction so
it looks like I'll need to multiply that
reaction by 2 in order to have my
electrons cancel
so when I do that multiplying through by 2
2
for all my coefficients I end up with 2
copper two plus plus two electrons gives
me two copper Plus
and now I want to rewrite my reactions
and I'll notice that the electrons
should now cancel and then I write um
um
all right now I can see that it is
balanced both in terms of the number of
atoms or ions right I have two copper
ions on the left and two on the right I
have only one ion of 10 on the left and
the right but now it's also balanced in
terms of charge so on the left hand side
I have 2 times plus 2 plus 4 from the
copper and then another plus two from
the tin so a total charge of plus six on
the left hand side and when I look at
the right hand side I have 10 and a 4
plus State and I add on another 2 plus
so that gives me a total charge of plus
6 on the right as well all right let's
just do one last example to really drive
home how we balance these redox
reactions so I have another
skeletal reaction that needs to be
balanced we can see that this one is not
balanced in terms of the atoms or the
charge so we're going to go through and
follow our steps by first writing the
half reactions so I'm going to write a
all right my first step after this is to
balance in terms of the atoms I only
have one iron ion on both sides for that
half reaction so that one is okay for my
iodine equation I have I2 on the right
hand side which tells me that I'll need
two I minus on the left hand side
so now we are balanced in terms of the
atoms or ions and so now we can move to
balance in terms of the charge by adding
in electrons I can see that for my iron
half reaction I have a three plus on the
left so that looks like it needs to be
reduced by adding in one electron on the
left hand side
for my iodine reaction I have a charge
of 2 minus on the left hand side so it
looks I will need to add in two
electrons to the right hand side so
these are my balanced half reactions I
can see that the iron's oxidation number
is decreasing right so decrease is a reduction
reduction
and the iodides oxidation number is
increasing as it goes from negative one
to zero in I2 so this is my oxidation
the next step in my process for
balancing these would be to make sure
that I have the same number of electrons
on both sides so I have 2 for this half
reaction I have only one for my iron
half reaction so it looks like I'll need
to multiply this one by two
so let's do that we multiply through our
coefficients by 2 which gives me two
iron three plus two electrons giving
rise to two iron two plus
and then I'll also write in my half
and then I'll need to add these together
my electrons cancel and I can add up my
reactants and products to give my
all right so now you are all experts on
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