Balancing Redox Reactions using Half Reaction Method Video Lecture | Chemistry Class 11 - NEET

Oh a gif never eat sauteed worms my very
educated mother just served us nine
pizzas except since we got rid of Cluedo
I guess that one can be sure a little
bit too many very educated men just
screwed up nature oil rig or even leo
says girl now most of these things are
mnemonic devices or things that are
gonna help you memorize some things in
science but the important thing that I
want you to realise out of this is that
just because we have pneumonic devices
it doesn't mean that the order of
planets was based on these pneumonic
devices no rather what we've done is we
have put together these memorizing
devices to help us remember the order of
certain things or to remember how to
perform a particular task in science the
important thing as always is that nature
does not conform to our understanding
rather what we do is we try and fit our
understandings to what we observe in
nature so we're going to take a look at
something that helps us balance and deal
with oxidation and reduction reactions
and it's important to note that this is
not the way that the reactions occur but
rather the way that we have come up with
to help us balance these equations so
let's take a look at how we balance
redox equations now first off you'll
have to excuse my voice a little bit I
am feeling a bit under the weather so
hopefully you can understand me as you
watch this video let's quickly review
redox now redox has two halves to it we
have the reduction half and we have the
oxidation half what I want you to
remember is that reduction is due to a
gain in electrons through the course of
a reaction and oxidation is due to a
loss of electrons through this same
reaction and so if we want to get into
what oxidation reduction means to in
terms of oxidation numbers because now
we understand oxidation numbers we can
explain reduction as a reduction in
oxidation number so the oxidation number
because the substance is gaining
electrons is reduced
hence the reduction if something is
losing electrons is becoming positive so
it's oxidation number
is increasing and we say that substance
is oxidized so ultimately again
reduction is a decrease in the oxidation
number and oxidation is an increase in
the oxidation number and if we take a
look at a really straightforward
reaction involving solid sodium and
gaseous chlorine which produces sodium
chloride or what we commonly call table
salt
this can be split up into oxidation and
reduction as well so first let's assign
an oxidation number for sodium of course
sodium and chlorine both being in their
elemental state we can assign both of
the monoxide amber of 0 and if we take a
look at the sodium chloride that's
formed we have a sodium cation or
positively charged ion and we have a
chloride anion or negatively charged ion
and we can of course assign these
according to those oxidation number
rules that we have sodium being plus 1
and the chloride ion being minus 1 and
then if we wanted to take this a step
further and split it apart into what we
call their half reactions we have an
oxidation half-reaction and of course we
can see that sodium is being oxidized
its oxidation number is increasing and
it's increasing because sodium is losing
those electrons it's becoming less
negative or more positive then the other
half of the reaction is going to involve
chlorine because as we can see chlorine
goes from an oxidation number of 0 to an
oxidation number of minus 1 so it's
oxidation number is reduced
it's the reduction portion of that
because it is gaining electrons gaining
negative charge throughout this reaction
so it's becoming more negative now what
I want you to notice is that there is a
way to balance this out because even
though the number of sodiums and
chlorines are equal on either side of
this particular equation there is
something that isn't equal and that is
the charge and so in order to balance
out the charge we need to put electrons
on either side of the arrow to help us
balance out the charge in an oxidation
reaction you're going to put electrons
on the product or right side of the
arrow as you can see here we need that
electron to balance out that positive
charge of the sodium cation
and in the reduction half-reaction
you're going to find those electrons on
the left side to balance out in this
case the negatively charged chloride
ions that we have on the right so these
are what we call half reactions and in a
neutral solution like this you probably
look at this and say I can just balance
this reaction by inspection why do I
have to split it up into half reactions
well that becomes relevant to us when we
put it under certain conditions a lot of
these reactions occur under acidic and
basic conditions were a cynic or basic
conditions and so we have to consider
other things that are present in the
solution like hydrogen ions and
hydroxide ions so the best way to take a
look at applying these rules is to work
through an example so we can see here
that the perchlorate ion and nitrogen
dioxide react to produce the chloride
ion in the nitrate ion I'm going to say
that this occurs under acidic conditions
so that is something that we will have
to take note of and you'll see that as
we work through this particular example
so the first thing that we want to do is
figure out what's being oxidized in
what's being reduced now if you're
comfortable with oxidation numbers at
this point this is the best time to use
them so let's go through and I'm not
gonna spend a lot of time on this but
let's just go through and assign these
elements their oxidation numbers there's
nothing really tricky here in assigning
oxidation numbers for this particular
example so I'm not too concerned about
this one so we can see here that the
chlorine is going from plus 7/2 plus 5
so it's oxidation number is going down
it is being reduced we can see that the
nitrogen goes from plus 4 to plus 5 so
it is being oxidized so our half
reactions here involve nitrogen being
oxidized and chlorine being reduced so
if we were to split this up we can see
that we would have our oxidation
half-reaction and our reduction
half-reaction involving those two
compounds or those two elements and the
compounds that they're involved with in
this particular reaction so what we're
going to do is we're going to take a
look at each half reaction and we have
steps for each one of those half
reactions so what we've done is we've
split up the oxidation and reduction in
two halves if you're arranging this on a
piece of paper for a note or you want to
prepare for a test or an exam or some
like that it's probably best to put it
into two columns one being oxidation one
being reduction for both the oxidation
reduction these steps will be the same
so you'll notice that we're gonna sort
of mirror these steps with our oxidation
reduction approach here and the first
step is very similar to all of the
balancing by inspection you've done to
this point in your science career and
that is you're just going to take a look
at the non oxygen on hydrogen and try
and balance those so in the case if
there was one nitrogen on one side and
two nitrogen on the other you would be
using a coefficient that's a number that
goes ahead of an atom or a molecule in
order to balance out those values so
we're not gonna worry about oxygen or
hydrogen as I said but anything else we
balanced that first now in our examples
for both our oxidation and reduction
they're both balanced so we don't have
to worry about that and we'll move on to
the next step which is we're going to
add water to balance out the oxygens so
obviously these reactions being an
acidic or basic solutions are going to
be taking place in water so there's lots
of water around for us to balance this
out again remember this is not the way
that these reactions occur this is just
a method that we use to balance these
types of reactions so the first step
we're going to do is balance the oxygens
by adding water so I'm just going to
take a look at our oxidation example
here we've got two nitrogen on the left
or reactant side we've got sorry two
oxygen on the Left or reactant side
we've got three oxygen on the right so
we need to add one water each water a
member contains one oxygen to the left
side in order to balance out the number
of oxygen on either side giving us a
total of three and again for the
reduction reaction you're going to do
the same thing to balance out the oxygen
the next step is we are going to add
hydrogen ions and this step is important
even if you're in a basic solution the
next process from the next step in the
process is still to add hydrogen ions to
balance out all the hydrogen's even the
hydrogen's that are not in water we're
still using the hydrogen ions to balance
that up so again in the case of our
oxidation we can see that the two
hydrogen's contributed by the water that
we just added we have to balance them
out by putting two hydrogen ions on the
right now if the solution is acidic we
can continue forward to what I'm going
to call step six but if the solution is
basic and this one is not but if the
solution is basic what would
we would add hydroxide ions to balance
out the number of hydrogen ions so if we
take a look at our reduction reaction I
have there eight hydrogen ions in the
reactant side I would put hydroxide ions
on both the reactant side and on the
product side so you're gonna add
hydroxide ions eight hydroxide ions on
the right and on the left in order to
balance each other out because if you
didn't put them on both sides then again
your equation would be unbalanced and
that sort of negates the whole purpose
so we're gonna add hydroxides to both
sides to balance and again if it's a
basic solution now what you'll notice is
if that you add hydroxide ions to
hydrogen ions those two things will
reactor will combine to form water so
anything that you add in terms of
hydroxide and hydrogen
you're gonna combine those things into
water and then you're gonna reduce as
necessary what I mean by that is if you
had six water on one side and four water
on the other then sort of those four
water molecules that really didn't do
anything so you're gonna reduce that six
to two and you're gonna eliminate that
four on the other side completely so
then we move on to step 6 and step six
is the same whether we were using an
acidic solution or a basic solution so
we're gonna use electrons to balance the
net charges so you remember everything
has to balance here even the charges so
we're gonna use electrons to balance out
the charge to make sure that the charge
on the left and the charge on the right
are equal and so in our oxidation you
can see that the left side or reactant
side is neutral on the product side
before we add those electrons you can
see that we have an excess proton plus
one charge so we're going to add that
electron there and again you'll notice
the oxidation gets the electrons on the
right side or the product side the
reduction gets them on the left side or
the reactant side now these electrons
are important because they're ultimately
what are going to help us balance this
equation so we have to come up with some
lowest common multiple that are gonna
help us figure out how to balance these
electrons so since our reduction
reaction has eight electrons and our
oxidation only has one we're going to
multiply our entire oxidation half way
action by 8 so that the number of
electrons on the product side or on the
right side of the arrow is equal to the
number of electrons on the left side of
the arrow now why is this important well
if we take these two half-reactions to
combine them together in one total
reaction that's now balanced under
acidic conditions we have to combine
these oxidation and reduction
half-reactions and in doing so what
we're going to do is we're gonna put
these reactions together we're going to
take everything that's on the left side
of that arrow combined it together we're
gonna take everything else on the right
side of that arrow combine it together
and then we're going to reduce anything
that's sort of redundant so if we take a
look at our examples we can see the
eight electrons that are on the left the
eight electrons that are on the right so
we can almost picture that as those
eight electrons not really doing
anything not really participating in the
reaction aside from balancing it out and
so we can eliminate those again you will
notice that the hydrogen's are on both
sides of that arrow and so one side we
have eight hydrogen the other side we
have 16 so again we can reduce that
eliminate them on the left side of the
arrow and have eight remaining on the
right and again if you're keen you'll
notice that you see water there as well
appearing on both sides of that arrow
and so we can reduce the total number of
water to just four on the left and then
ultimately you can add all of these
things up together and arrive at your
total balanced redox equation or redox
reaction under acidic conditions now I
know this might look like a lot and it
might even be a bit confusing at first
but most things are when you're learning
the first time around but I promise you
if you practice this a few times this
half-reaction method will become very
straightforward because