Balancing Redox Reactions using Oxidation Number Method Video Lecture - Chemistry Class 11 - NEET

welcome to balancing redox by oxidation
numbers in the last video we looked at
how to write half reactions to write a
half reaction we took a redox equation
and looked at what was being oxidized
and reduced and we wrote a half reaction
for the oxidation process and half
reaction for the reduction process the
half reactions also showed us how many
electrons were gained or lost we're now
going to look at how to balance redox
equations and we're going to use
oxidation numbers to keep track of those
electrons that are moving around so
let's look at the oxidation of copper by
silver so I have this chemical equation
representing this redox reaction and I
need to balance it
now there's one atom of every element
present here I have one copper on the
left one copper on the right one silver
on left one silver on the right so why
do I need to balance this if it looks
like I have the same number of
everything on each side because the
atoms are balanced we could say this
equation is balanced for mass however is
not balanced for charge there's a total
of a plus one charge on the left side
and a total +2 charge on the right side
now that can't be equations have to be
balanced for mass and charge so I need
to find a way to balance this equation
so the masses are balanced and the
charge is balanced now this is a pretty
simple reaction to balance but we're
going to use it to demonstrate a process
called balancing by oxidation number so
if I'm going to be balancing by
oxidation number I need to identify the
oxidation numbers of everything involved
so for copper the copper starts as a
zero
this one's zero and it ends up as a two
plus right here the silver on the other
hand starts as a plus one we can see
that right here and ends up as a zero
the solid silver on the product side is
a pure element so it has an oxidation
number of zero so what do these changes
in oxidation numbers tell us well for
copper it tells us that we lost two
electrons that's how we went from a zero
to a positive two we lost two negative
electrons for silver
we went from plus 1 to zero it was
reduced so we have to gain an electron
and here's where the discrepancy in the
reaction comes from the reason I don't
have balanced charges in the reaction is
because I so far have copper losing two
electrons but silver only gaining one
and we know that for redox the gain has
to be equal to the loss so I have to do
something to these two amounts of
electrons to get them equal to each
other slow for this case I'm going to
multiply the first one by one I'm going
to multiply the second one by two and
this times one applies to the copper and
the times two applies to the silver but
how do I actually use at times one times
two well if I go to the actual reaction
everywhere copper shows up I give it a
coefficient of one so they just stay the
same but everywhere that silver shows up
I give it a coefficient of two and now
this reaction is balanced for mass and
for charge I have two silvers on each
side I have one copper on each side I
have a total of plus two because there's
two positive ones on the left and I have
total plus two on the right because the
copper has a positive two charge so now
this reaction is balanced for mass and
charge let's try another example in the
last video we had this reaction and we
wrote these two half reactions if you're
at all unsure about how to get these
half reactions from this original
equation watch the last video because we
go through that process and this
reactions a little bit different than
the last one
I don't have a charge imbalance both
sides look neutral to start with but
this can be a tricky redox reaction to
balance because I have oxygen showing up
in two places on the product side
because I already have the half
reactions written I can see that my
issue comes from the seven electrons
being lost in the oxidation and four
electrons being gained in the reduction
there is my imbalance so how am I going
to get these to be the same as each
other well I can multiply the first one
by four and the other one by seven that
means I'll have four of these four of
these and 28 electrons for the oxygens
I'll have seven oh two 28 electrons and
1402 minus
these coefficients I wrote in here to
represent multiplying by 4 and multiply
by 7
these are going to directly match the
coefficients of the actual balanced
reaction so let's see what that looks
like this 4 n minus 3 comes from the
first ammonia molecule here that means
the ammonia molecule is in the Attic
coefficient of 4 the same one that's
right here this next nitrogen also is a
coefficient of 4 but that one comes from
this molecule so this one's going to get
a coefficient of 4 as well now I look at
the oxygens
I have 702 on the reactant side so
that's my 7 right here and I have 14
oxygens total on the right side now this
is where it's a little bit tricky here
remember we already used 8 oxygens here
8 oxygens were already used in this but
I should have a total of 14 according to
this I should have a total of 14 oxygens
on the product side so if I have 8 in
this little group here that means there
needs to be 6 in this one and now this
reaction is fully balanced for mass and
charge
balancing redox the oxidation numbers is
a simple method it's only really keeping
track of the oxidation number changes
and using that to balance the reaction
it's not that helpful when you are
dealing with a redox reaction that takes
place with an acid or a base and it also
is not as useful for the more
complicated redox reactions for all of
those we have a second method that we're
going to look at in the next video
called balancing by half reactions but
that's wraps up our lesson on balancing
redox by oxidation numbers write down
any questions you have in your notes and
bring them with you to class