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advanced videos Hello Friends  in the last
section we have studied all we have did
individually which was based on Faraday's
first law so now what you are going to
do is we are going to do the same kind
of numerical but with of different data
so let us see how can we solve it
so as the question is given on the
screen it says that an electrolytic cell
consists of alcl3 solution while the
other consists of Zn SFO solution and
then the both are connected in seeds and
here the world is in series so means
it's a very logical thing that we are
talking about the Faraday's second law
of electrolysis so if same amount of
electricity is passed through the cell
calculate the amount of zip deposited if
1.2 gram of aluminium is deposited in
the cell of alcl3 and here the data has
also they have provided that is molecule
bit of aluminum and monitor weight of
zinc is also being provided to us so let
us first of all do we have to determine
what are the data that are present in
the given motion so we have to extract
the data and then only if we could solve
the further part so the given data has
the values where the amount of e in the
amount of aluminum is be present so
therefore the amount that is weight mass
of aluminum is being provided as 1.2 by
that of mass of zinc is not provided and
we have to calculate the mass of zinc
and the Bolliger mass of aluminum is
given as 27 gram per mode by that of
molar mass of zinc is given as 65.4 gram
per mole so this is the molar mass and
the mass of the aluminum that they have
provided to us so with the help of this
data the first thing that we have going
to do is we are going to calculate what
would be the number of moles of copper
on what would be the number of moles of
and we mean about whatever the data that
they have provided us so this case they
have provided us the mass of aluminium
so the as well as the molar mass of
aluminum so by this data we could
calculate what would be the molar mass
of what would be the number of moles of
aluminium present in this amount of it
so let us calculate the first what would
be the number of moles of aluminium so
therefore the number of moles of
aluminium can be calculated as I would
write it as n of n this n represents the
number of moles of aluminium which is
equal to V or mass of aluminium divided
by the molar mass of aluminum so the
weight of aluminium which was provided
to us was 1.2 again when that of the
molar mass of aluminium which was
provided as 27 so let us see what would
be the answer so the answer is zero
point zero four four moles of aluminium
that is present in a weight of one point
two gram of aluminium so this is has
been calculated but we cannot calculate
what will be the number of moles of zinc
because the mass is not being provided
to us and that is what we have to
calculate so now because of this data
that they have provided us we could
estimate what would be the number of
moles of zinc by applying the law of
Faraday second know
so let us see how can we apply so before
that we have to know what is mole ratio
and how can we calculate the mole ratio
so like this so for mole ratio
it has a formula the mode of substance
deposited divided by the number of
electrons required during the process to
get deposited before the ions that would
be content basically so therefore the
you've ever seen number of moles
involved in that half-asian so in this
case we could calculate the mole ratio
for aluminum also as that is the mole
ratio for the zinc also so first of all
let us see what would be the mole ratio
of aluminum and since aluminum is big
present our aluminum is being extracted
from the solution which contains al say
in three so therefore the Al Syrians
will accept few electrons to get thicker
to get deposited on the electrode so
therefore if they simply get dissociated
it forms a n 3 plus and this n 3 plus
will accept 3 electrons to form solid
aluminum so in this case for producing
the one boon of aluminum the three
electrons are being required in this
process so therefore the mole ratio of
the aluminum can be written as
the mole of substance that is been
deposited that is one more in this case
and the number of electron that has been
required in the this half-cell reaction
is nothing but 3 or 3 moles of it so by
the help of this we could calculate what
will be the mole ratio for zinc also so
let us calculate what would be the mole
ratio of zinc but let me tell you
zinc is present in ZN s for solution so
if Z and s 4 gets dissociated it will
form ZN 2 plus and now this Z 2 plus
will accept 2 electron to form solid
zinc that would get deposited on the
particularly so therefore the mole ratio
of zinc is nothing but the number of
moles of saying that is been produced in
one mode that is one mole of zinc
divided by the number of electrons that
are being involved in this reaction so
in this section the number of electrons
that have been involved is 2 so
therefore so I could relate so this are
the data that we have got so now let's
see what do Boonen necessary so now
let's see how can we calculate the total
number of zinc
sorry total number of moles of zinc that
are being involved in this reaction so
for 22 Faraday's second law the mood
ratio or sorry the number of moles of
zinc or the number of moles of aluminium
I could write in this way also divided
by the number of moles of
Zink is equal to mole ratio of aluminium
divided by the mole ratio of sink so by
letting this thing what we are going to
get is the mole ratio of aluminum which
was of the number of moles of aluminium
that we have calculated was found to be
zero point four zero point zero four
four moles divided by the number of
moles of zinc is not being calculated
yet so I would write it as number of
moles of zinc in which the mole ratio of
aluminium was was being calculated as 1
mole of zinc that is one one one mole of
aluminum divided by three electrons the
whole divided by four think I could say
as one mole of zinc involved too late so
this holy equation can be written as 1
mole of aluminum divided by 3 into 2
divided by 1 mole of so therefore 2
divided by 1 mole of zinc so therefore
the answer that we put get is since one
mole of aluminum will be having the same
amount of the monument of aluminum and
that of one mole of zinc will also
contains the boundary weight of zinc so
therefore what the thing that we could
write it as you can write it in this way
that is zero point zero 4 4 into the
number of moles of zinc into 2/3 so
therefore the number of moles of zinc
will be equals to zero point zero four
four
into 3/2 so let us see what would be the
answer for this
it is 0.06 six moles of zinc that has
been deposited on the cell when ZN s o4
is present but because of this data we
could calculate what would be then
number or what could be the weight or
what could be the mass of the zinc that
has been deposited since we know the
formula and observe is nothing but
weight of Zn nearby morning the weight
of Z so therefore the number of moles of
zinc that has been we have calculated
right now is zero point zero six six
into the beat which is not known to us
right now and the moniker weight of zinc
which is found to be sixty five point
four so therefore the weight of Zn is
equals to zero point zero six six into
sixty five point four which means it
will give an answer of four point three
one six four grade so this is how we
have calculated the beat of zinc which
is connected in series
when ends in three solutions and s for
solution of the rating series and by the
help of fine a second of me have
calculated what is the weight of seeing
that has been deposited so it was so
much simple to calculate this kind of
answer so thank you friends for watching
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