L30 : Faraday's Law: Numerical - Electrochemistry, Chemistry, Class 12 Video Lecture

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exam we'll take some more numerical
storm it says if the current of 0.5
ampere flows through a metallic wire for
two hours how many electrons would flow
through the wire see current is given
and time is given in ours we know Q is
equal to I into T what is the value of
0.5 ampere and T in seconds you have to
write so two R's in to $60 minutes into
60 you get these seconds you solve it
you get three six double zero Coulomb
that is the charge is transferred to the
wire in this seconds now the question is
how many electrons are there in this
charge the number of electrons if you
want you can actually find the total
charge which we have passed by the
charge of one electron right simple
maths that is nothing but three six
double zero Coulomb by the charge of an
electron is nothing but 1 point 6 into
10 to power minus 19 almost
approximately this is also cool
unclogged it's canceled you solve this
you get 2 point 2 5 10 to the power 22
electrons so that is a number of
electrons that get passed through this
wire if you are passing this point 5
ampere current for 2 hours so these when
electrons will actually pass huge number
right the next question is suggest a
list of method metals that are extracted
electrolytically as I have told that
lithium sodium potassium magnesium
aluminum they are the metals they are in
a copper they are extracted electorally
why because it is difficult to reduce
them using chemical process so
electrical process is used to you do
them the next question is how much
charge is required to reduce the
following reaction 1 wool of al three
plus two al see for this one mole you
have seen that for example
X
two x NF charge required right for one
mole so in this case three F charge will
be required and three M is nothing but 3
into 9 6 for 8 7 Coulomb in solvent
you'll get this file similarly for this
copper to to copper to F charge choir to
Faraday you can just solve this 9 6 over
8 7 cool right you can just solve this
becomes 1 9 2 9 7 4 here also if you see
I made a 4-2 mn 2 plus oxidation number
of MN is 2 hair hair if you find
oxidation number MN oxidation number
will be little X X plus oxygen therefore
minus 2 into 4 because oxidation number
of oxygen is minus 2 total charge is
minus 1 you solve the value of X X is
nothing but minus 1 plus 8 that is 7 so
oxidation number of a minus plus 7 so it
is going from plus 7/2 plus 2 so the
change is what 7 minus 2 that is 5 so
this is 5 electron chain that means 5 F
charge is required to reduce 1 mole of
mno4 right there is nothing but 5 left
with nothing but 5 into 9 6 4 8 7
Coulomb
solve this you get 4 8 2 4 3 5 this much
all is required to reduce 1 mole of mno2
membrane 2 plus the question is how much
electricity in terms of faraday is
required to produce 20 gram of calcium
from molten ccl2 that is ca plus 2 and
40 gram of aluminum from molten aluminum
al2o3 there is a elastic charge correct
wire rod plus 2n plus 3 here there is
oxidation number of calcium ammonium in
this we have doubts you can watch the
chakra where we explain this oxidation
number so let's see the first question I
have CA 2 plus
it will take two electrons to become
calcium so that means two F charge
required to produce one mole of calcium
one mole of calcium is 140 gram cancel
because there is oxidation number so 2 F
charge is required to produce 40 gram of
calcium so if you want instead of 40
gram I 120 gram calcium what I'll do is
I'll say 2 F divided by 40 into 20 what
I will get is f F charge is required to
produce 20 gram of calcium it easy let's
take the next example 40 gram of
ammonium from al plus 3 so if you see L
3 plus 2 L means 3 electron so I will
say that one mole of L R is nothing but
20 7 gram needs 3 F charge so 27 grams
need 3 a charge 40 grams will need how
much charge 40 gram in 3 F charge by 27
into 40 we solve this you get 4 point 4
5 means you need 4.0 for F charge to
produce 40 gram of a solution of Ni no.3
- that is NI plus 2 is electrolyzed
between platinum electrodes using a
current of 5 ampere for 20 minutes how
much n is positive I am in the current I
can easily find time on top he cut
across will see current is nothing but I
into T what is the value of I 5 MP n T's
in 20 minutes I will make in seconds to
18 to 60 seconds so although this is
nothing what 6000 block so I have got
six thousand Coulomb charge I know that
what I know I know that - faraday
because this is ni 2 plus will become
Nicoll addictive electron so to file a
charge is required for one mole of
nickel and there is nothing but 58.7
time nickel
correct so I have to have required a 2 F
charge will give 2 F charge will give
58.7 graph y equal to f charge is
nothing but 2 into 9 6 4 8 7 Coulomb
charge will give 58.7 gram of nickel how
much I have 6,000
so if 6,000 Coulomb charge I'll be
putting I will be getting 58.7 y 2 into
9 6 4 8 7 into 6,000 Coulomb Coulomb
cancel and this gram is there so all
this you get 1 point 8 to 5 grams or 1
point 8 to 5 gram of nickel you'll get
if we will connect the circuit for 20
minutes using 5 ampere cut 3 electrolyte
Excel ABC containing solution of zinc
sulphate del n plus 2 silver nitrate eg
plus 1 copper sulfate CU plus 2
respectively up connected in a series
example we have 3 things so let us
suppose this is a this is B and as you
see their current it in series and this
is zinc plus alpha this is silver glass
and this is copper 2 plus a steady
current of 1 point 5 ampere is passed
until 1 point 4 5 gram of silver is
deposits here we are getting in this 1
point 4 pi gram silver because this man
owed is my cathode is my anode this is
cathode anode cathode anode Katter right
at cathode I will be getting this 1.05
gram of silver for the cell be
now question is how long did the current
flow and also we define the mass copper
and zinc people so first let's find how
long did the current flow see for this
reaction we know is we have AG plus is
AG one electron correct that means 1
Faraday of charge will give me 1 mole of
Ag and mamala which is nothing but 1 0 8
gram of gram eg 1 Faraday is nothing but
9 6 4 8 7 Coulomb so 9 6 4 it's an
Coulomb charge will give me holiday glam
of somebody right 100 gram of silver
needs 9 6 4 8 7 Coulomb of charge but
the deposit is 1 point 4 5 gram silver
so 1 point 4 5 gram silver needs how
much nine six four eight seven five
hundred eight into one point four five
so this how this Graham and Graham will
get cancel and we'll have Coulomb here
right you solve this you get one two
nine five point four three Coulomb this
is the charge that is required but I
know charge is nothing but I into T
charge we know current we know 1 point 5
ampere I can find the time T is nothing
but Q by put the values one two nine
five point four three hula by I is 1
point 5 ampere you will get in seconds
right you saw all this you get eight
sixty three point six second you can
convert this into minute divided by
sixty will get fourteen point four
minutes this is the time that was
required it's a current flew for this
time now the next question is how much
mass of
copper and zinc were deposited pretty
easy now since I know the amount of
charge that whew I can easily find c4
zinc if you see Forge ink for gingka
have Jaden to + give me a Jaden correct
so I can see that to fire any charge is
required to produce one mole of zinc
there is nothing but a 65.4 glam object
- Valda charge is 465 gram so my this
much charge is nothing but one two nine
five point four three Coulomb discharge
leave Oh for what 65.4 by 2/5 it is
nothing but 2 into 9 6 4 8 7 Coulomb and
this into 1 1 2 9 5 point 4 3 Paula mrs.
Graham you saw this good point 4 3 9
crap so this will be the gram of zinc
I'll be getting similarly we can do for
copper copper we know copper we know
that Cu 2 plus will take 2 electrons to
give copper so I can write that for
copper for copper I can say that 2f
charge is required to prepare again one
coil of copper which in this reaction
Omura copper is nothing but sixty three
point five gram of copper
so again if 2f is giving sixty three
point five copper so this is my charge
this is one two nine five point four
three Coulomb will give me what sixty
three point five gram by 2 F it is 2
into again 9 6 4 8 7 Coulomb into this
mess charge one two nine five point four
three coolin coolin club cancel you
solve this you get point four two six
gram copper and that is massive so
copper is here point four two six
okay let's see the application of little
Isis we have seen this we have seen the
refining of metals we have seen
animation for this it is used also for
the extraction of active metals like
sodium it is also used for the
production of water oxygen chlorine we
have seen that in so many numericals
that water sometimes hydrogen gas comes
out sometimes green Gaskell tone effect
if you do hydrolysis of pure water
Fedewa hydrolysis of NaCl Acworth you
see that chlorine gas comes out it so it
is used for hydrolysis preparation of
gases as I explained is also used for
electroplating of metals we have shown
you the animation for this and is also
used for electro typing to create old
monofins which needs a lot of human
level to create those kind of monuments
really fast to replicate those these are
the pretty good examples of applications
of electrochemistry electrolysis process
now let's first a recap and see the
difference between the galvanic and the
electrolytic cell if you see the
galvanic cell it is to convert chemical
energy to electrical energy electrolytic
cell is just reverse it is used to
convert electrical energy to chemical
energy
so since galvanic energy you convert
chemical to electrical energy work is
done by the cell but here in this case
you provide extra battery here and the
work is done on the cell since work is
done on by the cell the grips energy the
streets energy the system decreases here
the free energy of the system increases
because you provide extra energy
generally here we have two electrodes
they are in two different half shell
here we have genndy two electrodes and
there in the same cell since generally
not always here reduction is at cathode
oxidation anode the rule is same here
also because red cat and an ox is rule
an ox red cat this rule is true for all
philips formation whether it is galvanic
cell or
like Excel so anode oxidation production
cathode here cathode be has positive
charge here cathode we have generally
negative charge that is as I told the
negative and positive charge actually
you can't specify here to see the cell
and then you define where you are
getting electrons so here is generally
be a positive charge in cathode and
negative charge here
cathode has this is used for energy
generation it is uses battery also and
this is using electrolysis
electroplating defining gas generation
all those stuffs electrolyte six is used
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