Nernst Equation: Application & Numerical (Part - 13)- Electrochemistry, Chemistry, Class 12 Video Lecture

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exam on last equation let's assume I
have a reaction here I have a nickel
here and I have a silver here for this
reaction is nickel solid becomes nickel
2 plus Atmos then I have AG plus AK was
leaves aging I have to find the last
equation for this assuming they are not
one molar first write the balanced
equation of the cell the palace equation
is nickel solid reacts with or if you
want to find us first half the electrode
reactionary water cathode nickel becomes
we get to do us he gets two electron and
here AG plus will take two electrons to
become street correct to reaction this
is at anode oxidation is that quatre
production it's some of this reaction
you get nickel plus two AG class will
give you nickel two plus plus to each
that is my reaction right cell reaction
that is my cell reaction I get and two
electrons are used I can just use my
nurse to formula e cell is nothing but e
naught cell minus RT by NF and is to f
ln of concentration of NH 2 class
concentration of AG is 1 because it is
solid divided by concentration of ni
solid one and concentration of AG plus
they are 2 AG plus it will be square
because this is nothing but Ln of K that
is the Nernst equation for this reaction
correct
let's see some application of Gnostic
patient is used to find EMS of the cell
at any given concentration that is we
have seen the most important role of
Nernst equation is also used to find the
equilibrium constant you see the Nernst
equation has a equilibrium constant e is
equal to e naught minus RT by NF Ln K
right so this equilibrium constant we
can find you can also use to find the
Gibbs energy of the reaction right
because you have the K can find up gives
energy of the reaction minus NF e right
Gibbs energy n you can find the e then
find that gives energy reaction we'll
see some miracles on this aspect so let
us take one example here first we will
represent the seven is the following
reaction take place mg is getting
oxidized to mg 2 plus and AG is getting
reduced from AG plus 2 this is oxidation
then the other is reduction first you
represent the same so we can represent
in this fashion mg is becoming mg 2 plus
oxidation the concentration of are mg 2
plus is point 1 3 M and then AG plus
concentration is point 0 0 0 1 M is
becoming H that is my self generation
now I define the e cell if
standard ie naught is 3.17 be right so
we know that e cell is nothing but e
naught cell minus RT by NF Ln K so let's
put the value what is e cell inert cm3
3.1 7 volt minus RT by NF or we gain
this since these are happening at 298
Kelvin
assuming at solid temperature this
becomes point zero five nine by NF sorry
by any
and n is what two electrons because two
electrons are getting transfer here
right mg becomes n h2 plus two electrons
to AG becomes a T two electrons by 2
into this becomes log and K is what
concentration of mg 2 plus by
concentration of AG plus 4 square I tell
right here K is concentration of this my
concentration leave glass whole square
because there are two here so let us put
the value here so mg 2 plus is point 1 3
by this is point zero zero zero 1 and
square collect becomes log because we
have seen the formula easy not point
zero five nine by n log K that is also
there right so we have used that formula
here so this becomes you solve this this
is three point 1 7 V you solve this this
is 0.2 1 V and the answer is 2 point 9 6
so you see this is a drop in voltage the
standard voltage is three point one
seven but the actual road is two point
nine six because the concentration is
not unwell see as I told that the Nernst
equation can be used to find equilibrium
possible right so the equation we have e
cell is nothing but e naught cell minus
two point three zero three RT by NF log
of Q
there's a mistake here is Q here correct
now this at equilibrium what would
happen she will become K and at
equilibrium the cell will become dead so
e cell become zero
please note B naught cell the Lord
becomes 0 this will become zero because
the cell become dead right so at
equilibrium this will come zero e naught
cell is a standard value we know we
cannot change this - 2.303 RT by NF q
becomes okay so with this log K is
nothing but e naught cell in to NF by
2.303
that is the formula with this you can
easily find the K equilibrium for a cell
okay see for Daniel sale let's find this
for Daniel cell it's like this for dies
or Daniel sell log K is equal to e
naught cell we know is 1 point 1 volt
for Daniel 7 and 1/2 electrons F we know
is 9 6 4 8 7 Coulomb by 2 point 3 0 3 r
we know 8 point 3 1 4 and T is 298
Kelvin let's assume to the normal
temperature we solve this you get 2 into
10 to the power 4 you will get the value
of K 10 to power because it's log came
to solves you get some value and this
actually you get this is thirty-seven
point two eight eight yes Loki is
thirty-seven point two eight eight zero
K becomes 2 into 10 to the power 37 at
2:19 so we have to find the equilibrium
constant for this reaction so we know
that each cell is nothing but e naught
cell let's assume that this is having an
298 Kelvin user shortcut formula - point
zero five nine by n log of Q right so at
a given this is 0 and Q is K I can say
this is 0 0 equal to e naught cell - and
this for s equation n is two eight
copper becomes 2 u 2 plus R is 2
electron point zero five nine by two
o'clock and E cell value is given this
point four six volt
right this is also in both volt vodcast
so if you saw law of K is nothing but
point four six into two by five zero
five nine that is nothing but fifteen
point six that is K equilibrium is
nothing but three point nine two three
to 10 to the power 15 and that is -
correct
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