Gibbs Free Energy Video Lecture | Additional Study Material for Class 12

hello friends this video on
electrochemistry part 14 is brought to
you by exam fear calm
no more fear from exam now let's see the
Gibbs free energy for the last equation
we have seen that Delta G is nothing but
minus and F and E is nothing but e of
cell adding in given concentration this
can be found by Nernst equation because
this is the last is now we can find this
e cell in an easy fine Gibbs free energy
also correct
so the stand electric potential Daniel
set is given we have to calculate the
Gibbs energy for this reaction you know
that Delta G is nothing but and is not
this is e naught is given I need to find
standard Gibbs free energy with it when
Delta G naught just put the values here
n is equal to 2 F is we know 9 6 4 8 7
Coulomb per mole and II not is 1 point 1
volt so 1 point 1 1 I can write as 1
point 1 Joule per kilo right 1.1 into
Joule per Coulomb Coulomb it's cancel
you solve this you get is a minus here
minus NFE minus 2 1 triple 2 Joule per
mole for you can say minus 21 point dip
into kilo Joule but that is my Gibbs
free energy correct
the next question is why voltage of
battery we use does not drop after use
so that's a good question Natalie the
battery generally we use the pencil
battery right or lithium-ion battery so
in that case it is not in the liquid
form the electrolytes are not in the
wait for their form in the Sun in the
solid form we almost solid form it drops
with the cross very less because it's
almost in the solid form and we have
seen that when you talk about Q solid
concentration we always almost take s1
and that's why the Ln here is almost 1
why because we are dealing with solid
scene it's almost 1 and thus the voltage
doesn't drop here because we know that e
is equal to e naught minus T naught
minus 2.303
R T by n log of Q right this is almost 1
because we are generally dealing the
soil solids solids right so this is 0 so
the voltage doesn't drop for period of
time but the normal battery we use in
our life
the next question is try the Nernst
equation and the EMF of the following
cell at this temperature C you write the
Nernst equation the first thing is we
have to write the balance equation so
let us write the balance equation for
this mg becomes mg 2 plus 2 electrons
are exchanged copper 2 plus becomes
copper solid ok so for this equation I
can write as e cell is nothing but e
naught cell - since it is happening at
Point 298 Kelvin you can use this
shortcut equation point zero five nine
by n log of K correct that is what we
can write and log of K is nothing but if
you see o the concentration of I mean
here for this equation I write k k has
nothing but sorry log of Q U is nothing
but concentration of MD 2 plus
concentration of copper solid that is
almost 1 divided by concentration or C 2
plus concentration of M so this is 1
this is 1 so that is my Q and E not 7 we
can actually find write e naught cell
for this reaction is nothing but he
writes - he left that is e naught for
copper 2 plus 2 copper minus e naught of
mg 2 plus 2 MV let's find the sky he not
cell is he right - he left he right is
nothing but copper 2 plus 2 copper - e
he left is e mg 2 plus Tony please note
this mg 2 plus 2 energy both am writing
reduction so from the table I can say
that copper 2 plus 2 copper is 0.34 and
the reduction potential of MD 2 plus 2
MD is minus 2 point 3 7 volt total s 2
point 7 volt so I know the value of ki
not self-correct let me put the values
over here so e cell will be nothing but
e naught cell that is two point seven
volt minus point zero five nine one my n
is two electrons log of concentration of
mg plus that is nothing but this guy
point zero zero 1 by concentration of
copper plus that is point zero zero zero
1 so this becomes log of 10 log of 10 is
1 so you solve this difference 2 point 7
V is V 8 minus point zero 5 9 1 by 2
into log of 1 is 1 law of 10 is 1 so you
solve this is equal to 2 points and 1 v
minus point zero 3 v you solve this
becomes 2 point 6 8 that is my EMF for
this equation the first equation let us
do the same thing for the second
equation as we have done first you write
the balanced equation Fe
becomes fe2 plus two electrons and then
h+ is reduced to hydrogen gas their h+
required this is my balanced equation
correct I know that he cell is nothing
but e naught cell minus point zero five
nine y by M log of Q correct so let's
see the Q value use a liquid
concentration of Fe 2 plus my
concentration of s2 is almost 1 divided
by concentration Fe is almost 1 because
it's solid my concentration of h+ the
power 2 because they are 2 h plus here
right please not get to h plus so i am
saying
concentration h plus to the power 2 he
not cell we can find easily using the
chart we have the resulting burned your
right - your left what is in the right
right we have this guy that is your
reduction potential of h plus 2 H -
reduction potential of AP 2 plus 2 this
is 0 this is we know is - point 4 4 4
that is nothing but plus 0.45 so I know
you know so let's put everything this
equation any cell will be come in or
this point 4 4 4 minus point zero 5 9 1
by n is 2 here 2 electrons log of Q Q is
nothing but concentration equal to plus
that is nothing but point 0 0 1 M by a
concentration of h+ that is 1m its
square right this is you need not put em
in not here 2000 1 point 1 and if it is
small enough you solve this you'll get
0.44 volt plus actually because this is
point 0 1 right becomes minus minus 1 is
plus plus points 0
my volt and that is nothing but point
five three and that is my answer so this
becomes log of 10 to power minus three
this whole thing becomes minus 3 so
minus 3 into minus becomes plus when you
multiply this point zero five nine by 2
thank you
visit exam viacom to watch more videos
attend free online test
get free study materials find tutors and
mentors and much more thanks once again