Nernst Equation Problem Video Lecture | Additional Study Material for Class 12

today we are going to determine the EMF
alpha galvanic cell when the
concentrations in the anode compartment
and the cathode compartment are
different and for that we are going to
use the Nernst equation here is the
problem determine the EMF of the
following electrochemical cell I have
chosen the galvanic cell the data that
is provided are the molar concentration
of zinc is point zero zero one mole per
liter and the concentration of copper
ions is zero point one so the cell
notation has been given to you with a
different concentration from the
standard States you are also provided
standard reduction potentials of zinc
which is minus 0.76 and standard
reduction potential values for copper
which is zero point three for wool
let's see how we can solve it there are
certain things that you have to do in
sequence before you even start applying
the Nernst equation let's see what they
are we need to know which electrode
represents the anode and in this case it
is zinc and how do we determine that the
standard reduction potential value for
singers given a smaller value in
comparison between the two electrodes
determines which electrode would be the
anode in the galvanic cell the sink is
the anode because the standard reduction
potential value for sink is minus 0.76
and when you compare it to the cathode
which is going to be copper it has got a
larger standard reduction potential
value the standard reduction potential
value for copper is 0.34 therefore
copper tends to be reduced and sink
tends to be oxidized therefore sink is
the anode and copper is a cathode that's
how you determine the anode and the
cathode once you determine that you have
to go into the next level of calculation
let's look at the cell reaction the
galvanic cell looks something like this
you have the anode sink insync sulfate
or sig nitrate as the case may be the
cathode is copper in copper sulphate
solution and according to convention
usually you write the anode on the left
and the cathode on the right and the two
solutions are connected by a salt bridge
that's what you see here and we are
going to look at the reaction since we
already know that the anode is going to
be sink the reaction that takes place at
the anode is always oxidation so the
oxidation half-reaction is going to be
represented like this
which is taking place in the left
container or in the anode compartment
zinc changes into zinc to positive plus
2 electrons this is the oxidation half
reaction that is taking place at the
anode the reduction half-reaction
and the cathode would be the copper ions
gets reduced to copper adding at the two
equations we will get the net cell
reaction note that in the oxidation
half-reaction two electrons are produced
and in the reduction half-reaction
two electrons are consumed so the number
of moles of electron per deuced
and consumed are the same so apparently
the equation is balanced now if you add
the two you will get the net cell
reaction if in case the number of
electrons produced in the oxidation
half-reaction and reduction
half-reaction are not the same then you
balance them by using appropriate
coefficients so that the electrons lost
will be equal to electrons gained and
this value for the number of moles of
electrons transferred will be
represented by the letter N
we will need that while solving Nernst
equation so here is the cell reaction or
the net cell reaction and n represents
the number of moles of electrons that
are transferred in the balanced equation
the standard cell potential now notice
the difference here we are trying to
calculate the EMF of the cell if the
standard reduction potential values are
given you are given two numbers there
were the standard reduction potentials
for the cathode and the anode and in
order to calculate the standard cell
potential that is if the molar
concentration in the anode compartment
and the molar concentration in the
cathode compartment where one mole per
liter then we could directly calculate
the standard cell potential the e zero
cell the zero represents a standard
condition usually is calculated by the
equation e zero cell is equals to e zero
cathode minus e 0 a node for this the
standard reduction potential values will
be given for you and these are the two
values for copper it is 0.32 and was
seeing it as 0.76 that's a negative
value on applying these numbers to the
equation on the left you will get the
standard cell potential 0.34 volt minus
of minus 0.76 would give you 1 point 1 0
volt so if the molar concentrations of
the solutions were 1m and 1m then the
cell potential would I mean one point
one zero volt which is what you saw in
the diagram that was represented earlier
now in this case in this problem we are
actually given the concentration which
is different from the standard condition
that is why we will have to use the
Nernst equation it has been
serve that changing the concentrations
in the anode and the cathode compartment
will change the EMF of the cell now this
is the Nernst equation e cell is equals
to e zero cell minus 2.303 RT by NF log
KC KC can be determined from the net
cell reaction where you can write the
equilibrium constant expression R
represents the gas constant 8.314 joules
per Kelvin per mole F represents
Faraday's constant and Faraday's
constant is equals to nine thousand or
ninety six thousand four hundred and
eighty five coulombs the rounded values
96,500 coulombs is equals to 1 Faraday
temperature for all of these cases are
25 degrees Celsius or 298 Kelvin we will
be using Kelvin for calculations now if
you substitute these numbers in the
equation the equation reduces to e cell
is equal to e 0 cell minus point zero
five nine by n log KC now in this
equation you can see two e cells one of
them is the EMF of the cell when the
concentration is different that's what
we are going to find out
he zero cell is called the standard cell
EMF and KC represents the equilibrium
constant expression for the net cell
reaction now this equation further
reduces to you can change the KC value
to the concentration of the anode by
cathode so by writing the equilibrium
constant expression you will always find
that the anode compartment in this case
will be represented by sink ions and the
cathode compartment will be represented
by the copper ions so if you substitute
those numbers and put the appropriate
exponents you should be able to get the
case expression so what we're going to
do right now is substitute the values
for that we need to know the number of
moles electrons transferred we have
already calculated n is 2 e0 cell was
calculated 1 point 1 0 volt
case the expression would be either n 2
plus raised to 1 Cu 2 plus raised to 1
the 1 is not indicated when the value is
1 now substitute in the numbers kc 0
0.0001 m by 0.1 m with all this
information when you substitute it in
the Nernst equation you should be able
to get KC or you should be able to get
the EMF of the cell a zero cell is
calculated by a zero cathode minus a
zero a node concentrations of anode and
cathode I have been substituted here so
the equation reduces to e cell is equal
to one point one zero volt minus point
zero five nine by two when you divide Oh
point zero zero one by 0.1 you get zero
point zero zero one again a cell is
equals to one point one zero volt minus
zero point five zero five nine by 2 log
point zero zero one is negative three
now substituting the values you'll find
each cell is equals to one point one
zero plus the negative sign in the above
equation actually changes to positive
number so plus zero point zero eight
eight five adding them will give you one
point one eight eight five volts for the
concentration that has been suggested
the EMF of the cell would be one point
one eight eight five volts now if you
actually look at the concentrations you
will find that the anode compartment has
got a very low concentration and the
cathodic compartment has got a
relatively higher concentration or you
will say the delusion of the cat the
anions in the anodic compartment is one
thousand times less than the
concentration in the cathodic
compartment and you will find that
EMF has gone up or the cell potential
has increased or in general if the
concentration in the anode anode
compartment is low and the concentration
in the cathode compartment is high it
should always increase the cell
potential if the opposite is true
then there will be a decrease in cell
potential that's it for now if you liked
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