Balancing of Equations by Ion electron method (Half Cell Method) Video Lecture - Class 11

balancing of equations involving redox
reactions there are two different
methods today we are going to see how we
can balance the equations using ion
electron method or half reaction method
sometimes called have cell method
step one of the reaction find the
elements whose oxidation numbers have
changed
choose a substance that acts as an
oxidizing agent and one that acts as a
reducing agent today we are going to see
the example of the reaction between Fe
to positive with manganate ion mno4
minus reacting in an acidic medium to
produce Fe 3 positive and MN to positive
along with some water
first thing to do is write the equation
identify those elements that are
undergoing change in oxidation write
down the oxidation numbers of the
reactants and products 2 is oxidation
number of iron 7 is the oxidation number
of manganese in manganate ion 3 is the
oxidation number of fe 3 positive and
positive 2 is the oxidation number of MN
2 plus in order to determine oxidation
numbers you may refer to my earlier
video on how to assign oxidation numbers
whenever there is an increase in
oxidation number the reaction is
considered to be oxidation Fe 2 plus
changes to 3 plus so this is an
oxidation half-reaction in this case one
electron is lost from Fe 2 plus 2 change
in to Fe 3 plus manganese in manganate
ion changes into M in to positive the
oxidation number decreased by 5 units
which means it needs to gain electrons
or it is a reduction half-reaction and
we are to provide 5 electrons for it
so we have identified the oxidation
half-reaction and reduction
half-reaction in this reaction
fe gets oxidized and MN gets reduced Fe
to positive is the reducing agent as it
gets oxidized so all oxidizing agents
gets reduced in a redox reaction which
implies mno4 minus is oxidizing agent
and it is getting reduced in the process
the next thing to do is normally
separate the equations into two
half-reactions
identify the oxidation half-reaction
identify the reduction half-reaction in
this case
Fe 2 plus changing to 3 plus plus 1
electron reduction half-reaction mno4
minus plus hydrogen ions plus 5
electrons giving you em in toothless we
may also look at the same reaction in a
slightly different way just to get an
idea of what's going to happen in order
to do that we will see what we should be
doing balance the half reactions
balanced all atoms other than hydrogen
and oxygen calculate the oxidation
numbers on both sides of the equation
add electrons to whichever side it is
necessary to make up the difference and
make sure both half reactions have
gained equal number of electrons or lost
equal number of electrons
balance both sides so that the total
charge on the left and the right are the
same to balance oxygen atoms add water
molecules on the side deficient in
oxygen and on the side deficient in
hydrogen add H+ ions because the
reaction is being balanced in an acidic
medium this is the reduction
half-reaction
oxidation half reaction oxidation half
reaction produces one electron and we
need five electron in the first equation
that we multiply the oxidation
half-reaction by five so that we have
five moles of electrons add the two
reactions you have a balanced equation
now we can take a slightly different
look or slightly different approach to
the same reaction we have done this
earlier identify the oxidation numbers
oxidation half-reaction
reduction half-reaction if you multiply
the oxidation half-reaction by five it
will produce five electrons the
reduction half-reaction requires five
electrons so we have generated five
moles of electrons by taking five moles
of Fe 2 plus or we have to add a
coefficient for Fe 2 plus that's the
five moles of electrons that we need if
you have five moles of Fe 2 plus on the
reactant side we also need five moles of
Fe 3 plus on the product side since the
number of manganese ions on the left and
right is same we're not changing the
coefficient since it's one remains the
same next we have to balance oxygen on
the left side we have four oxygen atoms
in manganate ion according to the ion
electron method we need to have four
moles of water on the products side so
we will add four moles of water what
remains to be a balance at this time is
hydrogen ions since we have eight
hydrogen ions hydrogen atoms on the
right side we need to have eight
hydrogen ions
on the reactant side so add the required
quantity of acid now we have a balanced
equation so fundamentally we have
separated the oxidation half-reaction
reduction half-reaction and added them
up together and this is very similar to
the oxidation number method but when you
write it we tend to separate the two
half-reactions write them separately and
add them up
if on the reactant side on the product
side you have similar elements or
compounds cancel them in the above
example we don't have any such atoms or
molecules therefore the equation is
actually balanced so 5 fe 2 plus + mno4
minus + 8 h+ gives you 5 Fe 3 plus + M +
2 + + 4 h2o that's a balanced equation
since we have hydrogen ions in the
reactant side in the case of the
equation has been balanced in the acidic
medium here are some examples for you
write down the questions balance them
and check whether you're getting the
right answers pause the video to write
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video if you have any questions or
comments please do post them thank you
and have a great day