okay let's practice balancing redox reactions consider the following
oxidation reduction reaction we have MnO4 minus plus Fe 2 plus yields Mn 2
plus plus Fe 3 plus now we want to balance this reaction under acidic
conditions so if you're not sure how to approach this go ahead and check out my
tutorial on this subject and when you're ready give it a try.
so let's put our
reaction up top now what are we going to do first the first thing we have to do
always is we're gonna split this up into half reactions so the oxidation
half-reaction is Fe 2 plus to Fe 3 plus that is the oxidation half-reaction
because iron is oxidized we go from iron 2 to iron 3 so iron has lost an electron
and the reduction half reaction is MnO4 minus 2 Mn 2 plus that is a reduction
because we're going from manganese 7 to manganese 2 if you find the oxidation
number for manganese in mno4 minus that is going to be 7 and so we're going from
7 to 2 that is a reduction now step 2 we're going to have to balance oxygen
with water molecules so the oxidation half-reaction doesn't have any oxygen so
we don't have to worry about that but the reduction half-reaction does so we
have 4 oxygen atoms on the left because we have MnO4 so if we have four oxygen
atoms we're going to need 4 water molecules on the product side because
each water has one oxygen so let's put four waters there now step three is to
balance hydrogen with protons so once again the oxidation half-reaction we
don't have any hydrogen we don't have to worry about it
but we have just put four waters for the reduction half-reaction so we're going
to have to balance those hydrogen's so if we have eight hydrogen's on the
right because there are two per water molecule and there are four waters
we need eight H+ ions on the left so that's the end of step three let's bring
this up to the top so we have some more room to finish up now step four we're
going to have to balance charge with electrons so on the Left we have a plus
seven charge total because we have a minus one charge and then we have eight
plus one charges because we have eight protons so minus one plus eight is plus
seven and on the right we just have a plus two charge from that one mn 2 plus
ion so we have plus seven on the left and we have plus two on the right that
means we're going to need five electrons on the left to balance the charge
because if we have a plus seven charge we need five negatively charged
particles to bring that charge from seven down to two so that we have a plus
two charge on both sides now at this step we're finally going to have to
bring back the oxidation half-reaction because that charge is unbalanced and so
we are going to need one electron on the right to bring the three plus charge
down to two plus so that we have two plus on both sides now the next step is
to match the number of electrons and then combine the half reactions so we
need the number of electrons in these two half-reactions to be the same so
that we can combine them and on the top we have five electrons as reactant and
on the other one we have one electron as a product so what we need to do is
multiply this entire oxidation half-reaction by five which means we
have five Fe 2 plus yields five Fe 3 plus plus five electrons and now we are
ready to combine these which means we're just going to add everything up and we
altered the number of electrons so that we could cancel them we have five
electrons on the left and five electrons on the right those will cancel and now
we just add up everything else so let's take all of the reactants we have five
Fe 2 plus plus MnO4 minus plus eight protons
and then on the product side we have five Fe 3 plus + Mn 2 plus plus four
waters and this equation is now balanced under acidic conditions.