Hey it’s Professor Dave, let’s learn how to balance complex redox reactions.
We know that redox reactions are reactions in which electron transfer occurs.
One substance is oxidized, which means it loses electrons, and another substance is
reduced, meaning it gains electrons, which will occur when these electrons are transferred
from one substance to another.
Sometimes redox reactions are very simple, where a neutral metal atom is oxidized to
become a metal cation, when it transfers electrons to some other metal cation to form a different
neutral metal.
These reactions are easy to balance, because we simply need to make sure that the charges
are balanced, by adding electrons and coefficients if necessary.
But redox reactions can become more complicated if molecules of solvent are involved in the
chemical reaction, as water molecules can be if the reaction is carried out in aqueous
solution.
Hydrogen atoms or oxygen atoms from water molecules can be incorporated into these substances,
which will make it a little trickier to balance the corresponding equations.
Let’s learn how to balance these kinds of redox reactions now, considering both acidic
and basic conditions, as the approach will differ slightly for these scenarios.
For example, in aqueous solution, the dichromate ion will react with iron two to yield chromium
three and iron three.
How do we balance this?
The first step is to split this up into half reactions.
For the oxidation half reaction, iron two becomes iron three.
This is the oxidation half-reaction because the ion has lost another electron.
For the reduction half-reaction, dichromate becomes chromium three.
We can see by looking at the oxidation numbers for chromium that it has been reduced.
Now we must balance each half-reaction, starting with elements other than hydrogen or oxygen.
We can see that for the reduction, we need to put a two here to get two chromium atoms
on each side.
Next, we balance oxygen by adding water molecules.
We have seven oxygen atoms on the left, so we need to add seven water molecules to the
right, since there is one oxygen per molecule.
Now we can balance hydrogen atoms by adding hydrogen ions.
Since there are fourteen hydrogen atoms on the right, from the seven water molecules,
we will need fourteen hydrogen ions on the left.
This means all of the elements are balanced.
Now we can balance the electrical charge for each half-reaction by adding electrons.
In the oxidation, in order for iron two to become iron three, it must lose an electron,
so there must be one electron as a product here.
For the reduction, it’s a little trickier.
We have 2- on the dichromate and 14 H plusses, for a total of 12+ on the left.
On the right we have two chromium three ions for a total of 6+.
In order to balance this, we will need six electrons on the left to bring the charge
down to 6+ and balance the charge.
Now we will need to combine these in such a way so as to cancel out the electrons.
That means we will have to multiply this half-reaction by six in order to get six electrons to match
the other half-reaction.
So six of each iron ion and the six electrons we need.
Now we can finally combine these half reactions.
Once they are combined, we can cancel out any species that appear on both sides, which
in this case is just the electrons.
So we can now see the exact number of ions that must react in order to produce this particular
redox reaction.
Let’s make sure to realize that this is the algorithm we can use under acidic conditions,
which is why hydrogen ions are present.
If this is occurring under basic conditions, the algorithm will be just a little bit different,
but it will start out by balancing as though we are under acidic conditions, so let’s
do that first, just as we have learned.
Let’s say we have solid silver reacting with zinc two plus to become silver oxide
and solid zinc.
Again, we will start by separating this into half reactions.
Silver has been oxidized, going from zero to plus one, and zinc has been reduced, going
from plus two to zero.
Then, just as before, we balance elements other than hydrogen and oxygen.
That will involve placing a two here before solid silver.
Next, we balance oxygen with water molecules, that means one on the left here.
Then we balance hydrogens with protons, which means two on the right here.
Then we balance charge by including electrons.
That means two on the right for this one, and two on the left for this one.
Because the number of electrons in each half reaction is equal, we can go ahead and combine
them, cancel out the electrons, and we get this.
If conditions were acidic, we would be finished, but since they are basic, there is a little
more work to do, because it doesn’t make sense to say that protons exist in basic solution.
First, we note how many protons there are, and we add that number of hydroxides to both
sides, so that’s two hydroxides on each side.
Then, anywhere we see protons and hydroxides, we can combine these to form water molecules,
so that makes two waters on the right.
Then, we cancel any like terms that remain.
In this case, we can eliminate one water molecule from both sides.
And this will be the balanced redox reaction under basic conditions, with hydroxides listed
instead of protons.
We can see that in either case, we are involving solvent molecules in order to supply the hydrogens
and oxygens necessary for the reaction, and we simply follow a series of steps in order
to balance all the components involved.
Here is that list of steps summarized one more time, in case you need these instructions
for reference, with instructions for acidic conditions listed first, and then the continuation
if conditions are basic.
Let’s check comprehension.