Here we have an oxidation-reduction reaction
It takes place in acidic solution
And it's suggested that we use the half reaction
method to write down the complete equation
We only have three species here, so what
this means is that this ion containing iodine
is gonna go to I3 minus but also
I minus is going to I three minus
so here you can see the real advantage
of using this half reaction method. We have
two half reactions and then we want to follow a systematic procedure
The first is we have already done the half reactions
The second is to balance the oxygen by adding water
This means on the right side we would need that
three waters to balance that equation
and then we have to balance hydrogens by
adding H plus, so on this side we need to add six hydrogens
And now that's balanced
We need to balance the iodine. So I was a little
too quick to do this. First we should have balanced the iodine
and so we can see we have three iodines on each side
So actually nine oxygen atoms to balance both sides
and this says there are now 18 hydrogen ions
Likewise, the second reaction I will balance the iodine
A third step in this procedure is to balance the charge
So balancing the charge means adding electrons
In the second reaction I have minus one charge
on the right side, I have minus three on the left side
So wee need to add two electrons
The first reaction, minus one on the right side
plus 18 here, minus three, plus 15
So this would say 16 electrons
So we have balanced charge on both sides
Our next step is to recombine these two equations
So we combine the half reactions
And we combine them so that charges
Electrons balance, which means we would take
this equation times one and this equation times 8
so we have 16 electrons on both sides and electrons drop out
So I have combined the two equations
multiplying the second by eight
Now the electrons cancel
be careful here, I forgot to add the 9 waters here
Now I have a complete equation
And the last step is to verify that this is correct
So we can do that. I have 18 hydrogen atoms
on the left side. I have 18 on the right
I have nine oxygen atoms on the left, nine on the right
I have 24, no 27 iodine atoms on the left
And I have 27 iodines on the right
And the last thing is the charge balance
so plus 18, minus three, minus 24 left
And on the right, minus one, minus 8. This is minus nine
Also minus nine, so we have the correct balanced equation
And you can see it's unlikely we would have
been able to balance this equation by inspection
Following this systematic procedure, we directly get
the balanced equation for this oxidation-reduction reaction
I was a little to quick to write down that answer because
I forgot to check, I can divide this entire equation by 3
So I have six hydrogens ions, plus IO3 minus plus eight Iodines
Three I3 minus, and three water
So for this, we always want to have the
smallest possible integers in my balanced equation
So this is the correct balanced equation