Leah here from leah4sci.com/mcat and in this video we're going to continue our discussion
on balancing redox equations looking at a complicated example in acid.
In the last video we introduced balancing redox equations including how to utilize this
checklist.
You can find that video along with this checklist and more on my website leah4sci.com/reactions.
Say you're told to balance the following equations that takes place in acid, Ag solid + O2 gas
gives you Ag+ aqueous.
Remember that in an acidic solution we expect to have an excess of H+ but we also expect
to have water because the acid is dissolved in water.
Taking that into consideration, let's begin.
Remember the checklist, first we identify the oxidation number of everything involved,
Ag solid by itself has an oxidation number of 0, O2 which is in its elemental state has
an oxidation number of 0, Ag+ is a monoatomic ion with a charge of +1 giving me an oxidation
state of +1.
So far so good.
Next we identify what is oxidized and reduced.
I like to pair up the similar species in the reactants and products so we can identify
what's going on.
But we don’t appear to have a complete reaction because if I pair Ag solid with Ag+, I can
identify that Ag zero goes up or op telling me that it is the oxidation half reaction,
but what do I pair up the oxygen with there's nothing else here.
This is where I'm reminding you this is an unbalanced reaction taking place in an aqueous
acidic environment.
This means we have water and we have acid floating around and if its oxygen or hydrogen
not shown we are going to add that in to complete the steps for balancing.
So what do we do? we can either add an H2O here, recognize that the O in oxygen is -2
letting me know it's reduction, or we can do it this way instead: Let's start with Ag
as our oxidation half reaction.
Line up your arrows down the middle and what do we have?
We have Ag solid omitting phases.
For simplicity, go to Ag+ and our reduction half reaction.
Well let’s only write what we have.
We have O2 that has to get reduced but we don't know yet what it goes to so let's leave
it as is.
Now we go into balancing starting with the oxidation half reaction.
We balance the non H's and O's which is good because we have Ag we have1 Ag on the left
1 Ag on the right the atoms are balanced we have to have no oxygen, we're good to go.
For the reduction half reaction the only atom we have present is oxygen, so I have no non
H's and O's to balance.
Next to go to oxygen, and to balance oxygen we have water.
We had two oxygens on the left in the O2 molecule and nothing on the right.
But to balance oxygen, we add in water.
So if we have 2 oxygens in O2 we add two molecules of H2O.
Now the oxygens are balanced, two times one oxygen is two oxygen.
The last thing we balance is hydrogen and now you see why we left it for last because
if we first balance hydrogen then balance oxygen to get water, everytime we add water
we add hydrogen which would give us more work unless we save it for the very end.
We have to times H2O which is 4 hydrogen and so we add 4 H+ on the reactant side to balance.
Now everything is balanced . 4 hydrogen, 4 hydrogen, 2 oxygen, 2 oxygen.
Our atoms are balanced so let's move on to charge . For the oxidation reaction, sum of
the reactants zero, Sum of the products + 1.
We want to add electrons to the higher number to bring it down, 1 minus what number gives
me 0, 1 -1 = 0 so we add plus one electron to the products.
Now for reduction, sum of the reactants is 4 from the positive hydrogen, 0 for oxygen
for a total of +4.
Sum of the products zero.
How do we balance? we bring the 4 down to 0.
4 -4 = 0 so we add 4 electrons to the reactants side.
Our atoms are balanced, our electrons are balanced, now we wanna make sure that we have
the same number of electrons in the reactants and the products, so let's see.
Four electrons in the reduction half-reaction, 1 electron in the oxidation half reaction,
so we'll multiply the entire oxidation half by 4, giving me 4Ag goes to 4Ag+ plus four
electrons.
Now that the electrons are balanced we can cancel them out and now we can sum it all
up.
On the reactant side, we have 4Ag plus 4H+ plus O2.
On the product side we have 4Ag+ and 2H2O.
It looks like we're done but we're not.
We check for duplicates to cancel, we don't have any.
Next we'll check all the atoms.
4Ag, 4Ag, 4 hydrogens, don't worry about the charge yet.
2 x 2 is 4 hydrogens.
O2 is 2 oxygen, 2 times H2O is 2 Oxygen.
The atoms are balance, next we'll check to charge 0 + 4, zero gives us + 4 in reactants.
4 x + 1 is + 4 + 0 gives us + 4 on the products and we're good to go.
Let's add phases back in and here we have it.
4 Ag solid + 4 protons + O2 gas gives us 4Ag+ plus 2 H2O liquid.
Let's try another example.
Say you're told to complete and balance the following reaction where MnO4- aqueous + Cl-
aqueous goes to Mn2+ aqueous + Cl2 aqueous.
You can immediately identify this as a redox reaction because you have Cl2 as an element
by itself and Cl- on the other side.
Anytime something goes from an anion to a non ion that's your clue for a redox, the
first step is finding the oxidation number of anything that gets changed from the reactants
to the products.
Let's start with a simple one, Cl-, it's charge is the oxidation number of -1, Mn2+ is a monatomic
ion so it is oxidation number is a charge of + 2 and Cl2 is a diatomic molecule with
a charge and oxidation number of 0.
That leaves us with the complicated MnO4-.
We know that oxygen is -2 unless by itself or a peroxide.
But Mn is the unknown, so we're going to solve this using the method I teach in the finding
oxidation numbers video earlier in the series by setting it up as X for the Mn, +4 for the
4 oxygen times -2 for the oxidation number of oxygen equals the total charge which is
-1.
Or we can simplify that and say four times -2 is -8.
X-8 = -1, add 8 to both sides allowing this to cancel plus eight on the products gives
me X for the Mn is equal to -1 + 8 which is +7.
Next, let's figure out what gets oxidized and reduced by pairing up like species, Mn+7
goes to Mn+2, 7 goes down or is reduced to 2 giving me the reduction half reaction.
Chlorine -1 goes up or op to chlorine 0 giving us the oxidation half reaction.
We've identified what gets oxidized and reduced let's put the arrow down the middle then write
out our half reactions.
For the oxidation half reaction, we have Cl- goes to Cl2 (omitting phases for simplicity).
For the reduction half-reaction we have MnO4- going to Mn2+.
Let's balance the half reaction starting with the oxidation.
First, the non H's and O's in this case Cl, we have Cl2 in the products and 1Cl- in the
reactants so we simply put a 2 before the Cl-.
Two chloride ions gives us one Cl2, two on both sides.
Next we balance the H's and O's which we have none, so we're good to go.
For the reduction half reaction, let's start by balancing the non H's and O's.
We have one Mn in the reactants, one in the products.
Next we balance the oxygen by adding H2O, we have 4 oxygen in the reactants so we add
4 water molecules to introduce 4 oxygens to the products.
Next we balance the H's, none in the reactants but 8 in the products.
4H2 gives me 8 hydrogen so I had 8 H+ on the reactants side and the reduction half is balanced.
We have 8 hydrogen, 8 hydrogen, 1 manganese, 1 manganese, 4 oxygen, 4 oxygen.
Our atoms are balanced, let's move on to charge.
For the oxidation half reaction we have 2*-1 is -2 in the reactants, 0 in the products.
We want to bring the higher number down that means 0 + what number is equal to -2?
0-2 is equal to -2, this means we add 2 electrons to the products for a total of -2 to balance
the -2.
This is different from what we’ve seen before where we had 0 and positive but it doesn't
matter if it's positive or negative.
The higher number must be brought down by negative electrons as we saw here.
For the reduction we have +8 -1 = +7 on the reactants and just+2 in the products.
7 is a higher number so we bring it down, 7 minus what number is going to give us +
2?
7-5 is equal to 2, so we add +5 electrons to the reactant side of the reduction reaction.
Our atoms are balanced, our charge is balanced, now we want to make sure we have the same
number of electrons so we can cancel.
Up to this point we had one and another number so it is very easy to multiply 1 to get the
other.
In this case we have 5 electrons and 2 electrons.
One is odd one is even so the only way to get them close to each other is to find that
least common multiple, in this case 5*2 give me 10.
That means I have to multiply the entire oxidation half reaction by 5 and the entire reduction
half-reaction by 2.
Let's distribute the numbers, 5*2 is 10Cl-.
5*1 is 5 Cl2, 5*2 is 10 electrons.
2*5 is 10 electrons, 2*8 is 16H+, 2* invisible 1 (when there is nothing given), 2 * invisible
1 and 2*4 which is 8.
Now that the electrons are the same we cancel them out.
10 electrons on the reduction half.
10 electrons on the oxidation half.
Almost there, let's sum it all up.
Again this is why it's so important to keep that arrow right down the middle.
What do we have on the reactant side?
10 Cl- for oxidation, reacting with 16H+ and 2 MnO4-.
On the products we have 5 Cl2 from oxidation, 2 Mn2+, and 8 H2O from reduction, do we have
any duplicates?
Nope everything good.
Check everything.
Let's start with the atoms, 10 chlorines, 5*2 is 10 chlorine, 16 hydrogens 8*2 is 16
hydrogen, 2 manganese, 2 manganese, 2*4 is 8 oxygen, 8*1 is 8 oxygen.
Atoms are balanced, let's check on charges.
In the reactants we have -10 from 10 Cl-, +16 from the H+. -2 from the 2 Mn, 10-10+16-2
which is equal to +4.
On the products we have 0+2* +2 is + 4 and 0.
+ 4 = + 4, our atoms are balanced, our charges are balanced, our redox reaction is balanced.
Whew! that's a mouthful!
If you thought this was exciting (meaning annoying), wait till you see what happens
when you have to balance in base which is exactly what we'll do in the next video.
First give this video a thumbs up, make sure to subscribe to my channel so that you don't
miss out on any new MCAT videos, then visit my website leah4sci.com/reactions for the
MCAT stoichiometry and reactions practice quiz and cheat sheet.The link again leah4sci.com/reactions.