Leah here from leah4sci.com/mcat and in this video we're going to wrap up my discussion of balancing
redox equations by looking how to balance a reaction in basic conditions. In the previous
videos we looked at how to utilize this checklist for balancing redox equations under acidic
conditions. You can find this checklist along with the entire video series by visiting my website
leah4sci.com/redox. As a reminder an aqueous acidic solution has H plusses and an aqueous
basic solution has OH minus. Many professors will teach you balancing in basic conditions
by changing this step so that you balance the hydrogen by adding an OH minus. I learned it this
way but the problem is everytime you introduce an OH minus you introduce another oxygen and
find yourself continuously going back and forth. So instead, what we're going to do when balancing
in basic conditions is do everything exactly as we see here where the only difference is at the
very end after we sum up all our reaction we're going to find every H+ and add OH minus to both
sides accordingly. What you're doing here is taking an H+, adding an OH minus and turning
it into water. The H+ is what gets cancelled out but the other side of the reaction is going to
have OH minus turning your acidic situation into a basic situation. But enough hypotheticals, let's
see what it looks like in a practice problem. Let's take a look at this example where we want to
complete and balance the reaction under basic conditions. Nickel 2 plus aqueous reacts with
chromium hydroxide solid to give us nickel solid plus chromate aqueous. First step,
find the oxidation number. Let's start with the easy ones. A monatomic ion with a charge of plus 2
has an oxidation number of plus 2, a solid metal has an oxidation number of zero. This is where
it gets interesting. For the chromium hydroxide instead of looking at the O and the H, you should
recognize hydroxide as the basic polyatomic ion with the charge of minus 1. Knowing that the minus
1 comes from a minus 2 plus 1. I point this out because chromium is our unknown. We're going to
find the oxidation number of chromium using the method I teach in the earlier video on finding
oxidation numbers as follows. Chromium is our x plus 3 hydroxide which will simplify minus 2
plus 1 as minus 1 is equal to the charge of this molecule, in this case zero. To simplify,
we have x minus 3 for 3 times negative 1 is equal to zero plus 3 to cancel out, plus 3, that gives
me x is equal to plus 3, chromium has an oxidation number of plus 3. Same thing for the product side.
Oxygen has an oxidation number of minus 2 and less by itself or in a peroxide. That leaves again as
our unknown. Chromium X plus 4 times negative 2 for the 4 oxygen atoms is equal to the total
charge of ion in this case minus 2. To simplify we have, X minus 8 for 4 times negative 2 is equal to
negative 2. Add 8 to both sides to cancel out, plus 8, plus 8, gives me x is equal to negative
2 plus 8 or positive 6 giving me an oxidation number for chromium of positive 6. We have our
oxidation numbers, let's use it to figure out what gets oxidized and what got reduced. Nickel plus 2
went down to zero or we can say nickel was reduced from 2 to zero for the reduction half reaction.
Oxygen hasn't change, hydrogen is not visible in the products but chromium went from plus 3 to plus
6. Chromium went up or up giving me oxidation. I know it got oxidized, I know it got reduced so
let's put our arrows down the middle and write out our half reactions. For the oxidation half,
we have Cr(OH)3, I made this for simplicity goes to CrO42 minus. For the reduction half reaction,
we have nickel 2 plus goes to nickel zero. Let's balance the half reaction starting with oxidation,
looking at the non-hydrogen or oxygen, in this case chromium. 1 chromium, 1 chromium,
it's balanced. Moving on the oxygen. We have 3 oxygen in the reactants from 3OH and 4 oxygens
in the products. Instead of balancing the oxygen as we have here, for example saying 3 times 4
is 12 we can't do that because if we change the number, the chromium's going to change. Instead
remember that once that your non-H's and O's are balanced you don't touch that. If you need
to change your oxygen or hydrogen number you do so for oxygen by adding h2o and hydrogen
adding H+. If we have 4 oxygens on the product and only 3 on the reactants we add 1 molecule
of H2O giving us the fourth oxygen. Oxygen is balanced so let's move on to hydrogen. We have
2 hydrogens in water and 3 hydrogens and 3 hydroxides for a total of 5 so we add plus
5 H+ in the products as a temporary acidic place holder for a reaction that is balanced
in basic conditions. Let's check our atoms. 2 hydrogens and 3 hydrogens is 5, 5 hydrogens,
1 chromium on each side, 1 oxygens and 3 oxygens give me 4, 4 oxygens our atoms are balanced. Let's
move on to the reduction half. Reduction is easy. We just have 1 nickel going to 1 nickel,
no oxygens, no hydrogens, atoms are balanced . Next we balance charge by adding electrons so
let's take the net charge. For oxidation, we have zero for the reactants and minus 2 plus
5 equals plus 3 for the products. We balance the charge by bringing the larger number down asking
3 minus what number is equal to the reactants zero. 3 minus 3 is equal to zero so we add plus
3 electrons to the products side. For reduction we have plus 2 goes to zero. How do we bring plus
2 down to zero? plus 2 minus 2 equals zero so we add plus 2 electrons to the reactants side. Our
reactants are balanced, now we wanna make sure that we have the same number of electrons in
the oxidation and reduction half reactions so they can cancel out. We have 3 electrons in oxidation,
2 electrons in reduction, let's find the least common multiple which is 6 because 3 times 2 is 6,
that means we multiply the entire oxidation half reaction by 2 and the reduction half reaction by
3. Let's distribute the numbers so we don't get confused. 2 times the invisible 1 is 2H2O and 2
chromium hydroxide to give us 2 chromates plus 2 times 5 is 10 each plus and 3 times 2 is 6
electrons. For reduction we multiply everything by 3. 3 times 2 is 6, 3 times the invisible 1 is 3,
and 3 again. Now that we have 6 electrons in the reactants on reduction and 6 electrons on
the products on oxidation, they can cancel out. Alright, we're kinda done, let's sum it all up.
Bring the arrow down the middle, what do we have on the reactant side? For oxidation we have 2
H2O plus 2 CrOH3. This is where the arrow is very important. We still have reactants but instead of
writing it here and having an arrow towards the right making it confusing, I'll just keep
writing everything on the left side, because what else do we have? 3 Nickels. Plus 3 Ni2+,
that's everything on the reactants, what do we have on the products? 2 chromate ions, 10
H pluses and 3 Nickel solid. Since we're dealing with basic conditions, don't forget to add this
step. The H+ is in our reaction tells us acidic conditions, H+ is a proton and proton is acid.
If we're dealing a base we have to neutralize every single proton with an OH minus and the
only way I'm justified adding an OH minus to every H+ is if I add to both side. That means,
for the 10 H plusses in my products I have to now add plus 10 OH minuses to the products and plus
10 OH minuses to the reactants. Now 10H+ and 10OH minus is actually 10 waters so let's combine them.
And this is where our notes to cancel duplicates comes into play because look at what you have.
On the reactants side we have 2 molecules, on the products side we have 10 water molecules. We can't
completely cancel it out but what we can say is 2 on the left will cancel out 2 of the 10 on the
right so 10 minus 2 is 8H2O remaining. Now, it's a little bit of a mess so if this is an exam rather
than the MCAT I highly recommend taking the next step and rewriting it one more time very neatly
very clearly for our final final reaction which we'll still check before our final reaction that
looks like this. 2 Chromium Hydroxide + 3 Nickel 2 + 10OH minus yield 2 chromate plus 3Nickel plus
8 H2O. It looks like we're done and we wanna be done but we have to check everything first.
Let's start with atoms, we have 2 chromiums, 2 chromiums. Next we have Oxygen so 3 hydroxides
times 2 is 6 Oxygen. Another 10 hydroxides gives me 16 Oxygen. The products we have 4 oxygen times
2 is 8 plus another 8 in water gives me 16 oxygen and sometimes when it's complicated I do write the
numbers down just to make sure I don't forget what I said. Hydrogen, we have 3 hydroxide times 2 is 6
plus another 10 hydroxides giving me 16, on the product side we have 2 times 8 which is also 16
and 3 Nickels, 3 Nickels. Our atoms are balanced, now let's check the charge. On the reactants
we have zero plus 3 times 2 is 6 plus 10 times negative 1 minus 10 for a total of plus 6 minus
10 which is equal to negative 4. On the products we have negative 2 times 2 which is negative 4,
nickel solid is zero, water is zero for a net of negative 4. Negative is equals negative 4 add
your phases back in and you're good to go. Be sure to join me in upcoming videos where
we look at limiting reagents percent and theoretically yields. You can find this
along with this entire reactions on stoichiometry video series practice quiz, and cheat sheet on my
website. First by giving this video a thumbs up, second write me on the comments what kind
of videos you want me to create next, and finally by visiting my website leah4sci.com/reactions.