Our goal is to balance this

redox reaction in acid. And before we get

into the steps, let’s talk about

the fact that this is a redox reaction by

assigning some oxidation states. And so we start over here

with the dichromate anion. And we know that oxygen has an

oxidation state of negative 2. We have seven oxygens. So negative 2 times 7 gives me

negative 14 as our total here. We know that the total

for the entire anion has to equal

negative 2, which is the charge on the

dichromate anion. Therefore, we must have plus 12

for all of our chromiums here. So plus 12 and minus

14 give us negative 2. Since we have two chromiums,

each one must be plus 6. And so that’s the oxidation

state for chromium here. We go over here to

the chloride anion. So the charge is negative

1, so our oxidation state is negative 1. Chromium ion over here. So plus 3. And then, finally, chlorine over

here, so oxidation state of 0. So if we look at

chlorine, chlorine went from an oxidation

state of negative 1 to an oxidation state of 0. That’s an increase in

the oxidation state. Therefore, chlorine

was oxidized here. Look at chromium. Chromium went from

plus 6 to plus 3. That’s a decrease in

the oxidation state, or a reduction in

the oxidation state. Therefore, chromium was reduced. And so this is a redox reaction

because something is oxidized and something is reduced. In terms of balancing

it, our first step is to write the

different half reactions. And so we’re going to break

those into an oxidation half reaction and a

reduction half reaction. So let’s go ahead and

get some space down here. And let’s go ahead and

write our half reactions. And so we had a chloride anion

going to chlorine like that. And we said that this was our

oxidation half reaction, so put that way over

here on the right. So that’s out oxidation. Our reduction half

reaction involved chromium. So we had the chromate anion

here, so Cr2O7 2 minus, going to chromium

3 plus like that. So this is our

reduction half reaction. So that’s step one. Write the different

half reactions. Step two. Balance the atoms other

than oxygen and hydrogen. And so, if you look at

our first half reaction, we have one chlorine on the left

and two chlorines on the right. So we need to balance it

by putting a 2 over here on the left like that. We go down here to the

reduction half reaction, and we have two chromiums on the

left and only one on the right. And so we have to put a 2

right here to balance it. So step two is done. Step three. Balance the oxygens

by adding water. So if I look at my

oxidation half reaction, there are no oxygens

so I don’t need to worry about doing anything

to this half reaction at the moment. I go down to the

reduction half reaction, and I do have to

balance my oxygens. So if I go over

here and I can see that I have seven

oxygens on the left side and none on the right. And so I need to do that

by adding the water. And since I have seven

oxygens on the left, I need seven oxygens

on the right. So I’m going to go ahead and

add seven water molecules. And that now gives

me seven oxygens on the right of

my half reaction. So that’s this step,

step three, right here. Step four. Balance the hydrogens

by adding some protons. So let me go ahead and

I’ll use red for this. So step four. Balance the hydrogens

by adding protons here. Once again, the

oxidation half reaction, we don’t have to do

anything, because we don’t have to balance

oxygen or hydrogen here. But, again, we go down to

our reduction half reaction and we have the oxygens

balanced by adding water, but by adding water, now

we have some hydrogens on the right side. So we can see we have a total of

14 hydrogens on the right side. So 7 times 2. And so we’re going to balance

that by adding protons. And so we need to add protons

to the left side of our half reaction. So we need to add 14. So 7 times 2 is 14. So we go ahead and add 14 H plus

to the left side of our half reaction. So step four is done. Step five. Balance the charges

by adding electrons. So let’s get some more

space here, first of all. OK. So we’re going to balance the

charges by adding electrons. So first let’s

analyze what kinds of charges that we have here. So we’ll start with the top

oxidation half reaction. So we have the chloride anion,

which is a negative 1 charge. And we have two of them. So we have two negative charges. Notice, these are

not oxidation states. That’s what gets people

confused sometimes. These are charges. Over here on the right,

we have no charges. So that’s a neutral

chlorine molecule here. So we have two negative

charges on the left and 0 for a charge on the right. So we need to figure out how to

balance those charges by adding electrons, and so it makes

sense that we would have to add two electrons

to the right over here, because that now gives us a

total charge of negative 2 on the right. So that’s one way to

think about it– just getting these

numbers equal here. This negative 2 right

here and this negative 2. Another way to do it

would be, of course, you know the electrons have

to go on the right side because this is the

oxidation half reaction. And one of the ways to

remember that– LEO the lion. So Loss of Electrons

is Oxidation. And so if you’re

losing electrons they must go on the product

side of your half reaction. And so our top half reaction,

our oxidation half reaction, is now balanced. Let’s go down to our

reduction half reaction. So LEO the lion goes GER. So Gain of Electrons

is Reduction. So we already know we’re

going to have to add electrons to the reactant side

of this half reaction. Let’s see if we can figure

out how many electrons we’re going to need to use. So we have 14 positive

charges from the protons. And then we have

two negative charges from the dichromate anion here. So we have 14 positive charges

and two negative charges which gives us a total

of 12 positive charges on the left side. On the right side, we

have a chromium ion. So this is a charge of 3

plus, and I have two of them. So 2 times positive 3

gives me positive 6. So I have positive 6 on the

right side of my half reaction. So I have positive 12 on the

left, positive 6 on the right. I need to add some electrons

to balance out that charge. I already know I’m going to

add them to the reactant side. I know that from LEO

the lion goes GER. Or I could just

think about the fact that, if I have 12

positive charges, I would need to add six

negative charges to get me to a total charge of plus 6. So I need to add six electrons

to the left side over here. So let me go plus six

electrons like that. And now, we have the

charges balanced. So this step is done. So step five is done. We move on to step six. So make the number

of electrons equal. So what does that mean? Well, let’s focus in

here on the electrons that we just added to

our half reactions. So if I go back up to here,

we added two electrons to the oxidation

half reaction, and we had six for the

reduction half reaction. But we know that that number

has to be the exact same number because the electrons that are

lost from our oxidation half reaction are the exact

same electrons that are gained in our

reduction half reaction. So that is why we have to make

these electrons equal in terms of the number. And so the way to do

that would, of course, be to multiply my first

half reaction by 3. Because if I multiply my

first half reaction by 3, that would give me a total

of six electrons, which is what we’re looking for,

making the number of electrons equal. So let’s go ahead and do that. I’m going to rewrite

our first half reaction. I’m going to multiply

everything in our half reaction through by 3, so

everything in parentheses. So if I take 3 and multiply

that by 2 chloride anions, I would, of course, get 6. So we have six chloride

anions like that. And then the three would go

in front of the chlorine. So I have 3Cl2, and then

3 times two electrons gives me six

electrons like that. OK. Let’s go ahead and rewrite

our reduction half reaction because we have a lot

of stuff going on here. Some I’m just going to

rewrite exactly what we have. Six electrons plus 14 protons

plus the dichromate anion like that. And then I have two chromium

ions and seven the waters. OK. So now I have my

two half reactions. I’ve made the number

of electrons equal. And we’re ready

for the last step. We just take our

two half reactions and we’re going to add

them back together, and that’s going to give us

our overall balanced redox reaction. So let’s go ahead

say we did this. And let’s go ahead and

get some more room here so we can add those

half reactions. So all I’m going to do

is just take everything on the reactant side

and add that together. So let’s go ahead

and just rewrite everything on our reactant side. So we have 6Cl minus and then

six electrons, 14 protons. We have the

dichromate anion here. And then I’m going to take

everything on my product side. So I’m going to take

all of this stuff right here and put that

on my product side. So I have 3Cl2 plus six

electrons plus 2 chromium ions plus 7 water

molecules like that. And so now it’s just a

little bit easier, I find, for students to see that your

electrons are going to cancel, right? You have these six electrons

on the reactant side. You have these six electrons

on the product side. So you can go ahead

and take those out. And we’re left with

our final answer. So you can go ahead

and rewrite it if you want to make

it look better. We have 6Cl minus plus

14 H plus plus Cr2O7 2 minus yields 3Cl2 plus

2Cr 3 plus plus 7 H2O. And this should be

our final answer. So I always like to

box my final answer, so it just makes it easier

for your instructor to grade. And the nice thing

about redox reactions is, when you’re finished, you

can always check yourself. Because you know that

you need to balance both the atoms and the charge. So let’s go ahead and

check that real fast. Let’s first start with chlorine. So we have six

chlorines on the left, then over here on the right

we have 3 times 2, which is 6. So the chlorines balance. Hydrogen. We have 14 on the left,

and then 7 times 2 gives us 14 on the right. The chromiums. We have two on the left

and then over here we have two on the right. Oxygens. Seven oxygens on the left

and then over here this seven applies to this oxygen. And so the atoms are

balanced properly. Let’s next check

charge because you have to have the correct charge. Let’s see, we have

6 negatives charges. I’ll just put minus 6 here. And then I have 14 positives. And then I have a

negative 2 right here. So when I add all those

up, I get a total of plus 6 on the left side. So 14 minus 8. Over here on the right

side, the only charge is this chromium ion. It’s 3 plus, and I

have two of them. So 2 times 3 gives me plus 6. And so the charge

balance as well. And so this is the

correct answer. It balances both in terms of

atoms and in terms of charge.

Will you be organizing these MCAT videos in a playlist?

Very Good video. Thanks!!!!!!!!!!

bro, you just saved my life.

its just not the same without sal's voice.

Thanks anyway

Boss boss boss that all what I can say !

OMG thank you! Btw I thought you have Pakistani ethnicity because of you're last name than I went on you're Wikipedia and it says that you're actually a Bengali, and I was like OMG, I am a bengali too and I am so proud because you're so smart and you're helping so many students and you are one of us. Haha.

thankx a lot.

Thanks a lot now i get it 😀

Skipper did an excellent job in the place of Salman

god bless you

What. The. Heck.

If cl (-1) goes to cl (0) isn't that a reduction and not an oxidation?????

Thank you! it was really clear and easy to understand 🙂

I needed this clarification. Makes a lot more sense now. I like this guy better than the other guy

Don't we have in the end on the left side 2*7 O=14O –> on the left side we have only 7*O=7O so don't we have to multiply the H2O and H+ with 2 that we get it completly balanced?

stop going soo fast dude!

I was out with my bio class on a field trip so I missed this in school, and the trip made me sick therefor i lost even more days of school. But thanks to you i know i can catch up again, this is my first time on your channel hope you have more vids like this one! thanx

Thank you!

Thank you so much!

Sir, how did the chemical equation Cr2O7 ^

~~2 —~~> Cr^3 become the reduction reaction? It also lost electrons, right?This is just what I needed!

why is that in medicine version ??

How did you get oxidation number 0 for Cl2 ??

This tutorial helped me alot. Thanks.

You aren't sal !

BRING SALL BACKK LMAOO

I dont understand, why didn't he come up with the charge of 7H2O in the reduction half reaction…. He just did got the charge of 2Cr wich was 6+, why not the one of h2o

guys you don't know how thankful I am right now . thank you so very much our book added two extra steps which made it complicated

This just clarified a lot for me, thanks for sharing!

Very helpful! Thank you

Thank you so much! My chemistry teacher confused me so much with redox reactions and you clarified!

OIL RIG

Oxidation is loss and reduction is gain. (Of electrons)

God bless y'all

Because Cl2 is covalently bonded, there is no exchange of electrons and the charge of each Cl is zero.

Awesome. khan Academy, u teach brilliant

you saved my life, thank you so so much!!!

The Best Explanation!!♥♥♥

Thanks a lot.

My teachers confused me in those last two steps!!! But now I got it!!!

Thank you sooooooooooo much!

nicely done and very simple, got it

If you pretend, it sounds like Taco from The League giving the lesson.

This is a really clear step-by-step explanation. Thank you!

The electron part was confusing me. Thanks for clearing it up!

wow you are the best

This was beautiful, I actua

This was brilliantly explained. Thank you! 😀

Why is cromime 12?

Thank god for this video, finally understand it!

Thanks a lot.

very good explanation, steady watching in 2016

Khan Academy is always there for me when my teacher isn't. Which is always!

Very helpful

gr..8 dude 🙂

Why don't the elements in the beginning have an oxidation number of 0?

omg i wish i saw this vedio earlier

I will be very thankful if you give a video on balancing the redox reaction using oxidation no. method of acids and bases.

please give me this as soon as possible.

Wow, every NCERT teacher seems to confuse students on the last two steps, judging from the comments..

Thank you so much.

thnkx sir

great explanation as all ways.

OIL RIG

Oxidation is Loss Reduction is Gain ( of electrons)

In the last step why do you have to balance the electrons?

That's really a wonderful job.keep going. Thank u soooo…. much

bless

This method is not working for me for all problems. Try

ClO3- + I2 = Cl- + IO3-

This method is useless to me for that problem

Oh my God thank you so much

THANK YOUUUU!!! T___T

Great video and explanation but the 360p hurts my eyes

SIR U ARE 2 G00D

Thank you

thanks a ton

Putting half equations together used to be pain in the ass. Now it's a bliss for my brain.

Never thought it was this simple. Thanks man

You explained this better than my chem teacher… subbed and liked

Thanks! Ur the best

Really cant thank you people enough!

wow you have the skill to break something down.thanks saved me

VERY NICE!!!!!

This is a gift for everyone who's interested to learn. Thank you so much.

I love you.

oh wow the method my teacher taught us is much much simpler, basically just shortcuts through most of the legwork, i'll try and write it out if you're looking for something more brief

1. write out oxidation states, as you do, as well as adding 1/2 in front of things like Cl2 etc

2. count how much each thing got reduced or oxidized; switch the numbers, and multiply them into the equation

3. count the total charge on each side, by counting, and add H+ or OH- to balance it

4. count the oxygens, and add water where appropriate, and then count the H's to check

Man u r awesome.Thanks to khan academy.

Thank you for saving my chemistry grade!!!!!!!!

You never let me down!!❤️

Khan academy is best way for learning from youtube i got first in class from khan academy thanks ☺️☺️☺️☺️

Its just LEO says Ger not Leo the Lion says Ger.

reduction is a reduction of charge (gaining electrons, gaining negative charges)

top shagger

ger –gold experience requiem