Finding Ecell for a Reaction

[@BboySuperFreeze]

This one video cleared everything up for me since you showed both sides of spontaneous and non spontaneous. Thanks a lot!

[@tanjiro57]

The best Chemistry teacher...You should teach chemistry to my teacher!

[@runekid2]

exam in 40 mins thanks for saving me

[@albertoalvarez8986]

How'd you do

[@abskamo7483]

Now it's been 4 years for you since that test do you even remember how it was 😂

[@masghar14]

To keep people from getting confused, you could also point out that if you left the Fe equation as is (instead of flipping it) you could have done 0.34-(-0.44) which would give the same number, because that's how I was taught it: Ereduction-Eoxidation

[@girishbhardwaj2137]

masghar14

[@20k54n4]

Thanx!

[@yushiliang1]

The equation E cell = E cathode - E anode works like this: 0.34 - (- 0.44) = 0.78 , which is the same as what CHEMIST Nate did. He just flipped the sign to make it the 'oxidation potential'

[@ailurophileluv]

so, if we flipped the sign, we just need to add it of? but if we dont, we just used the E cell=E cathode-E anode ??

[@IamYou26]

​@@ailurophileluvyes

[@658magnus]

Hey, I like your videos but I've got one question. Given the formula E cell = E cathode - E anode, shouldn't it have been 0.34 - 0.44 = - 0.1 V ?

[@NumairQureshi]

Thanks for the awesome video. I love your teaching style.

[@lrissunflower]

using copper and magnesium and both their reduction potentials, i did eCathode - eAnode : 0.34--2.37 = 2.71, spontaneous, is this correct?

[@mro112m]

Quick question; Are all galvanic cells spontaneous? Since we are basically guessing and if I chose something that is negative and make that my answer as the galvanic cell is not spontaneous; how can this be correct?

[@jackw1014]

Just a student here, so someone correct me if I'm wrong, but.... I think the reason that you always choose the spontaneous one is that a non-spontaneous reaction, by definition, doesn't happen.* Say we mix all this iron and copper and they all decompose in the solution. Now we have a solution that has Fe(s), Cu(s), Fe2+ and Cu2+. Only the spontaneous reaction is going to happen. The non-spontaneous reaction is non-spontaneous, so it's not going to happen. That's why Mr. Nate uses the spontaneity test to figure out which way to write the reactions. *Of course, if we were adding extra energy to the system, the nonspontaneous reaction could occur, but we aren't doing that; we're just mixing stuff and seeing what happens. Bonus: I also noticed that you don't actually have to guess. When you look at your table of standard reduction potentials and find your two values, you just have to ask yourself, "Which one of these values do I need to change the sign so that when I add them I get a positive value?" Then you know which half-reaction to flip.

[@MrMoonfirespam]

This video helped so much. Keep it up!

[@poseidon4587]

Cell potentials are intensive properties so how are you adding them like that?

[@ritrade]

Really nice explanation

[@siminling5678]

This video helps me a lot. Thank you!!

[@Scoxn]

Awesome drawing! Your videos are great :D

[@princezgeekchic3690]

I thought Ecell equalled Ecathod-Eanode. My professor didn't mention adding cell potentials.

[@MeCicci]

So helpful for studying for my chemistry exam! Thanks so much

[@alexbarrelet9149]

Really helpful, thanks for making it clear!

[@gon2407]

So simple so clear thank you

[@anotherdoseyt]

What could cause the cell potential to increase?

[@ShubhraShandilya29]

Wonderful explanation!! Thanks!

[@vinodkumarramana3668]

Sir why we are adding while physics use voltage difference

[@gon2407]

I had the same question

[@chemistNATE]

That’s because here we are *adding two equations together* like Hess’ Law … we purposefully flipped one equation from Oxidation to Reduction, then added. You COULD subtract the two oxidation half-reaction voltages from each other, as you suggest.

[@JohnSmith-zf1lq]

Why did you add the half reactions? I thought you subtracted them?

[@elizabethroberts4860]

You have to add them in order to make the overall equation. hope this helps x

[@B.A.Gondal]

So when do you use the equation, E_cell = E_red - E_oxi ?

[@BryanJL]

I use that equation instead of doing what he's doing which is kind of confusing.

[@resilience679]

exactly what im saying he's not doing it right the second one is suppose to 0.10 v he's wrong

[@B.A.Gondal]

ok I got it now. He's doing it right but just using a different method. You can either look at the table and flip the signs according to what you have been given or you could just look at the table and see which value is more positive and use that as a E_red in your equation i described above without worrying about flipping the signs. For example, the cooper/zinc example he did, you could have just looked at the values and see which is more positive. Obviously it is +0.34 which is more positive than -0.44 right. SO just use 0.34 as a E-red and -0.44 as E-oxi and you will get the same results. If the values were, say, -0.34 and -0.44 which is more positive? obviously -0.44 and now use that as your E_red. That's it.

[@resilience679]

yea what you have is correct because from your example you would still get the same +0.10 as i did but in his example it was completely off because he flipped the sign then he added and so instead of getting a positive results he gets a negative one

[@GoodNewsForStrangers]

Doesn't it have something to do with reactivity of atoms as well?

[@omaralami6580]

I think when voltage is negative the reaction is actually spontaneous and when the rxn is pos. It's non spontaneous. My notes say other wise but my exam study guide proves my point

[@alexanderlotozynski882]

+Omar Alami No, I think you are confusing this with Gibb's free energy, or deltaG. When deltaG is negative, your reaction is spontaneous, and vice versa.

[@sohanroy7833]

awesome explanation!!!

[@katikar1]

yes, you balance the reactions like you would any other balanced equation, and coefficient affects everything in the second equation. sorry this is a month late.

[@TamaEnergy]

but i thought u always drew the reductive cell one the right? or do exam Q's not do this

[@chemistNATE]

I've had some kids tell me in 1-on-1 tutoring that their teacher required the reduction on the right as well. I'm not aware of an official rule for this, so I did it whatever way felt natural at the time. But you're right that it would have been better (i.e. more helpful for everyone) to have done it that way ...

[@nadirbaitsaleem7270]

Thank you, this is a useful trick!

[@AcousticLvr22]

great video! really helpful

[@putturdilip5109]

Thank you so much guruji...🙏👍

[@peclesswonder]

thanks this was really helpful!!

[@manlinpan2705]

Thank you so much for helping me!!!!

[@nicolecusack2184]

omg this video was so helpful

[@abhimanyubhakt0407]

Love from India ❤❤

[@baybarsoral7491]

Very informative.

[@saharhdp702]

Very helpful. Thank you

[@electricfeel2012]

chemistNATE for prezidezzy of the United Stezzy!

[@faal1805]

very helpful, thank you !

[@fatimarahimi0861]

Thank you so much Mr :)

[@AaronT04]

Thank you 🙏

[@freddyoverflow3165]

Thanks men, that help me a lot of :3

[@kaleemadamjee2764]

but he doesnt explain which one to use of the two equations???

[@chemistNATE]

Do you mean, Which half cell reactions? That will depend on the chemicals you are told you are starting with.

[@marianaverdugo4377]

You're the best!

[@Mrsdespoina757]

I don't know what ages are your target group here(i mean if it is high school it could be fine),but your video is very confussing and utterly wrong considering the IUPAC directions.All (half)reactions conserning the Eo value are written as reductions in the international bibliography to avoid confusion and make clear that the value of E is just the voltage diffrence between the element in question and the SHE(Standard Hydrogen Electrode).My point is that by taking a ''practical'' approach of just flipping the half-reactions and changing your signs you are making more difficult for a student to understand the theory behind these excersises(e.g. the half-reactions are just a chemical equilibrium,the use of Eo = Ered - Eoc , etc) Anyway,this is not a hostile comment I am just trying to clear up some misunderstandings that took me a while to solve in the University exactly because of these kind of practices by my high school teachers.

[@DarkLord61636]

This had confused me for a while as well, as EVERYWHERE I'd look online, I'd see the method above, whereas in my lectures we do Ered-EOx. I've been trying to understand why there's such different info everywhere, and I realize why now x_x It took a concept I THOUGHT I understood just fine, and made it more confusing

[@MrBlash93]

Thank you sir

[@captivatingcurios]

THANK YOU .

[@omaralami6580]

Ecell < 0 spontaneous Ecell > 0 non spontaneous

[@madison2302]

thought exactly the same

[@mehrajudean3032]

thank u sir

[@hamdan4251]

yeah thx for that

[@weilin5534]

Dont u have to consider the E of water

[@alishah-gg8uh]

YOU CONFUSED ME

[@mohammadhanif6609]

answer for this emf is -0.10v. not -0.71v buddy

[@ma.jeremelougane4115]

I think something is wrong with the calculation of the spontaneous example. Cause 0.34-(0.44)= -0.1v

[@sflm9277]

Ma. Jereme Lou Gane The calculation is correct: you have to add the reactions and the corresponding potentials, not subtract them.

[@francischileshe8615]

Got an exam in 20min

[@chemistNATE]

Oof

[@coastalflourishing]

Electrochemistry is backassward

[@prantasaha4904]

soooooo wasting my time

[@rahibmehdi3548]

Samjh ni aaya

[@gon2407]

Thanks Sir