I’m a newly qualified chemistry teacher and I watch your videos to aid my teaching, I like it the fact that you can simplify a complicated concept and made it accessible to everyone.
Thanks man. Your videos are really awesome. I got a C in chemistry but without your videos I would have very likely got an E instead. Keep up the good work bro.
Hey man, I used your videos a lot throughout year 10-12 to study chemistry. I am now almost done with my first year in university, and I still find myself coming back here to revise. Thanks so much for your videos
My teacher said it is a good idea to think of the weakest reducing agent as the strongest oxidising agent; and to think of the weakest oxidising agent as the strongest reducing agent.
Thank you so much for your videos. My school teachers are amazing but there are some things I just can't get my head around (electrochem being one) and these really help. It's also really good that you go through example questions at the end!
Please do a video going through batteries and fuel cells, especially the lithium-ion battery and hydrogen-oxygen fuel cell, I cant make good notes and I have no idea what exactly I'm meant to know
At 8:09 Where you've done the conventional representation and you've written Br2, 2Br-, does it matter what order you put them in? Would you have to write the most reduced species first? Thank you!!
which factors affect emf values and why? I've seen questions about changing the concentration of species and their effect on emf values but I don't really understand them.
In the cell representation for hydrogen and magnesium , do we still have to keep hydrogen on the left. Or for conventional cell representation is it always oxidation on the left and reduction on the right.
@Eliot Rintoul is the equation not already balanced without having the 4Ag? like even if we remove the 4 on both the Ag ion and solid, the equation still remains balanced.
Emilia Au+ gets reduced bc the value of the electrode potential is more positive and therefore it is more likely to undergo reduction but bc its reduced it therefore acts as an oxidising agent on H2O causing it undergo oxidation - the value of the electrode potential for H2O is more negative therefore it’s more likely to undergo oxidation Baso the Au+ gets reduced by taking an electron from the H2O which then gets oxidised as a result so therefore the Au+ oxidises the H2O Hope this helps xx
They asked to show a reaction when Fe2+ reacts with O2, thus leaving us with two equations, that we can choose from, that he has already marked out. +1.23 is more positive than +0.77 and - 0.44, thus it's confirmed that O2 will be the one that'll get reduced. This means that Fe2+ has to be oxidised. Fe2+ + 2e - ---> Fe is reduction. We want oxidation to take place. So, Fe---> Fe2+ + 2e - would be fine. But then, they want us to show how Fe2+ reacts with O2, not Fe. So now we are obliged to take the equation Fe3+ + e - ---> Fe2+. But this is reduction again, and we've already conceived that Fe is going to be oxidised. So we reverse the equation, leading us to Fe2+ ---> Fe3+ + e -. Now this how you make it work. Write down the two equations first. 1. O2 + 4H+ + 4e - ---> 2H2O 2. Fe2+ ---> Fe3+ + e - These two equations have to be combined to give the equation of the overall reaction. O2 + 4H+ + 4e- + Fe2+ ---> 2H2O + Fe3+ + e-. As you can see, they're not balanced. In order to do that, we simply have to make sure the electrons (and hydrogens if present) on the reactant and product side both cancel out (priority to cancel out is given to electrons). 1. O2 + 4H+ + 4e - ---> 2H2O 2. 4Fe2+ ---> 4Fe3+ + 4e - Equation 1 has 4 electrons, thus we multiply equation 2 by 4 in order to completely cancel out the electrons. Hydrogens are only present on the reactant side and not the product side, thus they can't be cancelled. Adding up the equation after the cancellation of electrons, we get this: O2 + 4H+ + 4Fe2+ ---> 4Fe3+ + 2H2O. As you can see the equation is balanced. This explanation is as thorough as possible, thus it's lengthiness. If you still don't understand it, perhaps you need more practice on how to add up equations.
Where you've given the example of Zn and Mg2+ . Magnesium is more negative so you could say it's been oxidised therfore its on the left hand side. Therefore you'd say the zinc has been reduced so it's on the right hand side. Use EMF = RHS - LHS and get -0.76 - (-2.38) = +1.62 WHY IS THIS NOT FEASIBLE??
I would always work out EMF using reduction - oxidation. I'm not sure what you're saying though...? Are you asking why the reaction of the zinc ions with magnesium metal is feasible?
Zinc is unable to react with magnesium ions, however the opposite reaction will work. The reason is that the more negative electrode potential (in this case magnesium) will show oxidation, therefore magnesium being oxidised to magnesium ions and the zinc will show the reduction. The opposite way around cannot work as magnesium would be reduced but would have the more negative electrode potential.
When you're dealing with elements in the gas phase, liquid or aqueous phase! So only time you don't is when you have a solid metal that can act as an electrode!
Only put in state symbols if they ask you because if you get them wrong they will mark you down. I think the only time you're asked to put in state symbols is the transition metal stuff.
At 21:34 I don’t understand why you chose Cl2 (g) as the answer to 5c instead of HOCl (aq). Surely the weakest reducing agent is the strongest oxidising agent, and the strongest oxidising agent is the one that’s most likely to be reduced itself, and the one that’s most likely to be reduced itself is the one with the highest electrode potential? That would mean that the answer is either HOCl (aq) or Cl2 (g). Now like I said above, the weakest reducing agent is the one that’s most likely to be reduced itself, surely that would be HOCl (aq) right??? Someone please correct me if I’m wrong. Thank you!
A reducing agent has to donate electrons for it to be a reducing agent. The HOCl cannot donate any electrons, it can only accept them. Hence it acts as an oxidising agent. However the Cl2 can donate electrons and is hence the reducing agent in the reaction (dispite it not being a very good one). I'm not 100% sure if this is the correct explaination for this but that's how it makes sense to me... I hope this helps :)
half 67 half 150 I do apologise! Must be terrible for you having to listen to such drivel in a free video. But thanks for the advice, I seriously do appreciate it.
This guy is so funny, but more importantly is saving my chemistry grade!
I’m a newly qualified chemistry teacher and I watch your videos to aid my teaching, I like it the fact that you can simplify a complicated concept and made it accessible to everyone.
I love you man
Thanks man. Your videos are really awesome. I got a C in chemistry but without your videos I would have very likely got an E instead. Keep up the good work bro.
Hey man, I used your videos a lot throughout year 10-12 to study chemistry. I am now almost done with my first year in university, and I still find myself coming back here to revise. Thanks so much for your videos
My teacher said it is a good idea to think of the weakest reducing agent as the strongest oxidising agent; and to think of the weakest oxidising agent as the strongest reducing agent.
does it count as revision if I watch videos instead of reading textbooks? the info just doesn't stick when I read...
Matthew Briggs ofc it counts as revisions it’s what I do
A lot of the things aren’t included tho, like in this topic we need to know about the hydrogen fuel cell, for example
Use the CGP revision guide textbook as well to make sure u cover all the content, they break it down really well 👍
5:14 remember it as NO PRoblem
Thank you so much for your videos. My school teachers are amazing but there are some things I just can't get my head around (electrochem being one) and these really help. It's also really good that you go through example questions at the end!
you were sent from heaven
Your laugh is so cute! Also passed chemistry because of you. Thanks love you.
a bit sus
@@higherhoping trust me
Thanks just in time for the Christmas break
thankyou so much exams are coming uppp and this and the last video really helps!!
13:33
lmaoooo
Mate this is a life saver
this guy is a life saver
Please do a video going through batteries and fuel cells, especially the lithium-ion battery and hydrogen-oxygen fuel cell, I cant make good notes and I have no idea what exactly I'm meant to know
You're actually so fab!!!
Hi man, could you do a video going over fuel cells and batteries please?
Thank you
Meek Mill does Alevel chemistry? WOW, I didn't know that. :D
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Great!
At 8:09 Where you've done the conventional representation and you've written Br2, 2Br-, does it matter what order you put them in? Would you have to write the most reduced species first? Thank you!!
Naoki Gunan that's what I thought
you have to write the most oxidised species first, the one which lost the most electrons, in this case it was Br2
which factors affect emf values and why? I've seen questions about changing the concentration of species and their effect on emf values but I don't really understand them.
same i don't understand it either. please go over it
In the cell representation for hydrogen and magnesium , do we still have to keep hydrogen on the left. Or for conventional cell representation is it always oxidation on the left and reduction on the right.
In SHE you put the hydrogen ALWAYS on the legt
Thanks for this
Great video! Could you do the final part of this topic which is Electrochemical Cells about batteries etc please?
Thank youuu
You're a life saver I swearrr
At Q5 a) 19:04 In the conventional representation, shouldnt It be Pt | H2 | 2H+ || Fe2+ | Fe ? to balance it out?
Yes it would
No you don't need to balance in conventional cell representation
very unsure about the explanation of the last question.
Does anyone get the Zn plus Mg2+ thing?
@Eliot Rintoul is the equation not already balanced without having the 4Ag? like even if we remove the 4 on both the Ag ion and solid, the equation still remains balanced.
don't you need to put the atoms' and ions' state when writing their conventional representation?
you do need to!
28:30 why did he put Au+ oxidises H2O? Wouldn’t it reduce since it’s the more positive? I’m bit confused here, someone please help me
Emilia Au+ gets reduced bc the value of the electrode potential is more positive and therefore it is more likely to undergo reduction but bc its reduced it therefore acts as an oxidising agent on H2O causing it undergo oxidation - the value of the electrode potential for H2O is more negative therefore it’s more likely to undergo oxidation
Baso the Au+ gets reduced by taking an electron from the H2O which then gets oxidised as a result so therefore the Au+ oxidises the H2O
Hope this helps xx
Anna Murphy yes it does, thank you so much! :)
32:19
That doesn't really make sense to me at all. Fe loses two electrons and oxygen gains two electrons, so shouldn't it be Fe2+ + 2e- forming Fe?
They asked to show a reaction when Fe2+ reacts with O2, thus leaving us with two equations, that we can choose from, that he has already marked out. +1.23 is more positive than +0.77 and - 0.44, thus it's confirmed that O2 will be the one that'll get reduced. This means that Fe2+ has to be oxidised. Fe2+ + 2e - ---> Fe is reduction. We want oxidation to take place. So, Fe---> Fe2+ + 2e - would be fine. But then, they want us to show how Fe2+ reacts with O2, not Fe. So now we are obliged to take the equation Fe3+ + e - ---> Fe2+. But this is reduction again, and we've already conceived that Fe is going to be oxidised. So we reverse the equation, leading us to Fe2+ ---> Fe3+ + e -.
Now this how you make it work. Write down the two equations first.
1. O2 + 4H+ + 4e - ---> 2H2O
2. Fe2+ ---> Fe3+ + e -
These two equations have to be combined to give the equation of the overall reaction. O2 + 4H+ + 4e- + Fe2+ ---> 2H2O + Fe3+ + e-. As you can see, they're not balanced. In order to do that, we simply have to make sure the electrons (and hydrogens if present) on the reactant and product side both cancel out (priority to cancel out is given to electrons).
1. O2 + 4H+ + 4e - ---> 2H2O
2. 4Fe2+ ---> 4Fe3+ + 4e -
Equation 1 has 4 electrons, thus we multiply equation 2 by 4 in order to completely cancel out the electrons. Hydrogens are only present on the reactant side and not the product side, thus they can't be cancelled. Adding up the equation after the cancellation of electrons, we get this:
O2 + 4H+ + 4Fe2+ ---> 4Fe3+ + 2H2O.
As you can see the equation is balanced. This explanation is as thorough as possible, thus it's lengthiness. If you still don't understand it, perhaps you need more practice on how to add up equations.
@@stablewinner Thanks a lot man, I get it now:)
@@stablewinner I just didn't really read the question properly so I was a bit confused, but I really appreciate your explanation!
Where you've given the example of Zn and Mg2+ . Magnesium is more negative so you could say it's been oxidised therfore its on the left hand side. Therefore you'd say the zinc has been reduced so it's on the right hand side. Use EMF = RHS - LHS and get -0.76 - (-2.38) = +1.62 WHY IS THIS NOT FEASIBLE??
Time 12:17
yeah Zn and Mg will react, but Zn and Mg2+ won't
hmm I dont get it :/
I would always work out EMF using reduction - oxidation. I'm not sure what you're saying though...? Are you asking why the reaction of the zinc ions with magnesium metal is feasible?
Zinc is unable to react with magnesium ions, however the opposite reaction will work. The reason is that the more negative electrode potential (in this case magnesium) will show oxidation, therefore magnesium being oxidised to magnesium ions and the zinc will show the reduction. The opposite way around cannot work as magnesium would be reduced but would have the more negative electrode potential.
really dont understand the last question. But great video
How do you know when to add platinum electrode
When you're dealing with elements in the gas phase, liquid or aqueous phase! So only time you don't is when you have a solid metal that can act as an electrode!
Kehkashan Ansari ahhh okay so it's needed when there's no solid involved ? cheers !
surely the weakest oxidising agent at 23:43 would be Fe as Fe2+ would want to be reduced to Fe
Agree
Joey kelliher just checked the mark scheme, Fe2+ is the right answer wtaf
@@jakehill2603 thinks it's because fe is not an oxidising agent at all, whereas fe2+ is an oxidising agent, it's just the weakest one.
18:00
Do we get marked down if we don't include state symbols?
Only put in state symbols if they ask you because if you get them wrong they will mark you down. I think the only time you're asked to put in state symbols is the transition metal stuff.
At 21:34 I don’t understand why you chose Cl2 (g) as the answer to 5c instead of HOCl (aq).
Surely the weakest reducing agent is the strongest oxidising agent, and the strongest oxidising agent is the one that’s most likely to be reduced itself, and the one that’s most likely to be reduced itself is the one with the highest electrode potential? That would mean that the answer is either HOCl (aq) or Cl2 (g).
Now like I said above, the weakest reducing agent is the one that’s most likely to be reduced itself, surely that would be HOCl (aq) right???
Someone please correct me if I’m wrong. Thank you!
A reducing agent has to donate electrons for it to be a reducing agent. The HOCl cannot donate any electrons, it can only accept them. Hence it acts as an oxidising agent.
However the Cl2 can donate electrons and is hence the reducing agent in the reaction (dispite it not being a very good one).
I'm not 100% sure if this is the correct explaination for this but that's how it makes sense to me... I hope this helps :)
Jodie Green Sounds about right, thanks!
9:19 Why does he write Mg/Mg2+ but in the table it is written Mg2+/Mg.
I am very confused.
And same for Ag/Ag+.
Overwatch Moments - TH3 B4NDIT it's all part of conventional representation. The most oxidised specie goes nearest the salt bridge.
Ahhh okay. I just found your channel and got my exam tomorrow. You need to put some ads out. I got all my mates watching you.
Why wouldn't Zn and Mg work? it has a +ve emf value so in theory it should work no?
nevermind its the magnesium ions that wouldn't be reduced so zinc metal + magnesium ions isn't feasible but what about zinc ions and magnesium metal?
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you babble alot in the beginning, get to the point. A tip for advice
half 67 half 150 I do apologise! Must be terrible for you having to listen to such drivel in a free video. But thanks for the advice, I seriously do appreciate it.
he teaches a2 topics that take a couple weeks to learn in school in less than an hour and ur tryna say he doesnt get to the point ur stupid af