This is one of the topics I find hardest to understand, with all the half equations and full equations, but I feel more confident after watching this. I'll need to try some examples to get fully comfortable with it.
I owe you my entire education oh my goodness, I’m home schooling AS & A2 level and your videos have me confident on things I didn’t even understand in GCSES when I was in school, thankyou so much Eliot, I couldn’t have done it without all this help 🙏
Thanks for your excellent work of teaching Chemistry via Skye. I have one remarks in this lesson is that you forgot to emphasise that the half reactions of redox involved equilibrium. You wrote the half equations of redox as if they were one-way reactions. Rather then reversible. This simple mistake can make the whole topic much more difficult to understand than is necessary. All the best, Dr Youssef El Aziz
Thank you so much for these amazing videos! Has really helped me get through a bulk of revision! Although it isn't related to redox questions in particular, I have noticed in past papers that they ask you to write an 'overall equation'. Do you use redox reactions or simply cancel out like compounds? Thank you!
Thank You for this video , i just have 2 questions. Oxidation is loss of electrons which will form a positive ion right . 15 mins into the video cr2o7(2-) is reduced to 2cr(3+). Can you please explain to me how this is reduction and not oxidation? Also isnt it oxidation as the oxidation number increased from -2 to +3?
Ishtiaq Hussain Hi! You are correct that oxidation is loss of electrons. And you are correct that SOMETIMES this will lead to a positive ion being formed. The better way to look at oxidation, and reduction, is to think about the terms in relation to the oxidation state of the elements in question. This brings me to your point... You are looking at the overall charge of the dichromate ion being 2- and asking how, when it is forming a 3+ ion, this is reduction and not oxidation? Well, we aren't using the entire dichromate ion. We don't care so much for the oxygen, instead we are looking at the chromium. In the dichomate ion, EACH chromium has an oxidation state of +7 and when the REDUCTION occurs, this chromium is REDUCED to +3. Does that make any sense?
Ishtiaq Hussain E Rintoul In the dichromate ion, doesn't EACH chromium have an oxidation state of +6, rather than +7? As the formula is Cr2o7^2-, so the oxygen, represented by an 'o', has an oxidation state of -14, since the overall ion is -2, the Cr2 must also be +12 as this will equal -2 overall. But because we want to know each Cr we divide 12 by 2 to give +6. Please can you tell me where you have got the +7 from?
Thank you very much this is helpful! But suppose you have to balance something like this: MnO4- + H+ + no2- ----> Mn2+ + h20 + No3- You can't use the same method by adding the h2o to balance the oxygens right?
As usual these videos are so helpful however, I am confused at an oxidation state, at 7:22 The Nitrogen in NH4 +. I believed that when the hydrogen was not named first in the compound it had a charge of -1 not +1 like in NaH. Then the nitrogen would have an oxidation state of +5 not -3. This is probably just me, however if you could clear this up it would be much appreciated.
Hello sir, fantastic video as always! i just have one question though, you stated the definition of reduction is: where an element/compound gains electron(s). So,at minute 14:42 why is the di-chromate ion undergoing reduction when it's actually losing electrons (because it's going from 2- to 3+). Thank you very much for your time and effort sir!
Rangahatimuhmon Thanks for the kind words! I think you might be a bit confused - I shall try and clear some things up. I'm going to assume that you've got the -2 from the fact the the dichromate ion has a charge of -2. This would sort of make sense, but remember that the dichromate ion contains oxygen too. Because of that, we look to work out the oxidation state of just the chromium. The overall charge of the dichromate is -2. Each oxygen has an oxidation state of -2. There are 7 oxygens. 7 x -2 = -14. Because the overall charge is -2, the 2 chromiums must account for the +12 that's missing. That means that each chromium is in fact +6. In that sense, it is the CHROMIUM that is being reduced, from +6 to +3. Does that make sense at all?
E Rintoul Yes sir, thanks a lot, So would it be correct to say that cr2 =12 and therefore you divide it by 2 to get a single cr which would give you +6, whereas you do not divide the 2cr as it is already on its own (meaning cr will be +3)? P.s i have some questions regarding the video you did on halogens, can i fire away? ;) Thanks a lot my man
E Rintoul You know my whole unit 2 book are notes derived from your videos, I haven't listened in class (because i don't like my teachers teaching method), nor have i read the aqa book. Your videos are extremely beneficial! Ok, on to my questions: In your Halogens video, you stated as you go down group 7, the oxidising power decreases. In other words, the ability to lose an electron decreases in strength as you go down group 7. Why is that? if the size of group 7 atoms become larger as you go down the group, wouldn't that mean an increase in distance from and nucleus and therefore and increase in shells and shielding? Wouldn't that therefore mean the ability to lose an electron will be EASIER as the outer electrons are further away from the nucleus. Edit: I just realized right now as i was typing this to you that the reason for this could be that the amount of protons that are increasing (as you go down the group) overpower the increase in shells and shielding? Either way, please clarify!
Rangahatimuhmon Well thank you again for the kind words! Out of curiosity, what is it about your teacher's methods that you don't like? Right I see what you're saying. And what you're saying kind of makes sense. Except for a tiny, tiny detail. Oxidation is loss electrons. Correct. As size increases, the outer shell electrons are further away and so the attraction is weaker. This means that the outer electrons are lost more easily. Correct. But that isn't what I said. I said that the oxidising power decreases. That is the ability to oxidise. As such, an oxidizing agent is reduced. Another way to look at it would be to say that down group 7, the elements are less easily reduced. Does that help at all?
Hello Sir, Im going to be retaking Chem2 this year however your videos are now in new specifications. Would there be a big effect for me by revising your videos??
+Bartholomew Pang Nope. The videos have stayed almost exactly the same - the only real difference at Year 1 (what was AS) is that they now do more on equilibria (what you do at A2 in fact) and mass spec is different.
Alex Ruding Unfortunately, you are expected to know them. To be fair, there aren't a huge number (hydrogen, oxygen, group 7, group 1 etc.). They normally give you compounds and then expect you to work out the oxidation state of an element in that compound.
Thank you for another great video, prior to watching the vid and even after it, I get +3e on both sides rather than +6e, Is this also correct or have I fallen short somewhere?
F.J Vinzenz I honestly don't know. I would stay WELL away from writing a minus. The reason I did it was to show what is happening. In terms of convention, electrons are written as being gained on the opposite side as this makes it much easier to then combine half-equations.
mr biggle might be too late but I think he’s accounting for a single chromium atom, oxygen(-2*7)= -14 Potassium(1*2)=2 the total of these would give you -12 for a chromium molecule divide by two for a single chromium atom.. think and the charge would be +6 to balance the charges ... I think x
AM I INCORRECT Oxidation is loss of electrons Reduction is gain of electrons So at 12:10 aren't they both oxidation. Hear me out. The first one is oxidation because of the reasons he stated. HOWEVER the second one goes from a 2- charge to a 3+ charge. This means it went from 2 additional electrons to 3 less electrons. This means that electrons were lost. So it must be oxidation. If I am incorrect please explain why as I am confused.
Hi, I have my AS Chemistry EMPA coming up after the Easter holidays...I don't suppose theres any chance you could make a video tutorial to help us prepare for these practicals and empa papers?
xxpinkladiesxx In all honesty, probably not. Due to the sporadic nature of the papers, it would be very difficult to make a "one size fits all" video, if you get what I mean. If you have any questions though, please fire away!
Your videos are a great help, and I have them on my iPod so that I can watch them on the bus before school. You could team up with myGCSEscience and so an A level version of everything if you teamed up with some other teachers.
pstasiw I'm glad to hear it! Yeah, I guess that's a possibility. I thought that I had seen that the GCSEscience guy had started charging for his videos though...
Could you do/do you have a video of balancing more complex redox reactions? With additional reactants and products that aren’t necessarily subject to redox?
Amazing video again Just to clarify will we be expected to know the oxidation states or do you get given them in the exam. PS. Have you completed a lesson for mass spectrometry and infrared spectroscopy Thanks really very helpful
moneyhoneyhoney You are expected to know some oxidation states - oxygen (remember that it is -1 in peroxide), hydrogen, group 1 metals, group 2 metals, halogens etc. The questions will normally want you to work out the oxidation state of nitrogen or sulphur for example in compounds of hydrogen and oxygen etc. Hope that helps!
What ISA prep is that... I know that the AQA ISA deadline has long passed! In all honesty, as much as I would love to do everything, I've just not got the time! Tonight I'm going to try and get a couple of videos done, but I don't think I'm going to be able to squeeze anything ISA related in, I'm afraid! By all means ask any questions through my channel and I'll try to help!
Danyaal Qureshi Yup! The ones that you need to know are the ones of chlorine and water and chlorine and sodium hydroxide. There's no reason that they couldn't give you another equation that was disproportionation though and get you to work things out based on that.
Hi! I love your videos as always. However, I do have a question, when balancing the half equation you figured out the oxidation state of the compound but you didn’t use it. However, my teacher said that you use the oxidation states to determine how many electrons you would add to make them balance, and then to balance the charges you add the hydrogen ions, and then to balance the equation you add water molecules. I was wondering if that was right and just another way to balance redox reactions, or if that is actually wrong and my teacher is just confused. Thank you so much for this video as it seems much simpler than what she was explaining and now I’m a little confused because I don’t know what method is correct.
I don't know what to put as my name hi! I would always suggest balancing the specie being reduced/oxidised then oxygen using water, then the hydrogen using H+ ions and finally the charge using electrons. I guess there are other ways to do it but I've always found this method to work!
can someone please help me im confused between AQA and caie as i am a caie student and i want to learn from these videos is the content also same for me pls can someone explain
Of course! Potassium is +1 (this you just need to know - it's in group 1) Oxygen is -2 (again you need to know this) 2x1 (from the K2) + 7x-2 (from the oxygen) = -12 This means that the 2 chromiums must together add up to +12 (for a compound, the sum of the oxidation states must equal its charge). From this, each chromium must be +6. Does that make any sense?
Of course! Potassium is +1 (this you just need to know - it's in group 1) Oxygen is -2 (again you need to know this) 2x1 (from the K2) + 7x-2 (from the oxygen) = -12 This means that the 2 chromiums must together add up to +12 (for a compound, the sum of the oxidation states must equal its charge). From this, each chromium must be +6. Does that make any sense?
Boro Jen xX It perhaps wasn't mentioned. I can't cover everything and AQA will more than likely throw some bits in to trick you at points! In H2O2, oxygen is -1 so don't always think of it as -2!
Hi thanks for these videos!! When you balance the reduction equation, you balanced the oxygen first, what happens if there is no oxygen in the equation? Some other methods balance the electrons first but I don't really understand that.
Hi, Laura! It all depends on the half-equation given. For example, balancing Cr2O72- --> Cr3+ would require water, hydrogen ions and electrons. However, Fe3+ --> Fe2+ would only require electrons. Balance in this order... 1) balance the species present (Cr2O72- --> 2Cr3+ - note that the Cr has been balanced by doubling it) 2) deal with any unbalanced oxygens using water on the other side 3) balance hydrogens using hydrogen ions 4) balance charges using electrons If one of these steps is missing, it doesn't matter; just follow the others and you'll be fine! If you're still struggling, post me an example of what's confusing you and I'll try to help!
Any chance you could do Group 7, Halogens and Group 2 Alkaline E.M Thanks in advance, Mainly how to write the simplest ionic equations and what we need to remember for the exam!.
Great video, you're amazing. Many thanks. Vanadium is prepared by reducing vanadium(V) oxide with silicon. Write an equation for this reaction. (1 mark) Is it possible for you to help me with this question?
Thanks for the kind words! Without looking, I would guess that it's going to be a case where the silicon is causing displacement to occur thereby leaving the vanadium on its own, making silicon dioxide in the process. In terms of an equation... 2V2O5 + 5Si --> 5SiO2 + 4V That looks like it could be about right?
Thank you so much for the videos mate! I have a teacher who does one side of the course and I don't understand his teaching methods at all :/ so cheers for these
