could you explain how you found the equations for the lithium/lithium cobalt oxide electrodes , and why they were written in that way? I would have thought the negative lithium electrode would mean reduction is happening , hence Li+ + e- --> Li ? and the other way around for the top one but that's clearly incorrect.
The negative electrode is negatively charged because, at that electrode, electrons are being produced. The negative electrode is always where oxidation occurs, and that's the half equation that has the lower electrode potential. Conversely, the positive electrode is the half reaction with the larger electrode potential. This reaction proceeds in the reduction direction, 'using up' electrons which is a good way to remember why it's considered to be the positive electrode
@gracechen2412 Great question! In electrochemical cells, it depends on the type of cell you're looking at. For a galvanic (or voltaic) cell, the negative electrode is the anode because that's where oxidation occurs, and electrons flow out. Reduction happens at the positive electrode, which is the cathode. However, in an electrolytic cell, the negative electrode is the cathode (where reduction occurs), and the positive electrode is the anode (where oxidation occurs). For A-Level Chemistry (e.g. AQA), you only need to focus on galvanic/voltaic cells - electrolysis isn't required for most exam boards. Hope that clears it up!
for question 8d- did you add the two iron half equations?( so you used 3 half equations?) and how did you know that the final product would be fe3+ hexaquaion and not the fe2+ one?
You could do it that way, but if you know the starting chemical and the final chemical you can devise your own half equation, which is what I did. I knew which iron product it would be because the vanadium electrode Potential is large enough that it can oxidise the iron from 0 to + 2 And then from +2 to plus 3 (a total of 3 e lost) I knew this because vanadium's electrode potential is larger than the value for both iron half Equations
@Charlotte-hf5jz how do you mean? Which question? 8d? In a way you don't need to combine them... you need to know iron starts as Fe 0 and then turns into Fe3+. Then balance that with 3 e- on the right
I don't understand part 9a. Both electrode potentials of iodine and water is more positive than lithium's. So won't lithium react the same way with iodine?
That's true, but you were not asked to consider the bottom half equation. You were only asked why water wasn't used as the electrolyte. The job of the electrolyte is to complete the circuit and maintain charge balance. It should do this by allowing the ions dissolved in it to move and cancel the effect of charges that have been generated/used. If the electrolyte reacts, it will get used up and won't be able to perform the job it needs to.
Hi for 8d why did you write Fe (H2O6) 2 + as Fe + H2O for recatant side instead of Fe+ H20+ (like as an ion u knwo they are boths ions with positive charge) ?
@offlinetvfriendsfandom6991 the reactants are from the right-hand side of the top half equation that I've shown in blue. Fe and H2O are the reactants there. The complex ion is an intermediate chemical
@@chemistrytutor But isn’t +5 to + 4 still oxidation and not reduction as the charge has got more positive meaning a loss of electron and therefore oxidation?
I'm presuming you mean part (e) Since the substance the [Co(H2O)6]3+ is on the left hand side of its half equation (its an oxidising agent) then the Fe(H2O)6]2+ *has* to be on the right hand side of its half equation (and be acting as a reducing agent). If they were both on the same side they'd both be wanting to gain electrons
It’s because the reaction can further occur to break the product of the Fe2+ complex down into the other Fe3+ complex (both reactions go left ways) so you can combine them together as long as you balance the electrons, hope that helps!
If you meant 8d... I chose both the 1st and the 4th. The Fe in the first equation is oxidised to Fe2+. This is then a reactant in the 4th equation and can be oxidised further to Fe3+. So both oxidation occur and ultimately Fe loses 3 electrons to form Fe3+
could you explain how you found the equations for the lithium/lithium cobalt oxide electrodes , and why they were written in that way? I would have thought the negative lithium electrode would mean reduction is happening , hence Li+ + e- --> Li ? and the other way around for the top one but that's clearly incorrect.
The negative electrode is negatively charged because, at that electrode, electrons are being produced. The negative electrode is always where oxidation occurs, and that's the half equation that has the lower electrode potential. Conversely, the positive electrode is the half reaction with the larger electrode potential. This reaction proceeds in the reduction direction, 'using up' electrons which is a good way to remember why it's considered to be the positive electrode
@@chemistrytutor Does it mean "negative electrode" is an anode? (I thought it's cathode where reduction happened)
@gracechen2412 Great question! In electrochemical cells, it depends on the type of cell you're looking at. For a galvanic (or voltaic) cell, the negative electrode is the anode because that's where oxidation occurs, and electrons flow out. Reduction happens at the positive electrode, which is the cathode. However, in an electrolytic cell, the negative electrode is the cathode (where reduction occurs), and the positive electrode is the anode (where oxidation occurs). For A-Level Chemistry (e.g. AQA), you only need to focus on galvanic/voltaic cells - electrolysis isn't required for most exam boards. Hope that clears it up!
I had just done these questions and hour before this was uploaded ahaha ,, the explanations were really helpful, thank you!
What a coincidence!
Glad it was useful 👍
Thanks for this. One of my hardest topics!
Very welcome 😀
for question 8d- did you add the two iron half equations?( so you used 3 half equations?) and how did you know that the final product would be fe3+ hexaquaion and not the fe2+ one?
You could do it that way, but if you know the starting chemical and the final chemical you can devise your own half equation, which is what I did.
I knew which iron product it would be because the vanadium electrode Potential is large enough that it can oxidise the iron from 0 to + 2
And then from +2 to plus 3 (a total of 3 e lost)
I knew this because vanadium's electrode potential is larger than the value for both iron half Equations
@@chemistrytutor thank you!
@@dno.16 😀👍
@@chemistrytutori don’t understand this? how can i combine the two fe half equations?
@Charlotte-hf5jz how do you mean? Which question? 8d? In a way you don't need to combine them... you need to know iron starts as Fe 0 and then turns into Fe3+. Then balance that with 3 e- on the right
very educative can ya do one for inorganic too (mcq)
I've done some periodicity MCQs.
Here is a full links document:
drive.google.com/file/d/19yYWdu3bczjCyeSC-NFp6_V781IRwcvU/view?usp=drivesdk
I don't understand part 9a. Both electrode potentials of iodine and water is more positive than lithium's. So won't lithium react the same way with iodine?
That's true, but you were not asked to consider the bottom half equation. You were only asked why water wasn't used as the electrolyte. The job of the electrolyte is to complete the circuit and maintain charge balance. It should do this by allowing the ions dissolved in it to move and cancel the effect of charges that have been generated/used. If the electrolyte reacts, it will get used up and won't be able to perform the job it needs to.
@@chemistrytutor Oh okay, thank you!
Thank you sir, you give me hope about this topic and the subject overall. Thank you
Very kind of you to say so. Keep going at it!
@@chemistrytutor thank you
Really good video. Thanks sir
Thank you 😊
Hi sir, are you planning to do an explanation video on this topic?
Yes definitely. Probably not before January though
Hi for 8d why did you write Fe (H2O6) 2 + as Fe + H2O for recatant side instead of Fe+ H20+ (like as an ion u knwo they are boths ions with positive charge) ?
@offlinetvfriendsfandom6991 the reactants are from the right-hand side of the top half equation that I've shown in blue. Fe and H2O are the reactants there. The complex ion is an intermediate chemical
I’m confused on question 8d. If VO2+ goes to VO^2+ doesn’t that mean that it’s being oxidised and not reduced as it goes from a +1 to +2 charge
It's actually going from +5 to +4
V -2 -2 = +1 so V is +5 at first
V -2 = +2 so V is +4
@@chemistrytutor But isn’t +5 to + 4 still oxidation and not reduction as the charge has got more positive meaning a loss of electron and therefore oxidation?
@@noveawalters4339 if somethings charge gets less positive it must have gained a negative election... reduction
@@chemistrytutor oh yes, that makes sense! Sorry redox isn’t my best topic so I get easily confused. Thank you!!
@@noveawalters4339 no problem at all.
I've made a video about redox if that's useful?
th-cam.com/video/bjtO_PEgd64/w-d-xo.html
I don’t understand why you picked the fourth one down rather than the first one (the iron equation).
I'm presuming you mean part (e)
Since the substance the [Co(H2O)6]3+ is on the left hand side of its half equation (its an oxidising agent) then the Fe(H2O)6]2+ *has* to be on the right hand side of its half equation (and be acting as a reducing agent). If they were both on the same side they'd both be wanting to gain electrons
It’s because the reaction can further occur to break the product of the Fe2+ complex down into the other Fe3+ complex (both reactions go left ways) so you can combine them together as long as you balance the electrons, hope that helps!
If you meant 8d... I chose both the 1st and the 4th.
The Fe in the first equation is oxidised to Fe2+. This is then a reactant in the 4th equation and can be oxidised further to Fe3+.
So both oxidation occur and ultimately Fe loses 3 electrons to form Fe3+
Underrateddddd
Very welcome! 😀
Thanks
Very welcome 😃
can you do six mark questions please
A variety of exam questions are in my plans!