MaChemGuy Help! The question said from Fe -> Fe^2+ @ 5:26 I don't understand why you have gone from Fe^2+ --> Fe^3+ @ 8:44 How have you concluded that the Fe2+ would be oxidised further with the addition of MnO4-??
for example 2 the method u did i understood, i wanted to know is that the only way of doing it. What I did was paused the video and did the example 2 i managed to get 65.2%. by working moles of fe2+ doing mass/mr and then dividing 5 etc i still managed to get 65.2%
Very good video, but isn't the final percentage 65.2%. I did the exact working and put it into the calculator a good 3/4 times and I got the answer 65.2%. Thanks
Thank you for your beatifull videos ....one question please : due to the presence of fe+2 ion , shouldnt we have green solution after the piece of iron reacted with acid ?...Thx
I don't understand how in the second example the molar ratio between Fe and MnO4 is 5:1. Should it not be 5:2 as when you try to cancel e- out and create the overall equation, I'm left with 5Fe + 2MnO4- +16H+ ---> 2Mn2+ + 8H20 MnO4- + 8H+ 5e- --> Mn2+ +4H2O (X2) Fe ---> Fe2+ + 2e- (X5) = 5Fe + 2 MnO4- +16H+ ---> 2Mn2+ + 8H20 Please let me know where I am going wrong :/
Hi, surely for the last problem, the equation would be 2Mn04- , as you would have used Fe-> Fe2+ + 2e- to write the overall equation?? Please can you confirm.Thank you
Hello Sir, I have a massive favour to ask, I have been stuck on a transition metal/redox question all afternoon and I've even looked at the mark scheme and still have no idea how to do it! I'm totally stumped on it! Is there any possibility that I could post it and you could explain it to me please? Sorry for asking such a big favour but I can't find anybody else that could explain it to me!
The question is 'Vanadium can exist in a number of different oxidation states. One compound of vanadium is ammonium vanadate(V) and this contains the ion VO3-. This can be reduced to V2+ in several steps, using zinc metal and aqueous sulphuric acid. (a) 25.0 cm3 of 0.100 mol dm-3 ammonium vanadate(V) is completely reduced to V2+(aq) using zinc and aqueous sulphuric acid. The resulting solution is titrated with 0.0500 mol dm-3 MnO4-(aq) and 30.0cm3 is required to oxidise the V2+(aq) back to VO3-(aq). The half equation for acidified MnO4-acting as an oxidising agent is shown below. MnO4- + 8H+ + 5e- → Mn2+ + 4H2O Show that the vanadium has changed oxidation state from +2 to +5 in this titration. [4 marks] (b) Suggest an equation for the oxidation of V2+(aq) to VO3-(aq) by MnO4-(aq) under acid conditions. [2 marks]' I believe it is off the June 2006 OCR Transition Metals paper.
+Eleanor Porter hi there. I think I've seen that question before. I'm on the train tomorrow so it will give me something to do :) I'll take a photo of my answer and post on Twitter. Do you follow me on that? If so let me know who you are and I'll tag you in the post if you like?
+Eleanor Porter Hi there, I've done the question and will post on Twitter later on. Just a quick explanation for you which may not come out in the photo. You're basically trying to prove a 5:3 mole ratio between the V2+ and MnO4 - You know there are 5 electrons involved in the MnO4- --> Mn2+ reaction (given in Q) so to bring about the +2 --> +5 change in the vanadium, there needs to be 3 electrons involved. The electron ratio reflects the mole ratio so if you can prove a 5:3 mole ratio the changes in oxidation number must be happening. You're given the concentration and volume for both chemicals in the titration so you just do two n = c x v calcs and then ratio them as if it's an empirical formula question Hope that all makes sense!!
ct123098 The iron in the ore is converted to Fe2+ with an acid, then the resultant solution of Fe2+ is then converted to Fe3+ in the titration with KMnO4. Two different reactions, the product of the first is the reactant in the second.
The answer to the last calculation is wrong, sorry! It's actually 65.2% Thanks for spotting my deliberate mistake Sharjeel!
I’ll be honest I Almost tripped up but thanks for the clarification 😅
your advice on visualising what's going on by drawing the diagrams is actually priceless
It's the only way!
in the first example why did u use f2+ to FE3+ because the question said fe to fe2+ right?
MaChemGuy Help!
The question said from Fe -> Fe^2+ @ 5:26
I don't understand why you have gone from Fe^2+ --> Fe^3+ @ 8:44
How have you concluded that the Fe2+ would be oxidised further with the addition of MnO4-??
The acid converts the Fe to Fe2+ then the MnO4- oxidises the Fe2+ to Fe3+
for example 2 the method u did i understood, i wanted to know is that the only way of doing it. What I did was paused the video and did the example 2 i managed to get 65.2%. by working moles of fe2+ doing mass/mr and then dividing 5 etc i still managed to get 65.2%
Very good video, but isn't the final percentage 65.2%. I did the exact working and put it into the calculator a good 3/4 times and I got the answer 65.2%. Thanks
shut up
I learned so much today bro, thank you. I love you.
Steady on!
Thank you very much! Commented that I wanted this topic earlier and here it is, thanks!!
I think all students find this topic difficult so a good choice! Uploading more examples now.
Thank you for your beatifull videos ....one question please : due to the presence of fe+2 ion , shouldnt we have green solution after the piece of iron reacted with acid ?...Thx
Where does the Potassium go?
It's a spectator ion so doesn't change
I don't understand how in the second example the molar ratio between Fe and MnO4 is 5:1. Should it not be 5:2 as when you try to cancel e- out and create the overall equation, I'm left with 5Fe + 2MnO4- +16H+ ---> 2Mn2+ + 8H20
MnO4- + 8H+ 5e- --> Mn2+ +4H2O (X2)
Fe ---> Fe2+ + 2e- (X5)
= 5Fe + 2 MnO4- +16H+ ---> 2Mn2+ + 8H20
Please let me know where I am going wrong :/
It's Fe2+ that is reacting with the MnO4 - (not Fe) and it is oxidised to Fe3+ so only one electron involved in that half equation hence the 5:1 ratio
+MaChemGuy ahh!! That makes a lot more sense... Thanks you :))
i don't quiet understand the balancing of the iron in the second part.
how do you know the charge on KMnO4 and Mn?
You’d be told in a question or given enough information to work them out
@@MaChemGuy thank you!
Hi, surely for the last problem, the equation would be 2Mn04- , as you would have used Fe-> Fe2+ + 2e- to write the overall equation?? Please can you confirm.Thank you
Sir i dont understand why you have taken the of moles of KMNO4- as the moles for MNO4- are they he same?
They are the same since in solution each mole of KMnO4 contains a mole of the MnO4- ion
@@MaChemGuy thx
How comes you times the moles by 1/5 on the first calculation but on the second calculation you multiplied it by 5? Sorry, I'm really confused :/
why is it that the fe2+ goes to fe3+. i get why it goes from fe to fe2+.
Hello Sir, I have a massive favour to ask, I have been stuck on a transition metal/redox question all afternoon and I've even looked at the mark scheme and still have no idea how to do it! I'm totally stumped on it! Is there any possibility that I could post it and you could explain it to me please? Sorry for asking such a big favour but I can't find anybody else that could explain it to me!
The question is
'Vanadium can exist in a number of different oxidation states. One compound of vanadium is ammonium vanadate(V) and this contains the ion VO3-. This can be reduced to V2+ in several steps, using zinc metal and aqueous sulphuric acid.
(a) 25.0 cm3 of 0.100 mol dm-3 ammonium vanadate(V) is completely reduced to V2+(aq) using zinc and aqueous sulphuric acid. The resulting solution is titrated with 0.0500 mol dm-3 MnO4-(aq) and 30.0cm3 is required to oxidise the V2+(aq) back to VO3-(aq).
The half equation for acidified MnO4-acting as an oxidising agent is shown below.
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Show that the vanadium has changed oxidation state from +2 to +5 in this
titration. [4 marks]
(b) Suggest an equation for the oxidation of V2+(aq) to VO3-(aq) by MnO4-(aq) under acid conditions. [2 marks]'
I believe it is off the June 2006 OCR Transition Metals paper.
+Eleanor Porter hi there. I think I've seen that question before. I'm on the train tomorrow so it will give me something to do :) I'll take a photo of my answer and post on Twitter. Do you follow me on that? If so let me know who you are and I'll tag you in the post if you like?
oh my gosh you are a legend! Thank you sooo much! I don't have twitter but someone i know can log in and show me your post tomorrow, thanks!
+Eleanor Porter Cool - I'll put your name to it so your friend spots it. It'll be around lunchtime. Happy to help :)
+Eleanor Porter Hi there,
I've done the question and will post on Twitter later on. Just a quick explanation for you which may not come out in the photo.
You're basically trying to prove a 5:3 mole ratio between the V2+ and MnO4 -
You know there are 5 electrons involved in the MnO4- --> Mn2+ reaction (given in Q) so to bring about the +2 --> +5 change in the vanadium, there needs to be 3 electrons involved. The electron ratio reflects the mole ratio so if you can prove a 5:3 mole ratio the changes in oxidation number must be happening.
You're given the concentration and volume for both chemicals in the titration so you just do two n = c x v calcs and then ratio them as if it's an empirical formula question
Hope that all makes sense!!
I the last calculation it says that fe atoms are converted to fe2+ ions, yet in the redox equation fe2+ are converted to fe3+ ?
ct123098 The iron in the ore is converted to Fe2+ with an acid, then the resultant solution of Fe2+ is then converted to Fe3+ in the titration with KMnO4. Two different reactions, the product of the first is the reactant in the second.
Please i need your help i can calculate the concentration of the mixture of 2.5g of iodine dissolved in 1.67g of potassium iodide?
thanku sir. very helpful
Good video.
thanks man, you helped a lot!
Thank you
where the fuck beeen this video for god sake !?!?!?!? i wished i have found them earlier :((((((((((((((((((
hahaha you must be about 23 now
Very helpful! :)
I know a Ben Hodge, what a coincidence !
That's strange, my chemistry teacher sounds a lot like you too..
What are the chances!?!?!
+MaChemGuy +BenHodge woww LOOL what a crazy coincidence. you're so lucky to have/have had such a brilliant chemistry teacher, Ben.
+Umme Waseem I know :)