how did you, in 12 mins, take something so scary and intimidating and break it down into easy to understand steps that I can actually do in my exam and aaaaa my mind is blown thank you so much
I don't understand how you were able to construct the half equation. (before doing the + H20 then + H+ etc) simply how do we know to start with Mno4- ---------> Mn 2+ and Fe2+ -------------> Fe3+
Does the oxidation of Fe always goes from 2+ to 3+ if originally it was 2+ as it is with the sulphate solution? If not or in other situations where there is unknown ions, what is the procedure to work out the change in oxidation states for unknown ions which are being oxidised in order to balance the equation?
I have a question, if you could respond that will help me very greatly! How did you know MnO4- would be reduced to Mn2+ and what the half equations were? In the question it didn't state how much it got reduced. This really is the only bottleneck for me, I don't know what the half equations are since surely it could be oxidized or reduced further? Thank you for taking your time to reply back if you do so! :)
+Zift™ I reply to all comments I get through so that's lucky heh! That is a good question too. It always reduces to Mn2+ as this is the most stable reduced form of Mn. Mn4+ can form but is readily reduced to Mn2+ with any reducing agent present. The equation is a half equation and can be balanced using water, H+ ions and electrons in the normal way. I have a video on this that may be of use - ua-cam.com/video/uFNAPXQXXNc/v-deo.html Hope this helps!
one quick question, how do I know when to multiply the moles of chemical in flask by 10 and when not to? I know this comes up in exam questions but I never know when to multiply by 10 and when I shouldn't. Thanks
+Sam Kvon Yes if you are working out the moles or mass or concentration of a solution that was made in a flask and a portion of this was used in the titration then you multiply up. e.g. if you titrated 25cm3 of a solution that originally came from a solution in a 250cm3 flask, you need to multiply by 10 to find out the number of moles in 250cm3.
When writing the half equation for potassium manganate why don’t we add the potassium in the equation why is it just manganate and how do we know which is getting oxidised and which is getting reduced? Thanks
As a follow-up question can this be used to work out the concentration of any reducing agent? Provided you know the half equations so you know the molar ratios of the Manganate with the reducing agent?
Can you tell me why the color change for this reaction is colorless to pink ? I dont get it ...? Cause MnO4- ions are purple and they reduce to Mn2+ which is usually colorless... Am i missing something ?
Brilliant video!!!!!! sir I have a question. An steel wire (steel is mostly iron) of mass 3.225g was dissolved in dilute sulphuric acid and the solution made up to 250 cm3. A 25 cm3 portion of this solution was further acidified and titrated against potassium dichromate (VI) of concentration 0.031 moldm-3. The volume required was 31.0 cm3. Calculate the percentage of iron in the steel wire.
+Kother Alhumaidi Thanks! Your answer is 99.75%. I have the working written down on paper but I can't put the image on youtube. I can share it with you on twitter if you follow me @allerytutors Cheers!
how did you, in 12 mins, take something so scary and intimidating and break it down into easy to understand steps that I can actually do in my exam and aaaaa my mind is blown thank you so much
You are so awesome, I use these videos to study for Caribbean Advanced Proficient Exams, the syllabus is really similar and you're a great help
I don't understand how you were able to construct the half equation. (before doing the + H20 then + H+ etc) simply how do we know to start with Mno4- ---------> Mn 2+ and Fe2+ -------------> Fe3+
thank you so much this was very easy to understand and very helpful, keep doing what your doing.
You're welcome!
Does the oxidation of Fe always goes from 2+ to 3+ if originally it was 2+ as it is with the sulphate solution? If not or in other situations where there is unknown ions, what is the procedure to work out the change in oxidation states for unknown ions which are being oxidised in order to balance the equation?
Gosh, this topic is really tough! But with your videos I am slowly getting there! Any tips on clearing it up? Thanks!
Excellent Video Sir
+zannatul Zan Thanks very much! Please share the vid.
ur saving my alevels thank you
I have a question, if you could respond that will help me very greatly!
How did you know MnO4- would be reduced to Mn2+ and what the half equations were? In the question it didn't state how much it got reduced. This really is the only bottleneck for me, I don't know what the half equations are since surely it could be oxidized or reduced further? Thank you for taking your time to reply back if you do so! :)
+Zift™ I reply to all comments I get through so that's lucky heh! That is a good question too. It always reduces to Mn2+ as this is the most stable reduced form of Mn. Mn4+ can form but is readily reduced to Mn2+ with any reducing agent present. The equation is a half equation and can be balanced using water, H+ ions and electrons in the normal way. I have a video on this that may be of use - ua-cam.com/video/uFNAPXQXXNc/v-deo.html Hope this helps!
+A Level Chemistry Revision Videos by Allery Tutors Thank you so much! :)
+Zift™ No problem!
one quick question, how do I know when to multiply the moles of chemical in flask by 10 and when not to? I know this comes up in exam questions but I never know when to multiply by 10 and when I shouldn't. Thanks
+Sam Kvon Yes if you are working out the moles or mass or concentration of a solution that was made in a flask and a portion of this was used in the titration then you multiply up. e.g. if you titrated 25cm3 of a solution that originally came from a solution in a 250cm3 flask, you need to multiply by 10 to find out the number of moles in 250cm3.
Are all these videos for the AQA spec?
Hi, thanks for the video. How would the situation change with waters of crystallisation, because some exam qs ive seen use this, eg. FeSO4.xH20?
Same dude
Thank you.This was very helpful
legend
great video!!
How do we know which is oxidised and which is reduced??
Thank you for making this video!!
How would you know MnO4^- was the charge on the ion in the half equation?
Thanks
When writing the half equation for potassium manganate why don’t we add the potassium in the equation why is it just manganate and how do we know which is getting oxidised and which is getting reduced? Thanks
Or is potassium manganate always an oxidising agent
I love you, thanks for the amazing and very useful video m8!
how did you get mno4-
Hey Allery thank you for your videos :) Please could you tell me if iron is the reducing agent because potassium manganate is the oxidising agent?
You're welcome. Yes iron would be.
As a follow-up question can this be used to work out the concentration of any reducing agent? Provided you know the half equations so you know the molar ratios of the Manganate with the reducing agent?
Can you tell me why the color change for this reaction is colorless to pink ? I dont get it ...? Cause MnO4- ions are purple and they reduce to Mn2+ which is usually colorless... Am i missing something ?
Brilliant video!!!!!!
sir I have a question.
An steel wire (steel is mostly iron) of mass 3.225g was dissolved in dilute sulphuric acid and the solution made up to 250 cm3. A 25 cm3 portion of this solution was further acidified and titrated against potassium dichromate (VI) of concentration 0.031 moldm-3. The volume required was 31.0 cm3. Calculate the percentage of iron in the steel wire.
+Kother Alhumaidi Thanks! Your answer is 99.75%. I have the working written down on paper but I can't put the image on youtube. I can share it with you on twitter if you follow me @allerytutors Cheers!
@@AlleryChemistry Hi is there a way you could show me the answer too? I would really like to know.
Thank you, this was so helpful! Just one thing- what did you say the colour change is for this titration? I missed it
how can you tell whether it is MnO42- or MnO4-
I am 2 years late lol , but i am confused how did you know to use manganate ion and Fe , why not work out sulphate ion and potassium ?
cm3 is ml and dm3 is liters. moldm-3 is mol/L or M
Where did the potassium go ????
The potassium is a spectator ion and doesn't show in ionic equations as it doesn't react in the reaction. It's chemically unchanged
Very goooood
Thank you! Cheers!
You need a bigger white board
Thanks so much I finally understand it! Subbed :)
Thank you! :)
Thankyou soo much sir
I really thought you'ld actually demonstrate the experiment. Still helpful though. Thanks
are you mr.hegarty? :/