But if you have all of the positive ions disappearing and only half of the negative ones, then the process should stop rather quickly because the build up of negative charge should block further process ... or what am I missing here?
what is left behind? sulfate ions boned to Hydrogen ions ... so the contrentraions of H2SO4 increases until sulfate ions begin getting discharged at the anode?
shouldn’t the anode form peroxydisulfuric acid/peroxydisulfate ions? Which would then get hydrolized to sulfate ions and hydrogen peroxide? (Aswell as oxgen ofc) but I guess with dilute you mean less than 50% concentration.
In our syllabus we only deal with really dilute sulfuric acid where you only see oxygen as the product at the anode. Things change when stuff gets concentrated!
excellent presentation, I liked the chart. Im not a chemistry student, how did you get that 4 in that 4OH-. what does it mean. I often see numbers before the symbols.
Jonas Bolander - this is actually way more complicated than is required for gcse and based on the ions you know about you’re best off thinking about it as 2h+ and SO4(2-). In reality H2SO4 is a mixture of ionic and covalent character and is most likely to only lose one H+ to form H+ and HSO4(-) but since the sulphate ion is very unlikely to discharge it’s not really worth worrying about too much! Sorry to not really answer your question - but hope that justifies why I’ve done it this way! :)
This video is wrong. Ionization of water occured at very small rate (considered not happening) and by considering E° what actually happened at Anode(Oxidation) was 2H2O _> O2 + 4H+ + 4e- with E°red =1.23
The reality is that both the OH- and the H2O will be oxidised, but the IGCSE syllabus that we do focuses on the OH- (even though there is a higher concentration of H2O and so that process is more likely!)
The big numbers in front (the coefficients) just tell you the ratio in which things react. So for every one oxygen molecule that is made, 4 OH- ions are removed from the solution. :)
Thank you soooooooo much i have an exame tommorow i thought i will fail but after i saw ur videos now i am sure i will pass thank youuu❤️❤️❤️❤️❤️❤️
Thanks ma'am for this amazing explanation
Concept Crystal Cleared , Thanks Ma'am for this awesome explanation!!
You're welcome :)
Just got the concept . I am from India and I like your video, voice and the way you expressed all the reactions.❤️
effective class... Want more of them 🙂🙂🙂
very informative vid my dear
Amazing mam....
HELPED SOO MUCH THX
Very useful
Thanks mam😊 I’m from India 🇮🇳
ohhh tnxxx
Thanks for the Awesome explanation mam 😊
My pleasure 😊
Solved my doubt of fundamentals of Electrolysis
The concentration of the acid will increse or decrease
Thank-you
Is there any difference when we use Cu electrode for the electrolysis?
U r grt
Thanks
But if you have all of the positive ions disappearing and only half of the negative ones, then the process should stop rather quickly because the build up of negative charge should block further process ... or what am I missing here?
you saved me
Video was amazing and electrolysis became like a piece of cake 🎂
Thank you so much 😀
what is left behind? sulfate ions boned to Hydrogen ions ... so the contrentraions of H2SO4 increases until sulfate ions begin getting discharged at the anode?
w explanation
What should the ph value of diluted sulphuric acid
What should concentration of Sulphuric acid, how much diluted?
shouldn’t the anode form peroxydisulfuric acid/peroxydisulfate ions? Which would then get hydrolized to sulfate ions and hydrogen peroxide? (Aswell as oxgen ofc) but I guess with dilute you mean less than 50% concentration.
In our syllabus we only deal with really dilute sulfuric acid where you only see oxygen as the product at the anode. Things change when stuff gets concentrated!
Watching this 10 hours before the exam day on the exam🥲
So it's only the sulfate ions left in solution?
excellent presentation, I liked the chart. Im not a chemistry student, how did you get that 4 in that 4OH-. what does it mean. I often see numbers before the symbols.
there should be same H for the left and right of the reaction
@@nurrohmannurrohman1280 didnt get it
To balance the equations
Why the concentration of the sulfuric acid increases?
Concentrated sulphuric acid video?
@ChemJungle you didnt say anything about voltage or electrode type
Nice Explanation ❤️
Thanks!
i dont understand how 2e- were gained but 4e- were lost
EXPLAIN PLEEEASEE!!!!
Cathode adds 1 electron to the H+ so that it becomes H. So the H+ GAINS electron. The anode pulls 1 electron from OH-. So, the OH- LOSES 1 electron.
why does h2so4 become h+ and so4 - -, shouldn't it become h2+ and so4 - -?
Jonas Bolander - this is actually way
more complicated than is required for gcse and based on the ions you know about you’re best off thinking about it as 2h+ and SO4(2-). In reality H2SO4 is a mixture of ionic and covalent character and is most likely to only lose one H+ to form H+ and HSO4(-) but since the sulphate ion is very unlikely to discharge it’s not really worth worrying about too much! Sorry to not really answer your question - but hope that justifies why I’ve done it this way! :)
Because the 2 belongs to the sulfate and not the hydrogen
This video is wrong. Ionization of water occured at very small rate (considered not happening) and by considering E° what actually happened at Anode(Oxidation) was 2H2O _> O2 + 4H+ + 4e- with E°red =1.23
Kya English ha
can u come give me private lesson at home ? hehe
you're explanation genuinely lacks too many basics and details, why are you in a rush?
OH MY GOD! U give FuseSchool a beating with that concise a video.... Love it
ik, i just came here from one of their videos, understood nth there and im thorough now!
Thanks
Excellent explanation!
I didn't get why OH is oxidized instead of H2O? Why the anode half equation is not 2H2O>O2+ 4[H+] ？
The reality is that both the OH- and the H2O will be oxidised, but the IGCSE syllabus that we do focuses on the OH- (even though there is a higher concentration of H2O and so that process is more likely!)
I have a doubt madam.
Why H2SO4 can't dissosiate into
H+ and HSO4-
That would be partial disassociation, that occurs in weak acids, but sulfuric acid is a strong acid so it is completely broken
At anode why there is 4OH- and not only oh-?
The big numbers in front (the coefficients) just tell you the ratio in which things react. So for every one oxygen molecule that is made, 4 OH- ions are removed from the solution. :)
ChemJungle
Thank you