I don't know if you still monitor this video but I am really thankful for your very straight forward explanation of this topic. By far the best I have seen. Thank you so much
ambidex@med... after balancing in acidic solution as I've shown in the video, you could just add enough H2O to BOTH sides of the equation to convert all the H+ to H3O+. So in the example I show here, you would add 10 H2O to both sides ... resulting in 15 H2O on the left and 10 H3O+ on the right. When balancing in basic solution, just add enough OH- to neutralize all the H3O+ in your equation ... keeping in mind that OH- and H3O+ will make 2 H2O.
wow thank you! I mean that from the bottom of my heart this is so helpful to me it being midnight and my test is tomorrow morning! needless to say I am cramming and this is one small part of one chapter and its a four chapter test! YIKES!
Took me a while to apprehend it but completely worthwhile! You just explained to me something in 17 min which took my teacher almost a week.. well maybe because my teacher had taught be all the basics of oxidation and reduction state!😂
Hello Cameron. I’m not sure why you concluded that. The electrons must cancel out when we combine the 2 half-reactions. That means the number of electrons gained in the reduction half reaction must equal the number of electrons lost in the oxidation half reaction. This is achieved when you multiply each half reaction by a coefficient to make the electrons lost equal to the electrons gained. Right before you re-combine.
Hey Farhan ... are you referring to about 13:20 in the video? If so, I multiply the top half reaction by 2 so that the two half reactions have equal numbers of electrons
Neelam Arya ... for basic solutions I first balance as though it is acidic. Then I add OH- to both sides to neutralize the acid and form H2O. Look at the example in the video. :)
I don't know if you still monitor this video but I am really thankful for your very straight forward explanation of this topic. By far the best I have seen. Thank you so much
This is by far the best tutorial and I've been binge-watching a whole lot tonight. Helped me out so much. Thank you!
This was some pretty detailed info, but after watching this video about 7 times its easily understood. thanks
ambidex@med... after balancing in acidic solution as I've shown in the video, you could just add enough H2O to BOTH sides of the equation to convert all the H+ to H3O+. So in the example I show here, you would add 10 H2O to both sides ... resulting in 15 H2O on the left and 10 H3O+ on the right. When balancing in basic solution, just add enough OH- to neutralize all the H3O+ in your equation ... keeping in mind that OH- and H3O+ will make 2 H2O.
Thank you sir!
i viewed this just before my board exam , and i was sucessfull in balancing rxn of d block by this concept
wow thank you! I mean that from the bottom of my heart this is so helpful to me it being midnight and my test is tomorrow morning! needless to say I am cramming and this is one small part of one chapter and its a four chapter test! YIKES!
Took me a while to apprehend it but completely worthwhile! You just explained to me something in 17 min which took my teacher almost a week..
well maybe because my teacher had taught be all the basics of oxidation and reduction state!😂
That's why i was able to understand your examples^
This video help me so much in balancing .I only want say thanks for this video
from Tanzania.....this has helped me a lot...Hakuna matata tena.....God bless u
Thank you for the excellent example problems and explanation!
greatest video of all time!!!!
it's 2018 and this video is still helpful
I wish I could like this video ten times :'D
Hi. Is there a method for changing the equation into H30+ instead of H+? H3O+ is so much harder and my professor told me I had to use H3O+ method.
Nice but want more examples
Sir you missed 2 water molecules on right side of basic redox reaction
Thank you every much! Very satisfied with this video
Very helpful. Thank you for explaining this topic clearly. 😁
Outstanding, Thank you so much, never understood half reactions, your video is terrific
why do you need to balance it like this other than for tests?
So your electrons don’t have to be balanced at the end
Hello Cameron. I’m not sure why you concluded that. The electrons must cancel out when we combine the 2 half-reactions. That means the number of electrons gained in the reduction half reaction must equal the number of electrons lost in the oxidation half reaction. This is achieved when you multiply each half reaction by a coefficient to make the electrons lost equal to the electrons gained. Right before you re-combine.
Thank you so much, you hav been extremely helpful!
thankyou so much for giving us such a good idea for balancing
That's amazing, thank you so much!
This helped sooooo much, thank you!
in base ; how r u multiplying it by 2..
Hey Farhan ... are you referring to about 13:20 in the video? If so, I multiply the top half reaction by 2 so that the two half reactions have equal numbers of electrons
thanks...
Thank you so much!!
Man you're the bestt
very very helpful ...................
thanks man great help!!!
thank you sir,this video really helped.
thanks a lot sir. it was indeed really helpful.
nicely explained thanks a lot
Thanks.....u have great experience
thank you very much
Thank you so much☺
thank you this helped me alot
Thank you Sir!
Thank you so much ....
thanxxx sir
I was thinking that we balance basic by OH negative ions
Thanks Thanks Thanks !!!
thanks a lot for this good video
☺☺☺☺😊
thank u sir....
In basic , don' t we have to add OH-
Neelam Arya ... for basic solutions I first balance as though it is acidic. Then I add OH- to both sides to neutralize the acid and form H2O. Look at the example in the video. :)
@@chempatenaude OK
Yo im grade 10 and im learning this college stuff wtf