not all heroes wear capes

@AlleryChemistry

7 жыл бұрын

Heh. Just doing something I love.

This guy is a national treasure he deserved to be knighted

I'm sooooo glad I found this channel :D You guys are awesome ! Thank you !!

@AlleryChemistry

8 жыл бұрын

+Angelina D'Costa Thanks! Please spread the word, the more people they help the better. ☺

you're doing an amazing job...keep making such videos

I seriously can't thank you enough!! I actually understand!

It's finally clicked, thank you so much

I miss the old syllabus. maybe i am one of those who are the last batch of the old IAL spec. Thank you very Much for taking the time to care Mr. Harris.

@cha0spretzel-534

2 ай бұрын

why was the old syllabus better?

First video watched, brilliant explanation :) Thanks a lot! CGP isn't very clear on this topic so it was very useful to watch this

@AlleryChemistry

8 жыл бұрын

+MrEekaz Pleased it helped so much! This can be a tough topic the anticlockwise rule just clarifies what you have. Less chance of making a mistake.

Thank you for your help, Just a quick question, do we have to know about the anti-clockwise Rule for AQA ?

Hi , thanks for the video ! I have a question that according to anti-clockwise rule, if silver ion reacts with iodide ion that should give out iodine and silver metal, but why silver iodide is formed instead? Thank you!

this video is excellent but i wonder does the E value double if you go from Cr2+ to 2Cr2+? you doubled the rest of the equation so algebraically i guess it should but does it?

life saver thank you ! just wondering how is the anticlockwise rule effected by the concentrations? surely the same principle would apply ie/ put the negative one on the top and the more positive one on the bottom? do you know if you need to know this for AQA current spec?

@AlleryChemistry

7 жыл бұрын

Yes same thing. You will need to know this for Year 2 chemistry at AQA. Anticlockwise is just a method so you may not find this in your books.

Thank you so much for this video, I've been confused by the electrochemical series for months 😅

@AlleryChemistry

3 ай бұрын

You're welcome 😊

Thank you, brilliant

@AlleryChemistry

8 жыл бұрын

+DiLLZGFX No problem at all! Thanks for your support!

Hi , thanks for the video ! I have a question sorry if it's dumb but when doing titrations why do we need to acidify manganate anyways? What's the purpose of adding sulfuric acid etc ?

@AlleryChemistry

7 жыл бұрын

Yes. If you write the half equation for MnO4- being reduced to Mn2+ all will become clear (I have balanced in stages here) - MnO4- --> Mn2+ MnO4- --> Mn2+ + 4H2O MnO4- + 8H+ --> Mn2+ + 4H2O MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O In order for MnO4- to be reduced protons (from an acid) are needed to bond with the oxygens that are released from the MnO4-. Without this the reaction wouldn't go.

@vampette145

7 жыл бұрын

thanks a lot ^__^ your vids are super helpful specially for people studying on their own :[ btw sorry to keep asking qs but i was doing this experimental chem paper and they asked us to choose if we had to use a high resistance or low resistance volt meter.Could you pls explain why we need to use a high resistance voltmeter? :/

@AlleryChemistry

7 жыл бұрын

Yes. If you want to measure the voltage across, say, a resistor in a circuit, then that voltage will (mathematically) have a value given by V=IR. When you place your voltmeter (in parallel) across the resistor, it draws some current from the circuit; current which no longer goes through the resistor. The result is that you have decreased I in the resistor and as V=IR you have reduced the voltage. The act of trying to measure the voltage has caused that voltage to drop. So you haven't measured the "true" value. The way to minimise this is to have a voltmeter with a very high resistance. This means it draws very little current from the circuit and the voltage you are trying to measure doesn't drop.

Super helpful- thanks so much. It's sad I'm only watching this the day b4 paper 3 :(

thank you !!!

justttt great!!!

Thank you for the brilliant video! I didn't understand the explanation as to why one shouldn't use HCl to acidify KMnO4 solutions :/ Could you please explain it again? Thank you!

@AlleryChemistry

7 жыл бұрын

Yes. KMnO4 oxidises Cl- ions (produced by HCl) to Cl2. This would occur in the burette. This means you will have less MnO4- ions as these will be reduced. KMnO4 doesn't oxidise SO42- ions so H2SO4 is used to acidify KMnO4 instead. Hope this helps?

@intheblinkofaneye1912

7 жыл бұрын

Ahh, so KMnO4 doesn't oxidise SO42- because it's too difficult to oxidise because of it's -2 charge? That's the bit I didn't understand - why KMnO4 doesn't oxidise SO42- ions.. Sorry if I didn't make this clear in my last comment. Thank you for the fast reply!

@intheblinkofaneye1912

7 жыл бұрын

And I also don't know why YT just crossed out the text in my previous comment ^ :/ I didn't mean for it to be crossed out..

@AlleryChemistry

7 жыл бұрын

That's right! If you take the SO42- half equation and MnO4- half equation and put the most negative one on top. Draw your anti clockwise arrows as described in the vid and you will see that SO42- and MnO4- will appear on opposite sides of the arrows meaning they don't react! Yay!

@intheblinkofaneye1912

7 жыл бұрын

Thank you so much!!!

Thank you :)

@AlleryChemistry

7 жыл бұрын

You're welcome!

Sir just to clarify at 14:16 if the E0 cell becomes more negative is it more oxidising and if it becomes more positive is it more reducing?

@paddy6976

2 ай бұрын

i was thinking this too

Tq genius

Could you do a video on common ion effect in relation to electrochemistry? for instance MJ 15 9701 Varient 1 Q4ciii has Fe2+/Fe3+ with So4 as the -ve ion in one half cell and Ag/Ag+ with So4 again in the other. the examiner asks what effect addition of Na2So4 will have on both.

@AlleryChemistry

7 жыл бұрын

Yes. Could you send me the exam question via my twitter / facebook - @allerytutors and I'll take a look. Cheers!

MY SAVIOUR

It must have taken you a while to copy out all those half equations with Eo values!

Thank you for the video! Very helpful. My only question is, if you were to decrease the concentration of Pb^2+ ion, would the E value become less negative and where would the position of equilibrium shift in that situation?

@AlleryChemistry

9 жыл бұрын

BK29111 Thanks for your feedback! If Pb2+ was decreased then equilibrium will shift left to replace them, making the E value less negative or more positive. Hope this helps! Please share my channel to others who you think it will help to. You can follow me on twitter @allerytutors to get pre-release information on video releases in the future.

Hi Mr Allery Chemistry, For the Pb reaction, the E of Pb should be more positive, not more negative as equilibrium shifts to RHS!

Are all your videos AQA exam board?

@AlleryChemistry

Жыл бұрын

The whiteboard ones are generic and the black screen ones are exam specific. I do have AQA ones, check the home screen of the channel. 👍

12:10 incase i need this again

whos here for tmrw?