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IAL Edexcel A2 Chemistry Unit 6 (Old syllabus)

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Reply 60
need some explanation to this,

chem.png
I figured out we have to use KMnO4, and somehow get a positive Ecell value. Using the third equation makes sense, because vanadium v is being reduced to vanadium iv there, so we reverse the equation and use -1.00, then we add the two equations together? and their E values?
So in the end we get +1.51+-1.00, giving +0.51v?? (MS answer)
Or is it some other witchcraft because as far as i know Ecell = Ered - Eox, which gives (1.51--1.00) = 2.51v
(edited 4 years ago)
Reply 61
Here's the physics thread: https://www.thestudentroom.co.uk/showthread.php?t=5938010
Original post by Keka Ferdousi
IMG_4880.JPGIMG_4879.JPG Hey guys I had a question from physics unit 6.Why did we multiply the length with 4?
Ecell= Ered-Eoxd
=1.51-+1.00
u NEver change the sign of Evalues
u reverse the equation but values sign remains same

Original post by Khazard
need some explanation to this,

chem.png
I figured out we have to use KMnO4, and somehow get a positive Ecell value. Using the third equation makes sense, because vanadium v is being reduced to vanadium iv there, so we reverse the equation and use -1.00, then we add the two equations together? and their E values?
So in the end we get +1.51+-1.00, giving +0.51v?? (MS answer)
Or is it some other witchcraft because as far as i know Ecell = Ered - Eox, which gives (1.51--1.00) = 2.51v
Original post by Khazard
need some explanation to this,

chem.png
I figured out we have to use KMnO4, and somehow get a positive Ecell value. Using the third equation makes sense, because vanadium v is being reduced to vanadium iv there, so we reverse the equation and use -1.00, then we add the two equations together? and their E values?
So in the end we get +1.51+-1.00, giving +0.51v?? (MS answer)
Or is it some other witchcraft because as far as i know Ecell = Ered - Eox, which gives (1.51--1.00) = 2.51v


What paper is this question from?
Can someone please tell me if we get the Data booklet for unit 6 or not? Its mentioned in the unit 4 and 5 papers but not in the 6 papers
Original post by slightly_edited
Can someone please tell me if we get the Data booklet for unit 6 or not? Its mentioned in the unit 4 and 5 papers but not in the 6 papers


We do
Original post by slightly_edited
Can someone please tell me if we get the Data booklet for unit 6 or not? Its mentioned in the unit 4 and 5 papers but not in the 6 papers


No data booklet for paper 6. We will not need it. They won't ask for IR values and things like that. But they might ask what bonds might cause a peak in an IR spectrum.
Original post by saphira_jade
We do
Original post by CuriosityYay


Then for NMR?
we need the chemical shift values right? :frown:
Original post by saphira_jade
Then for NMR?


Original post by saphira_jade
we need the chemical shift values right? :frown:


I said we don't need it! Do a past paper and stop these questions.:bhangra:
Original post by CuriosityYay
I said we don't need it! Do a past paper and stop these questions.:bhangra:


Ok :smile:
Reply 72
do we have to know cobalt and manganese colours?
Original post by lali000
do we have to know cobalt and manganese colours?


Manganese is definitely. Cobalt is just to be sure.

You can remember Cobalt colors from the Cobalt(II) chloride paper used to test for water. Anhydrous is Blue, Hydrated is Pink.

Manganese(II) ions are very pale pink and can be described as colorless. (Don't use the word purple, it is reserved for manganate 7)
Manganate(VII) ions are purple.

If you add alkali like NaOH or NH3 to Mn2+ it will turn to white/off-white/beige/light brown/pale brown (use off-white) precipitate, which will darken (turn darker brown when left standing as it gets oxidised to (manganese(III) oxide Mn2O3 and then) manganese(IV) oxide MnO2
Original post by slightly_edited
Can someone please tell me if we get the Data booklet for unit 6 or not? Its mentioned in the unit 4 and 5 papers but not in the 6 papers


They will give if necessary
Reply 75
November 2018
Original post by saphira_jade
What paper is this question from?
Reply 76
Original post by sickomode
Ecell= Ered-Eoxd
=1.51-+1.00
u NEver change the sign of Evalues
u reverse the equation but values sign remains same

in the third equation, vanadium is getting reduced not oxidized, so how can one plug that value in Eoxd?
and im sure if the equation is reversed, e value reverses

I figured it out, you have to add the two equations to form a redox equation with the manganate ions and REVERSED vanadium equation(3rd), so even E values are added, so E ovr will be 1.51+(-1.00) = 0.51v
(edited 4 years ago)
Do you think May 2019 chemistry unit6 will be easy or hard?
Original post by Sandar456
They will give if necessary


They won't give. NO DATA BOOKLET for paper 6
Original post by Khazard
in the third equation, vanadium is getting reduced not oxidized, so how can one plug that value in Eoxd?
and im sure if the equation is reversed, e value reverses

I figured it out, you have to add the two equations to form a redox equation with the manganate ions and REVERSED vanadium equation(3rd), so even E values are added, so E ovr will be 1.51+(-1.00) = 0.51v


The sign of a standard electrode potential indicates the polarity of the electrode relative to standard hydrogen electrode. This sign is fixed and doesn't change if the equation for the half-cell is reversed. Standard electrode potential is described as being a 'sign invariant quantity'.

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