IAL Edexcel A2 Chemistry Unit 6 (Old syllabus) Watch

Khazard
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need some explanation to this,

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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
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amal29
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Here's the physics thread: https://www.thestudentroom.co.uk/sho....php?t=5938010
(Original post by Keka Ferdousi)
Attachment 817132Attachment 817134 Hey guys I had a question from physics unit 6.Why did we multiply the length with 4?
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sickomode
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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,

Name:  chem.png
Views: 18
Size:  55.3 KB
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
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saphira_jade
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(Original post by Khazard)
need some explanation to this,

Name:  chem.png
Views: 18
Size:  55.3 KB
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?
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slightly_edited
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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
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saphira_jade
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(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
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CuriosityYay
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(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
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saphira_jade
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Then for NMR?
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saphira_jade
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we need the chemical shift values right?
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CuriosityYay
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(Original post by saphira_jade)
Then for NMR?
(Original post by saphira_jade)
we need the chemical shift values right?
I said we don't need it! Do a past paper and stop these questions.:bhangra:
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saphira_jade
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(Original post by CuriosityYay)
I said we don't need it! Do a past paper and stop these questions.:bhangra:
Ok
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Keka Ferdousi
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Thank you
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lali000
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do we have to know cobalt and manganese colours?
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CuriosityYay
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(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
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Sandar456
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(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
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Khazard
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November 2018
(Original post by saphira_jade)
What paper is this question from?
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Khazard
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(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
Last edited by Khazard; 1 week ago
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Sandar456
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Do you think May 2019 chemistry unit6 will be easy or hard?
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CuriosityYay
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(Original post by Sandar456)
They will give if necessary
They won't give. NO DATA BOOKLET for paper 6
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CuriosityYay
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(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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