Aqa chem 4/ chem 5 june 2016 thread

When drawing the conventional cell representation for electrochemical cells, when do you include Pt on the outside of the lines? Also is it most negative E value on the left and most positive E value on the right??

Yes with E values

Platinum is used if a half-cell does not contain a metal than can act as an electrode, where the metal MUST be a SOLID. This is when Pt is indicated on the outside of conventional cell diagram.

E.g:

Fe2+ + e- -> Fe+
In this example, the metals in this half-cell are not solid because they are ions, hence you do not have a metal that can act as an electrode which is why a platinum electrode is necessary:
Pt | Fe+, Fe2+ ||

Zn2+ + 2e- -> Zn
In this example, there is a metal that can act as an electrode (Zn) because it is a solid, hence a platinum electrode is not necessary: Zn | Zn2+ ||

(edited 7 years ago)

Reply 1821

chzm

10

Original post by Jpw1097

Firstly you have to select the alkaline half equations which are equations 2 and 4 in the table, which are:

O₂ + 2H₂O + 4e^- ==> 4OH^-
H₂ + 2OH^- ==> 2H₂O + 2e^-

The hydrogen electrode goes on the left and the oxygen electrode goes on the right. Remember, the species with the lowest oxidation states go on the outside while species with higher oxidation states go on the inside.

The oxidation state of hydrogen in H₂ is 0, the oxidation state of hydrogen in OH^- and H₂O is +1, therefore H₂ will be on the outside while OH^- and H₂O will go on the inside.

The oxidation state of oxygen in O₂ is 0, the oxidation state of oxygen in H₂O and OH^- is -2, therefore O₂ will be on the inside while H₂O and OH^- will be on the outside.

Therefore the conventional representation for an alkaline hydrogen-oxygen fuel cell is:

Pt(s) | H₂(g) | OH^-(aq) | H₂O(l) || O₂(g) | OH^-(aq) | H₂O(l) | Pt(s)

Pt|H2(g)|OH–(aq),H2O(l)||O2(g)|H2O(l),OH–(aq)|Pt
is this the same thing even if you swap the OH and the H2O over at the end as this is what the mark scheme states

Reply 1822

thenextchemist

21

Someone help me with January 2013
Question 7a
Can someone please draw the diagram for me
Thank you

Posted from TSR Mobile

Reply 1823

thenextchemist

21

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-QP-JUN14.PDF

Help with question 5di
the answer is Cl but why can't it be oxygen because the oxidation state has also been reduced from 0 to -2?

Thank you

Posted from TSR Mobile

Reply 1824

username2280751

2

http://www.thestudentroom.co.uk/attachment.php?attachmentid=390319&stc=1

^ Easily the most simple table showing transition metal complex formulas & colors.

Reply 1825

Mriffin

1

Hardest Chem 5 paper to date?

Reply 1826

Suits101

2

Original post by Bloom77

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-QP-JUN14.PDF

Help with question 5di
the answer is Cl but why can't it be oxygen because the oxidation state has also been reduced from 0 to -2?

Thank you

Posted from TSR Mobile

In HOCl:

H = +1
O = -2
Cl = +1

Oxygen (in water) is -2 so there's no change in oxidation state/number of oxygen.

A good way to remember it is that oxygen is always -2 except in peroxides (H2O2) where's it's -1 and in compounds with F where it's +2 (OF2). Or, oxygen is more electronegative than Cl hence O is -2 and chlorine is +1, not -1 as it normally is.

(edited 7 years ago)

Reply 1827

Suits101

2

Original post by Mriffin

Hardest Chem 5 paper to date?

June 13/15 in my opinion.

Reply 1828

Cakey_101

7

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-QP-JUN15.PDF
Q1C)

Guys whats the best way to approach a born-haber cycle question? They've asked for electron affinity so which route would be the best to take?
I always seem to get these wrong :/