Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread

Hi guys, I'm also doing edexcel and am dreading exam period :\

Quick question:

In the textbook(blue edexcel textbook), it says if you react [Cr(H2O)6]3+ with excess sodium hydroxide, you get [Cr(H2O)2(OH)4]- on page 187 but on page 181, it says reaction of [Cr(H2O)6]3+ with a base gives [Cr(OH)6]3-. So I am slightly confused about what is produced when Cr 3+ reacts with excess sodium hydroxide here

It depends of the amount of NaOH you're adding. The first one is a result of adding it drop-wise until a precipitate forms and the second one is until an excess amount of NaOH is added and the precipitate dissolves to form a solution.

Hmm, not sure abt that because if you add it drop wise, u get

[Cr(H2O)3(OH)3] (s) precipitate forming. My question is if u add an excess of sodium hydroxide from that stage, do you get [Cr(H2O)2(OH)4]- or [Cr(OH)6]3- because both of them form solutions.

or is it simply that if you add NaOH in excess at once, you get one and if u add it dropwise in excess u get the other?

Oops, I'm sorry - I didn't read your question properly at first. You're right about the precipitate, that is the correct formula for it. But when you add excess, you get [Cr(OH)₆]^3- as per my knowledge. I can't see why the Edexcel textbook would disagree with that because the revision guide and George Facer say otherwise.

Yeah that's what I'm confused about, Chemguide, George Facer and CGP all say that you get [Cr(OH)6]3- but on page 187 of the edexcel table where it gives the table with the transition elements, it says you get [Cr(H2O)2(OH)4]- on the addition of excess NaOH to give you a solution after the precipitate dissolves.

Sorry to be a bit pedantic about this, just really make sure I know the it properly

Yeah that's what I'm confused about, Chemguide, George Facer and CGP all say that you get [Cr(OH)6]3- but on page 187 of the edexcel table where it gives the table with the transition elements, it says you get [Cr(H2O)2(OH)4]- on the addition of excess NaOH to give you a solution after the precipitate dissolves.

Sorry to be a bit pedantic about this, just really make sure I know the it properly

This question is in the Jan 2015 paper:
http://prntscr.com/aliwr7

The change in oxidation numbers is as follows:

For N, it is a decrease of -8. (reduction)
For Al, there is an increase of 3. (oxidation)

how do we relate this to the stiochiometric coefficients?

Any helpful revision notes for the transition metals and ligand exchange/ deprotonation section? I cannot understand these equations aghhh??

Think about what the half equations would be