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

For Edexcel, I've always learnt Fe2+ as green and Fe3+ as brown coloured.

Yeah

https://a5c076379da85d2c0b501a60f1a40f8eeaf5f812.googledrive.com/host/0B1ZiqBksUHNYdEFJbUVsVUl3Ums/June%202010%20QP%20-%20Unit%205%20Edexcel%20Chemistry.pdf

Could someone please explain how to do 24di,ei and f please :/ So confused
-thanks

okay thanks

according to the user guide fe2+ is a pale blue blue (basically turquoise by the looks of it) and fe3+ is red brown

Oh right okay...The Fe3+ makes sense but I never thought Fe2+ was any shade of blue XD

I am never seem to figure out the half equations anyone have any tips on how to do This


Posted from TSR Mobile

Alright well this is the technique I use for half equations. I'll show you with an example other than these two, so you can try and work it out yourself:

So let's say we want to write the half equation for the oxidation of Cr₂O₇^2- to Cr³⁺.

FIRST STEP:

Write reactant ---> product

Cr₂O₇^2- ---> 2Cr³⁺.

(We have a 2 in front of the Cr³⁺ because there are two chromium atoms on the left so we need to balance that on the right hand side.

SECOND STEP:

Balance the oxygen atoms using water molecules.

Now, we have 7 oxygen atoms on the left hand side, and none of the right hand side. So, we add 7 water molecules on the right hand side:

Cr₂O₇^2- ---> 2Cr³⁺ + 7H₂O

THIRD STEP:

Balance the hydrogen atoms using H⁺ ions.

We now have 14 hydrogen atoms on the right hand side, and none on the left hand side. So, we add 14 H⁺ ions on the left hand side:
Cr₂O₇^2- + 14H⁺ ---> 2Cr³⁺ + 7H₂O

FOURTH STEP:

Balance out the charges on both sides using electrons.

We have a combined charge of (-2 + 14) = +12 on the left hand side, and (2*3) = +6 on the left hand side. So, we need 6 electrons on the left hand side in order to get equal charges on both sides:

Cr₂O₇^2- + 14H⁺ + 6e^- ---> 2Cr³⁺ + 7H₂O

And you're done!

Now, try and see if you can figure out the half-equations for your question using this method!

Thank you so much I did it one second it wasn't hard at all


Posted from TSR Mobile

Thank you so much I did it one second it wasn't hard at all


Posted from TSR Mobile

Haha no problem! See, it's easy when you know the method

Just so you know that method only works if done in acid, it's very slightly different in alkali

It is? How so?

easiest way is to add alkali to both sides of the equation at the end to get rid of H+ ions

@samb1234 Can you explain how to do part ii of this question please? I didn't get it right and I don't understand the mark scheme xD

You have the amount of energy transferred for that number of moles, you need to find the energy for 1 mol, convert to KJ and stick the right sign in front since the temp increased

@samb1234 Can you check my reasoning behind the idea of water being neutral despite a decrease in pH?

So, from what I understand, in pure water, the concentration of [OH-] and [H+] is always the same. So, if [H+] was to increase, the pH would decrease, since pH = -log([H+]), but the water still remains neutral since the [OH-] also increases.

Is that right?