AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry

theres barely any synoptic in unit 5 chem ! its not like bio lol
i dont think there was any in the jan11 paper
i probs wont revise any synoptic

hey guys, good luck for the exam on friday
just have a little question to ask, if anyone could help, i would be grateful!

in redox reactions, when do you know when to flip one of the half equations around, when combining? because in some cases one is flipped, in other cases they are combined like a normal redox equation, and i'm confused as to how this comes about... thanks!

6+ for Cr2O72-

3+ for 2Cr3+ but counts as +6 due to the '2'

don't understand your last question

It's used when there isn't a solid involved.

E.g. Zn2+(aq) + 2e- ---> Zn(s)

doesn't need platinum as it has a solid (Zn) which acts as the electrode so it becomes:

Zn(s)|Zn2+(aq) ||

However,

Fe3+(aq) + e- ---> Fe2+(aq)

there is no solid involved here so a platinum electrode is needed to transfer electrons, and the cell becomes:

Pt | Fe2+(aq), Fe3+(aq) ||

So if you see a half cell equation with no solid involved, chuck in platinum

It can be left or right depending on if the half cell without the solid is more negative or more positive, and can appear on both if neither half cell involves a solid as it does with the hydrogen fuel cell:

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

Hope this helped

does anyone have answers to the NT book?

you only reverse the reactions when looking at spontaneous cell reactions , meaning , to see if they are feasible the question will usuallly specify by saying spontaneous or feeasible.
you dont flip them aroound when your looking for a overall reaction

thanks man
so just judging by the question, i should be able to know?

sorry where does 6e- come from i knw its silly que. but this is the only bit

Is everyone just memorizing the tons of equations from chapter 15/16? Cos I don't know what else to do about them.

Start with: Cr2O72- -> 2Cr3+ (Cr's balanced)

Balance oxygens by adding 7 H20 to right

Balance Hydrogens by adding 14 H+ to the left

The charge on the left is +12, the charge on the right is +6, so to make it +6 on both sides, add 6e- to the left, giving:

Cr2O72- + 14H+ + 6e- -> 2Cr3+ +7H2O

Hope this helps

I was thinking that, do we really have to remember every single one?

can some one explain to me how to find the charge outside the bracket, {Cr(NH3)6}3+ ?

for {Cr(OH)6}3- i guess cr is 3+ and OH is -6 :. 3-6=-3 am i right?

I thought this "Pt|H2(g)|OH-(aq), H2O(l) || H2O(l) | O2(g), OH-(aq)|Pt" should be.. Pt|H2(g)|OH-(aq), H2O(l) || O2(g) | H20(l), OH-(aq)|Pt
Because the O2 is in a different phase to the water and hydroxide...
Or maybe im wrong