Aqa chem 4/ chem 5 june 2016 thread

Some help with oxidising and reducing agents?
The strongest reducing agent has the most negative electrode potential and the weakest reducing agent has the most positive electrode potential
The strongest oxidising agent has the most positive electrode potential and the weakest oxidising agent has the most negative electrode potential.
Is that correct?
So in this table

The STRONGEST R.A would be H2O2 and the WEAKEST R.A would be H2O

The STRONGEST O.A would be H2O2
and the WEAKEST R.A would be O2

???


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Does anyone know why with [Co(H2O)6]2+ NH3 is able to fully displaced (by ligand substitution) all the H2O molecules however with [Cu(H2O)6]2+ only 4 H2O molecules will be displaced by NH3?

2(b) The straight line equation y = mx + c can be used to represent the the gibbs free energy equation where y = ΔG, m = -ΔS x = T and c = ΔH. To calculate the gradient you just do the change in y divided by the change in x. So take two points on the line and find the corresponding y an x values and do the same for a second point and do this calculation:

y₁ - y₂ / x₁ - x₂

The slope is -ΔS so the units are J K^-1 Mol^-1

2(d) is understanding that the rate of change of entropy changes for an element in it's different states. If the rate of change of entropy has changed then the slope (gradient) of your line has changed. So the ammonia has turned to liquid as the temperature is going lower.

can someone explain to me why when writing out cell representations using the SHE you always put H2 | H+ on the left hand side even if it has an E value more positive than the other species?? e.g. i did a question where i had to write the cell representation for Fe2+ being reduced to Fe (s) using SHE but it had an E value of -1.63 (or something like that)

but surely SHE= 0V which is more positive than -1.63 so it should be written on the right?

can someone explain to me why when writing out cell representations using the SHE you always put H2 | H+ on the left hand side even if it has an E value more positive than the other species?? e.g. i did a question where i had to write the cell representation for Fe2+ being reduced to Fe (s) using SHE but it had an E value of -1.63 (or something like that)

but surely SHE= 0V which is more positive than -1.63 so it should be written on the right?

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-QP-JAN13.PDF

can someone explain to me q3cii

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-W-MS-JAN13.PDF

I just dont understand why the reverse reaction takes place ?

I'm not sure but could it be because the forward reaction is exothermic?

Strictly speaking, you don't need to know them.

In the exam they will tell you which equations to use, such as 'write the half equation at the hydrogen electrode in a hydrogen-oxygen fuel cell in acidic/alkaline conditions' or if they don't then you can use either one.

I say you don't need to learn them because you need to know what happens at the electrodes, e.g you know hydrogen is a reactant so you can deduce how to get to hydroxide ions (if alkaline) or hydrogen ions (if acidic) and you know oxygen is a reactant so you can deduce how to get to hydroxide ions (if alkaline) or hydrogen ions (if acidic) once again.

can someone explain to me why when writing out cell representations using the SHE you always put H2 | H+ on the left hand side even if it has an E value more positive than the other species?? e.g. i did a question where i had to write the cell representation for Fe2+ being reduced to Fe (s) using SHE but it had an E value of -1.63 (or something like that)

but surely SHE= 0V which is more positive than -1.63 so it should be written on the right?

thank you !! how did you know m = -ΔS x = T
could they ever ask us to work out ΔH using a graph like this, and if so,how would u do it?
thanks once again!

Can someone help me figure with oxygen reaction to use for hydrogen-oyygen in alkaline conditions for 3c and 3d please thanks http://filestore.aqa.org.uk/subjects/AQA-CHEM5-W-QP-JAN10.PDF

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)

Does anyone know how to do this please???

find dy/dx so find two the difference between two different values of y then divide it by the difference between the corresponding values of x
The units are the y units divided by the x units
Then delta S is derived from the formula of deltaG= deltaH-TdeltaS using the same as above