OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

You are unable to determine the enthalpy changes that occur in the CO3 2- ion as it consists of more than one element

I think thats one way you can interpret it but im sure theres more answers

Cell potential increases*

Posted from TSR Mobile

Can anyone show how you would work out the enthalpy change of neutralisation for a reaction between 50cm3 of 0.1M H2SO4 and 50cm3 of of 0.1M NaOH with specific heat capacity of 4.18 and temp change from 20 to 30?

I know the equation would be H2S04 + 2NaOH -> NA2SO4+ 2H20 and Q = -4180J
What would you do next?

So would you need to work out the enthalpy change of formation of CO32- also?

why would oxidation increase if the cell is going in the reduction process?

A quick query about question 4 e (iii) of the June 2015 F325 paper
It tells you the Ka for propanoic acid is 1.35x10-5 and that both the concentration of the acid + its conjugate base are 1 mol dm-3

The student then adds 6.075g of Mg to 1.00dm3 of this buffer and you have to calculate the new pH.
If you write out the acid-metal reaction it is clear that the acid:Mg ratio is 1:2 so once the moles of Mg are calculated (6.075/24.3 =0.25) the actual moles of acid used are 0.25x2 so 0.5.

Am I correct in assuming the salt of the acid that is formed could also be considered the conjugate base of the acid and therefore the salts concentration would also be used on top of the original salt concentration in a buffer calculation? And if so as the salt is (C2H5COO-)2Mg if there were 0.25 mol of that then in 1dm3 there would be 0.5mol dm-3 of C2H5COO- from this salt + the original 1 mol dm-3 C2H5COO- already in the reaction?

With equilibrium questions for example - 2NO2 > N2O4
If pressure increases, why does the conc of NO2 increase more and then decreases?

Is it because initially when we increase pressure, the side with more moles with increase and those with the least moles decrease and the equilibrium shifting happens to increase pressure? Confused with this

Hello peeps, could someone please explain how in electrochemical cells we decide which half cell is negative electrode and which is positive?

So when drawing a diagram of two half cells, which one goes on left or right?

The question is from Jan 12

*not letting me upload pic*

I know many of you have asked this question before, but could someone elaborate on how to work this out ? [part (ii)]


I under standard that the Ag+ + Cl- --> agcl Calculating 2.2868/(107.9+35.5) I'd get 0.02 which is the mole for agcl. The mole ratio of cl- to agcl is 0.01 : 0.02(1:2). But how could there be 2Agcl and 1cl- ? It just doesn't make much sense to me as the equation doesn't balance up.

I know many of you have asked this question before, but could someone elaborate on how to work this out ? [part (ii)]


I under standard that the Ag+ + Cl- --> agcl Calculating 2.2868/(107.9+35.5) I'd get 0.02 which is the mole for agcl. The mole ratio of cl- to agcl is 0.01 : 0.02(1:2). But how could there be 2Agcl and 1cl- ? It just doesn't make much sense to me as the equation doesn't balance up.

Is the answer complex B?

You need to assume that silver nitrate is in excess. Therefore the ratio indicates that there must have been 2 chloride ions in the original complex

Cu2+ + 2e- <-> Cu (s)
That is the eq on the table.
Adding H2O, decreases conc of Cu2+.
As its eq, the eq will shift to left to compensate. Therefore its more:

Cu (s) -> Cu2+ + 2e-

Hence, more oxidiation.

But the thing to understand I think is that, less Cu2+,so eq shifts to the left.

This means, the EP value decreases and so overall cell potential between Ag and Cu will increase due to the larger difference in EP between the two.

Hope that makes sense. (Im literally explaining this from wha I figured from nervous' answer, literally just understood all this from him, thought id give him a break though )