ocr a f325 revision thread

First one correct.

Second one you'd divide by three and times by two which is equivalent to multiplying by 2/3.

oh yeah i meant multiplying.

an explanation of June 10 6b would be much appreciated. I just don't understand how the e- could get that way.

Each Cr is 6+ you are making 2Cr2+, therefore each one needs to gain 4 electrons, hence 4x2=8e-

jan 2013 the last question, how do you know n = +2 from the molar ratio?? thank u!!

Standard cell potentials can only be used to predict a reaction's feasibility under standard conditions - actual reactions are very rarely at RTP, standard states, unit ion concentrations etc. so may differ from the hypothetical cell reactions.

Furthermore, the cell potentials say nothing about how fast a reaction will be - it can be hypothetically feasible but so slow it effectively doesn't happen.

KMnO4 = 2.25x10-2 x 13.2/1000 = 2.97x10-4 mol
n(v) = 0.126/50.9 = 2.48x10-3 mol
with factor of 5 you do 2.48x10-3 / 5 = 4.96x10-4
ratio = vn+/mnO4- = 4.96x10-4/2.97x10-3 i just rounded them to 5:3 then wrote the equation

Thank you, but from the mark scheme, they worked out the value of n just from the molar ratio.

The moles of thiosulphate are 4.48x10^-3. You then use the ratios in the third equation (thiosuphate plus iodine) to get:
thiosulphate:iodine
2:1
4.48x10^-3:2.24x10^-3

The moles of iodine formed can then be used in the second equation (as the iodine formed from this reaction are then used in the third reaction which was used above) to get the ratio:
Iodine:copper(II)
1:2
2.24x10^-3:4.48x10^-3

Which means that the moles of Cu²⁺ are 4.48x10^-3, like the book says.

I think that what you did was try to use the ratio of thiosulphate:iodide(2:2) in the third equation, and then link that with the ratio in the second equation (Iodide:copper(II)(4:2)), which you can't do as the iodides in each of the reactions are not the same amounts.

no I think it's one to one as well.

For every 2 moles of S2O3^2- reacted, 1 mol I2 reacts with it.
Say all that 1mol was produced by the previous reaction, then 2 moles of Cu2+ produced that.
Say all 2 moles of that Cu2+ was produced by the previous reaction, that is produced by 2 moles of Cu.

So it's 1:1 thio : Cu

if electrode potential is negative does that mean it can lose electrons and if its positive it can gain electrons? sorry don't know how to word the question

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MY ORIGINAL POST: http://www.thestudentroom.co.uk/showthread.php?t=2286155&page=141&p=43103717#post43103717

I understand all that, but if you then compare the 2nd equation to the 1st, then dont we have to divide the moles of Cu(II) by 2 to get the original number of moles in the bronze sample?

if electrode potential is negative does that mean it can lose electrons and if its positive it can gain electrons? sorry don't know how to word the question

The concentration of the acid has halved