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

Hey, please can someone just explain how to get the mole ratio in this calculation? I'm really struggling!! Thanks

Cr²⁺ --> Cr⁶⁺ + 4e^-
Mn⁷⁺ + 5e^- --> Mn²⁺

therefore

5Cr²⁺ : 4Mn⁷⁺

Be careful in the calculation with the moles of chromium ion in solution from the chromium ethanoate

i think i understand why it would be A. Is it due to the auto ionisation of water being slightly endothermic meaning that an increase in temperature would favour ionisation? Causing an increase in Kw. The pH would still be neutral as number of H+ would be the same because H20 is in surplus?

Thanks in advance

i think i understand why it would be A. Is it due to the auto ionisation of water being slightly endothermic meaning that an increase in temperature would favour ionisation? Causing an increase in Kw. The pH would still be neutral as number of H+ would be the same because H20 is in surplus?

Thanks in advance

Yeah there's still a fair amount of time before unit 5 so you should be good. Are you going for the A or the A*?

Think about it this way - although your pH is decreasing due to an increase in temperature, the ratio of [OH^-]:[H⁺] will still remain the same. This is from chemguide: "Although the pH of pure water changes with temperature, it is important to realise that it is still neutral. In the case of pure water, there are always going to be the same number of hydrogen ions and hydroxide ions present. That means that the pure water remains neutral - even if its pH changes."



I'll be trying to do my best - I have U4 under control, it's just U5 that needs taming. I'll be happy with an A in chemistry, though - I won't need to do it university anyway. Which is a bad thing since I've really come to enjoy the subject.

Hey guys, I was wondering, what would happen to the electrode potential if current was allowed to pass through the cell? Because I know that normally you use a high-resistance voltmeter to prevent this from happening but I think I remember seeing a question asking what would happen if current was to pass through?

Thanks

You get electron flow from where there are lots of them to where there are less of them, so can you work out what will happen from there?

Think about it this way - although your pH is decreasing due to an increase in temperature, the ratio of [OH^-]:[H⁺] will still remain the same. This is from chemguide: "Although the pH of pure water changes with temperature, it is important to realise that it is still neutral. In the case of pure water, there are always going to be the same number of hydrogen ions and hydroxide ions present. That means that the pure water remains neutral - even if its pH changes."



I'll be trying to do my best - I have U4 under control, it's just U5 that needs taming. I'll be happy with an A in chemistry, though - I won't need to do it university anyway. Which is a bad thing since I've really come to enjoy the subject.

Ah okay soooo electrode potential would decrease!

Yes but there are more extreme things that happen as well. At each electrode you have the equilibrium X --->> X+ +e- (insert number of charges and electrons accordingly). When electrons move from the negative electrode (both are technically negative but voltages is a relative measurement) at that electrode there are now less electrons so by le chatelier's eq moves to the right, and the converse will happen at the positive electrode

Wait, I'm not quite sure what you mean by that :s

So, let's say you have Mg2+ + 2e- ---> Mg, for which the SEP = -2.37V.
and Al3+ + 3e- ---> Al, for which the SEP = +0.80V.

Could you try to explain what you're trying to explain (lol xD) in this context?

Of course. Essentially the standard electrode potential is a measurement of the position of equilibrium for the reaction for each electrode. this reaction here :

Mg2+ + 2e- ---> Mg, for which the SEP = -2.37V, is in reality an equilibrium, so at that electrode we have the equilibrium
Mg2+ + 2e- <---> Mg going on.
The negative electrode potential tells us that, relative to a SHE, this equilibrium lies more towards the side with the electrons on. In comparison at the aluminum electrode we have a similar equilibrium going on, except this time the position of equilibrium is much more on the side of the elemental Al, hence the positive SEP (it's important to realise that in reality all of these are technically negative, but a voltmeter measures the potential difference so being less negative than an SHE or other electrode means it is the 'positive' electrode). When we allow current to flow, the electrons will flow from the Mg side as there are more of them there, to the Al side where there are less. As each of these reactions are in equilibrium, by removing electrons from the Mg side the position of equilibrium is going to move even further to the left to produce more electrons, meaning it pretty much turns into the one way reaction
Mg---->Mg2+ +2e-

At the Al side, we are adding electrons so to counteract this the position of equilibrium is going to shift even further to the right so that the reaction essentially becomes one way and is Al3+ +3e- ---->Al

Holy crap you just explained that in the most masterful way ever. That makes perfect sense; thank you so much! You're sorted for chemistry aren't you hahaha xD!

We'll see lol, haven't quite finished notes for unit 5 yet because they take too long but still a while until the exam. And thanks, glad it helped.

One last thing, in exam questions, they say that a reaction won't be energetically feasible if E(cell) < 0V, such as here:



But in the textbook, we're told that if -0.6< E(cell) < 0, the reactants predominate and only when E(cell) < -0.6V is there no reaction.

Who do we trust, Edexcel's textbook or Edexcel's papers xD?!

Depends what you mean by no reaction. For exam purposes I would stick with negative means reaction isn't feasible, in reality there will be some negligible amounts of reaction taking place. This can relatively easily be shown assuming you do A2 maths, we know that Ecell is proportional to total entropy change which is proportional to ln(K), where K is equilibrium constant so if you say that Ecell=Alnk you can solve that for K and see that it will never actually be 0, even if the ecell was massively massively negative

Ah yeah makes sense haha but as you said I will just stick to what the mark scheme says, because I guess that's what matters at the end of the day (annoyingly, to be honest haha).

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