Edexcel Chemistry Unit 5 June 19 2012
Chemistry exam discussion - share revision tips in preparation for GCSE, A Level and other chemistry exams and discuss how they went afterwards.
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Re: Chemistry Unit 5 June 19 2012Man honestly its ''oxidizing agent'' - ''reducing agaent'' which is same as what u said.. but it depends on the question. Lets say u have two positive values for two half reactions, you reverse the least positive value and in doing that, the equation oxidiz. agent - reduci. agent becomes oxidising agent PLUS (change of sign) reducing agent.(Original post by arnab)
yhh calculations and whether e.m.f reaction will occur or not. My teacher just told just to the most positive E value - the most negative E value but i don't really understand how that helps to predict whether a reaction will take place or not
But if lets say a question comes up asking ''does bromine oxidize chloride ions''. Basically bromine is your oxidizing agent so itself gets reduced, and the chloride ions get oxidized to chlorine. In the data booklet it shows the formula for chlorine being reduced to chloride ions. In this case, you need to reverse the E value for that half equation. Add this reversed value to the E Value of bromine getting reduced to bromide, and u can see whether its spontaneous or not. hope this helps.
Plus the fact, the overall E total value HAS to be postive for a reaction to be spontaneous.Last edited by shahofiran; 15-06-2012 at 23:00. -
Re: Chemistry Unit 5 June 19 2012Yup, I remember you telling me that you're also aiming for an A*, hopefully unit 5 will go smooth!(Original post by Jukeboxing)
Dont worry about it, are you aiming for A*? -
Re: Chemistry Unit 5 June 19 2012Yh I am, I dont need it but I still want it.(Original post by This Honest)
Yup, I remember you telling me that you're also aiming for an A*, hopefully unit 5 will go smooth! -
Re: Chemistry Unit 5 June 19 2012Lol same.(Original post by Jukeboxing)
Yh I am, I dont need it but I still want it. -
Re: Chemistry Unit 5 June 19 2012I'll give you a quick example here, it doesnt require any formula(Original post by arnab)
yhh calculations and whether e.m.f reaction will occur or not. My teacher just told just to the most positive E value - the most negative E value but i don't really understand how that helps to predict whether a reaction will take place or not
Lets say I want to react Cu(s) with Fe3+(aq)
Heres the data i'm given.
Fe3+ +e- ----> Fe2+ E = +0.77V
Cu2+ +2e- -----> Cu(s) E = +0.34V
What I do is reverse the second equation, because I want Cu to react with Fe3+
Cu(s) -------> Cu2+ +2e- E = -0.34V
The sign of the emf changes because I reversed the equation, now all you have to do is add the emfs together and thats it.
Ecell = 0.77 + (-0.34) = +0.43V
Reaction is feasible if Ecell is positive, however it may not occur if the conditions are not standard or if the activation energy is too high. So we'd expect this reaction to occur, but it may not occur for reasons that I just mentioned.
I'm sorry If i'm not making it clear. -
Re: Chemistry Unit 5 June 19 2012
http://www.scriblink.com/index.jsp?a...331A26703886BA
Chem5 revision -
Re: Chemistry Unit 5 June 19 2012Thanks, i think i am getting the hang of it now(Original post by shahofiran)
Man honestly its ''oxidizing agent'' - ''reducing agaent'' which is same as what u said.. but it depends on the question. Lets say u have two positive values for two half reactions, you reverse the least positive value and in doing that, the equation oxidiz. agent - reduci. agent becomes oxidising agent PLUS (change of sign) reducing agent.
But if lets say a question comes up asking ''does bromine oxidize chloride ions''. Basically bromine is your oxidizing agent so itself gets reduced, and the chloride ions get oxidized to chlorine. In the data booklet it shows the formula for chlorine being reduced to chloride ions. In this case, you need to reverse the E value for that half equation. Add this reversed value to the E Value of bromine getting reduced to bromide, and u can see whether its spontaneous or not. hope this helps.
Plus the fact, the overall E total value HAS to be postive for a reaction to be spontaneous.
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Re: Chemistry Unit 5 June 19 2012so wait, am i right in understanding that which ever species you want to be reduced in the overall reaction, you switch equations of it?(Original post by Jukeboxing)
I'll give you a quick example here, it doesnt require any formula
Lets say I want to react Cu(s) with Fe3+(aq)
Heres the data i'm given.
Fe3+ +e- ----> Fe2+ E = +0.77V
Cu2+ +2e- -----> Cu(s) E = +0.34V
What I do is reverse the second equation, because I want Cu to react with Fe3+
Cu(s) -------> Cu2+ +2e- E = -0.34V
The sign of the emf changes because I reversed the equation, now all you have to do is add the emfs together and thats it.
Ecell = 0.77 + (-0.34) = +0.43V
Reaction is feasible if Ecell is positive, however it may not occur if the conditions are not standard or if the activation energy is too high. So we'd expect this reaction to occur, but it may not occur for reasons that I just mentioned.
I'm sorry If i'm not making it clear. -
Re: Chemistry Unit 5 June 19 2012Try it with a few questions and see if the method suits you.(Original post by arnab)
so wait, am i right in understanding that which ever species you want to be reduced in the overall reaction, you switch equations of it? -
Re: Chemistry Unit 5 June 19 2012are you sure this is right?(Original post by Jukeboxing)
x
For Cr with OH-
Blue-purple solution >>>>>>> Muddy green precipitate
[Cr(H2O)6]3+ + 3NH3 >>>>>>>> [Cr(H2O)3(OH)3] + 3NH4+
I thought the formula for the purple solution was [Cr(Nh3)6]3+?
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Re: Chemistry Unit 5 June 19 2012Yhh i will do tomorrow. Right now i am just going through the theory one more time, before i hit the questions tomorrow morning(Original post by Jukeboxing)
Try it with a few questions and see if the method suits you.
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Re: Chemistry Unit 5 June 19 2012Cheers that's really helpful, only isn't Cr 3+ Green?(Original post by Jukeboxing)
I just finished memorising the colours and the metals that dissolve in excess etc.
I made this PDF, and memorised it all.
I know some may not agree with the colours but my teacher did all of the reactions in front of the class, so I'm using it. -
Re: Chemistry Unit 5 June 19 2012I posted this to someones EMF question on the joint thread- hope it maybe helps?(Original post by arnab)
yhh calculations and whether e.m.f reaction will occur or not. My teacher just told just to the most positive E value - the most negative E value but i don't really understand how that helps to predict whether a reaction will take place or not

I'm not too keen on the most positive / most negative way - I can't seem to grasp it! you could think of it this way:
Either
E value of the one being reduced - E value of the one being oxidised
(to know which one is being reduced, look at the oxidation states of each chemical/element in the equation that they give you at the start of the question (the equation that they ask you if '-----' is feasible or not)
Or
you can think of it as E of oxidising agent - E of reducing agent
(which is essentially the same) - shahofiran and jukeboxing gave a really clear good example of this earlier
OR, and the most clear way to me: half-equation way
eg. Br2 + 2I- >>> 2Br- + I2 'Is this feasible?'
I2 + 2e- = 2I +0.54 V
Br2 + 2e- = 2Br- +1.07 V
To calculate E cell, just add the two E values together FOR THE DIRECTION YOU WANT THE REACTION TO GO (ie the direction it tells you in the equation) - so here you need to 'flip' the Iodine equation - therefore reverse the sign of it's E value so you'd get -0.54 + 1.07 = + 0.53 V which is +ve so its spontaneous (as long as standard conditions, and the activation energy isnt too high)
So as you can see if you want to think of it as a 'method' - you're reversing the equation (and flipping the sign of the E) of what's being oxidised, and then simply adding the two E values together.
You can try it the 1st way too - the Iodine is the one being oxidised so it's the reducing agent. Using E ox agent - E red agent gives 1.07 - 0.54 = +0.53 V again
The anticlockwise rule is just a 'quick' way of seeing if the reaction will be spontaneous or not. with our previous example
I2 + 2e- = 2I +0.54 V
Br2 + 2e- = 2Br- +1.07 V
you put the less +ve equation on top,
the more +ve equation at the bottom.
Then apply your anticlockwise arrows and see if the direction of spontaneous change matches what they show you in the theoretical equation.
Think of it this way, the less +ve (top) will have more tendency to go to << the left
and the more +ve (bottom) will have more tendency to go to the right >>
(that's where the 'anticlockwise' arrows come from)
<<<<<<<<<<<<<<<<<<<<<<
I2 + 2e- = 2I +0.54 V
Br2 + 2e- = 2Br- +1.07 V
>>>>>>>>>>>>>>>>>>>>>>>>>
the arrows match the direction of what the initial theoretical (full) equation gave us in the question (Iodide to iodine, bromine to bromide) so you know it is spontaneous.
I struggled loads with this too before I kind of grasped it, hope this helps?!Last edited by giraffegiraffe; 16-06-2012 at 06:46. -
Re: Chemistry Unit 5 June 19 2012LOL i literally spent like 40mins yesterday looking for that. Thank you!!(Original post by giraffegiraffe)
I posted this to someones EMF question on the joint thread- hope it maybe helps?
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Re: Chemistry Unit 5 June 19 2012no our teacher said we just need to know the electrophile is SO3 and the product is benezenesulfonic acid, but we don't need the mechanism(Original post by arnab)
guys are we expected to know the mechanism of the reaction between benzene and fuming sulfuric acid?
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Re: Chemistry Unit 5 June 19 2012i don't think need to but it's not that different from any other E+ substitution(Original post by arnab)
guys are we expected to know the mechanism of the reaction between benzene and fuming sulfuric acid?
http://www.chemguide.co.uk/mechanism...phonation.html
