Edexcel Chemistry Unit 5 June 19 2012

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

Dont worry about it, are you aiming for A*?

Yup, I remember you telling me that you're also aiming for an A*, hopefully unit 5 will go smooth!

Yh I am, I dont need it but I still want it.

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

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.

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.

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?

x

Try it with a few questions and see if the method suits you.

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.

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?!

I posted this to someones EMF question on the joint thread- hope it maybe helps?

guys are we expected to know the mechanism of the reaction between benzene and fuming sulfuric acid?

guys are we expected to know the mechanism of the reaction between benzene and fuming sulfuric acid?