AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry

here are the papers for chem5 jan 2011 ones
the mark scheme and the question paper

hey
can anyone answer my question please really confused about something
In redox equilibria in the textbook it says how if a species ABSORBS electrons it becomes more positive. how is this . because surely if it absorbs electorons it becomees more negativeSS:

What textbook do yu have ?

hey
can anyone answer my question please really confused about something
In redox equilibria in the textbook it says how if a species ABSORBS electrons it becomes more positive. how is this . because surely if it absorbs electorons it becomees more negativeSS:

misprint?

it says how Zn2+ + 2e- -------> Zn

then it says if the equilibrium lies to the right then a positive charge wil build up .. how is thisss ..
pleasee explain

hey guys..

im kinda confused when it comes to linking redox with equilibria..

like when a question comes up and asks 'what should be done do the Fe2+/Fe3+ (for example) cell to decrease the emf value'

i don't understand these sort of questions..

any help is much appreciated

As far as I know:

Emf is a measure for how far a reaction is from reaching equilibrium (hence why if a current is allowed to flow emf approaches 0 - equilibrium is reached).

Example:

Zn2+(aq) + 2e- --> Zn(s) E= - 0.76V
Fe2+(aq) + 2e- --> Fe(s) E= - 0.44V
The zinc half-cell is the more negative so becomes Zn(s) ---> Zn2+(aq) + 2e-

The overall reaction becomes: Zn + Fe2+ ---> Fe + Zn2+ E = +0.32

The fact that it is ---> rather than <---> illustrates that the reaction is approaching equilibrium but has not yet reached it.

Thus if you increase the concentration of Fe2+ or decrease the concentration of Zn2+ the equilibrium would shift further to the right (to decrease the concentration of Fe2+ or increase the concentration of Zn2 - Le Chatelier's Principle).

You can imagine it as the equilibrium shifting further to the right so that the reaction is further away from reaching it (the distance is greater) so the emf becomes more positive!

If you would increase the concentration of Zn2+ or decrease the concentration of Fe2+ then the equilibrium would shift left, the reaction would be closer to reaching it and emf would be less positive.

Exam question to show how you could be asked on it:

Deduce how the e.m.f. of the cell Mg(s)|Mg2+(aq)||Fe2+(aq)|Fe(s) changes when the concentration of Mg2+ is decreased. Explain your answer.

Mg(s) + Fe2+(aq) ---> Mg2+(aq) + Fe(s)
Equilibrium would shift right to replace the lost Mg2+
Thus the reaction is further from equilibrium
Thus the emf would increase.

You could also be asked about the effect of concentration on the electrode potentials of individual half cells, and the principle is similar but not the same, and depends on whether the half cell is negative or positive.

E.g.

S2O8(2-) (aq) + 2e– --> 2SO4(2-) (aq) E = + 2.01 V

State how, if at all, the electrode potential of the S2O8(2-) /SO4(2-) equilibrium would change if the concentration of SO4(2-) ions was increased.
Explain your answer.

the E is positive so this half cell takes in electrons when fitted to a hydrogen half cell. When SO4(2-) concentration increases the equilibrium shifts left to decrease the concentration and so fewer electrons can be accepted (or thought another way electrons are produced) so the Electrode potential becomes less positive.

With Zn2+(aq) + 2e- --> Zn(s) E= - 0.76V

E is negative so it donates electrons when attached a hydrogen half cell (the reaction is swapped to Zn ---> Zn2+ + 2e-).

Thus if you increase the concentration of Zn2+ the equilibrium shifts left (or right if you are working with the original equation and haven't had to swap it) and fewer electrons are given off so the Electrode potential becomes more positive.

Hope this is helpful!
Not sure this is 100% right but it seems to work with the questions I've come across so I'm sticking to it unless someone can correct me

bye bye CHEM4, hellooooo CHEM5
don't I just have the best life, the numbness of my brain has just radiated everywhere

Let's just hope CHEM5's better than the latter





.. However unlikely :/

anyone else envisaging a nightmare paper after hearing the horror stories of CHEM4 and the OCR paper today?

hey jimmy303

thanks alot for the explanation..

im still a bit confused though. i don't understand the red bits ^^^

could you explain that a bit more simply plz?

Hope you don't mind

I'll try my best!

The fact that it is ---> rather than <---> illustrates that the reaction is approaching equilibrium but has not yet reached it.

A reaction can either be one that goes to completion (complete reactions) or equilibrium (reversible reactions).

Complete reactions continue until one of the reactants is used up and the reaction stops (e.g. 2Mg + O2 --> MgO)

Reversible reactions (such as those involved in electrochemical cells) do not go to completion, but reach a balance (aka equilibrium).

At equilibrium:


For a reaction that is not yet at equilibrium to reach equilibrium, either the forward reaction or the backward reaction occurs at a greater rate than the other.

_1 mole____1 mole_________0 moles__0 moles
E.g. Zn(s) + Fe2+(aq)----------> Zn2+(aq) + Fe(s)
___________________<-

To begin with, there is little product (Zn2+ and Fe) so the rate of reaction in the backward direction is minimal, and the reation in the forward direction is much greater.

Thus we can substitutute the typical sign for a reversible reaction (two half arrows in opposite directions) for a full arrow pointing in the direction of the main product at that moment in time.

This has the benefit of showing where the equilibrium is, as the reaction will move towards the point of equilibrium

In this case the rate of the forward direction will gradually decline as the concentration or reactants decreases (number of collisions of the reactants decrease) and the rate of the backward direction will increase as the concentration of products increase (number of collisions of the products increase) until a balance is made (i.e. the number of successful collisions of reactants is the same as the number of sucessful collisions of the products).

This explains why the emf approaches zero as a current is allowed to run. The rate of the forward reaction decreases and the rate of the backward direction increases until equilibrium is reached, the rates are the same and there is no net exchange of electrons (e.g. the number of electrons donated by zinc would equal the number of electrons accepted by it).

This substitution of the arrow only applies in this situation as we are considering the initial reaction rates and not the reaction as it takes place. A high resisitance Voltmeter prevents the reaction from occurring and measures the emf (i.e. how far the reaction has to go before it reaches that balance).


You can imagine it as the equilibrium shifting further to the right so that the reaction is further away from reaching it (the distance is greater) so the emf becomes more positive!

imagine it like this:

one red ball + one green ball -----> one yellow ball + one blue ball
_______________________ <----

Initially there are 30 red balls and 30 green balls, whereas there are only 10 yellow balls and 10 blue balls.

To reach a balance (equilibrium) where the number of collisions of the red and green balls are the same as the number of collisions between the yellow and blue balls, the number of red and green balls must decrease and the number of yellow and blue balls must increase, so that each has a value of 20.

This would happen if the reaction was allowed to occur.

___30___________30______________10___________10
one red ball + one green ball -----> one yellow ball + one blue ball

Becomes

___20___________20______________20___________ 20
one red ball + one green ball -----> one yellow ball + one blue ball
_______________________<----

To reach equilibrium there must be a loss of 10 green and 10 red, and a gain of 10 yellow and 10 blue. This is the difference between the initial reaction and equilibrium.

If we were to increase the number of red and green balls, or decrease the number of yellow and blue balls, the difference would be greater, so the reaction would have further to go to reach equilibrium (i.e. the equilibrium would have shifted even more right to decrease the number of red and green balls) and so by extension emf would be more positive.

E.g.


__50____________50______________10____________10
one red ball + one green ball -----> one yellow ball + one blue ball

Becomes

__30____________30______________30____________30
one red ball + one green ball -----> one yellow ball + one blue ball
_______________________<----

When we increase the number of green and red balls, the equilibrium shifts further right to decrease the number of green and red balls and the distance between the initial reaction and equilibrium becomes 20 so the emf is greater.

Similarly,


___30____________30______________20 __________20
one red ball + one green ball -----> one yellow ball + one blue ball

Becomes

___25_____________25______________25__________25
one red ball + one green ball -----> one yellow ball + one blue ball
_______________________<----

We increase the number of yellow and blue balls, so the equilibrium shifts left slightly but the initial reaction still has to continue forward to reach equilibrium. However, the distance is lesser (5 balls) so the emf is by extension less positive.

Obviously the numbers on each side aren't necessarily equal on each side at equilibrium (due to conditions such as pressure and temperature) but they are in this scenario as it's not important in the point i'm trying to make.

Wow, sorry if I lost you there I tried to explain it in more simple terms and kind of failed... Maybe someone else can help lol.

Ignore the ___. Spaces and tabs don't really work

And now I'm going to bed!