Aqa a2 chem5 19th june 2012

Original post by Top Banana

Has anyone else done the specimen paper? Just did it and got 55% :cry2:

Did anyone else find it a really difficult and odd paper or was I just terrible?

Specimen papers are usually odd :smile:

The CHEM5 paper has an old style section B which is unlike what we'd be asked - our questions are split up into smaller marks.

They're good for practicing weird questions that are on the fringe of the specification but otherwise I wouldn't worry - what have you been getting in recent past papers? I'd rely on that instead :tongue:

Reply 102

Top Banana

10

Original post by wibletg

Specimen papers are usually odd :smile:

The CHEM5 paper has an old style section B which is unlike what we'd be asked - our questions are split up into smaller marks.

They're good for practicing weird questions that are on the fringe of the specification but otherwise I wouldn't worry - what have you been getting in recent past papers? I'd rely on that instead :tongue:

Thanks, ive usually been getting 5%-10% more, but a result like that does make you worry. Thanks again :smile:

Reply 103

Does anyone here know who do the conventional cell representation???

Reply 104

Alpha5

Good thing about Chem 5 is that the GB's are lower in comparison to Chem 4.

Reply 105

Can someone tell me what we need to know about the hydrogen - oxygen fuel cell???

Reply 106

MattRay94

1

@Shuaib

Conventional Cell Representation is done as follows (Eg. for the following cell...)

Zn²⁺_(aq) + 2e^- -> Zn_(s) = -0.76V
Fe³⁺_(aq) + e^- -> Fe²⁺_(aq) = +0.77V

1. Identify which of the two electrodes is the positive terminal of the cell. This will simply be the one with the more positive standard electrode potential value. This will be written on the right hand side of the cell representation. In this case, it is the iron 3+/iron 2+ electrode.

2. Write the reduced form on the outside. Identify which is the reduction product in each of the reactions (i.e. has the lower oxidation number). In this case it is Iron 2+ and zinc solid respectively.

3. Add a salt bridge between the two halves (written as a dashed line, or a double vertical line)

4. Add a single vertical line between chemicals in different states, and a comma between those in the same state.

5. If either of the sides only contains aqueous chemicals, add Pt(s) to show a platinum electrode is used.

So here we get
Zn_(s) | Zn²⁺_(aq) || Fe³⁺_(aq) , Fe²⁺_(aq) | Pt(s)

To check: Do the chemicals on the outsides have a lower oxidation state than their respective chemical in the centre?
Is the positive electrode on the right hand side?
Have I included appropriate notation to account for state, including (if required) a platinum electrode.

Job done. :biggrin:

(edited 11 years ago)

Reply 107

Original post by MattRay94

@Shuaib

Conventional Cell Representation is done as follows (Eg. for the following cell...)

Zn²⁺_(aq) + 2e^- -> Zn_(s) = -0.76V
Fe³⁺_(aq) + e^- -> Fe²⁺_(aq) = +0.77V

1. Identify which of the two electrodes is the positive terminal of the cell. This will simply be the one with the more positive standard electrode potential value. This will be written on the right hand side of the cell representation. In this case, it is the iron 3+/iron 2+ electrode.

2. Write the reduced form on the outside. Identify which is the reduction product in each of the reactions (i.e. has the lower oxidation number). In this case it is Iron 2+ and zinc solid respectively.

3. Add a salt bridge between the two halves (written as a dashed line, or a double vertical line)

4. Add a single vertical line between chemicals in different states, and a comma between those in the same state.

5. If either of the sides only contains aqueous chemicals, add Pt(s) to show a platinum electrode is used.

So here we get
Zn_(s) | Zn²⁺_(aq) || Fe³⁺_(aq) , Fe²⁺_(aq) | Pt(s)

To check: Do the chemicals on the outsides have a lower oxidation state than their respective chemical in the centre?
Is the positive electrode on the right hand side?
Have I included appropriate notation to account for state, including (if required) a platinum electrode.

Job done. :biggrin:

Thanks so much, but what happens if a gas is in on of the equations???

(edited 11 years ago)

Reply 108

wasting-time

9

Original post by shuaib786

Can someone tell me what we need to know about the hydrogen - oxygen fuel cell???

1. The electrode reactions, so:
H2 + 2OH- --> H2O + 2e-

O2 + 2H2O + 4e- --> 4OH-

2. Understand why the cell does not need to be recharged (the reagants are constantly beign fed into the cell so their concentrations don't decrease over time thus p.d. stays constant).

3.Benefits to society ie. less use of fossil fuels.

Reply 109

Original post by wasting-time

1. The electrode reactions, so:
H2 + 2OH- --> H2O + 2e-

O2 + 2H2O + 4e- --> 4OH-

2. Understand why the cell does not need to be recharged (the reagants are constantly beign fed into the cell so their concentrations don't decrease over time thus p.d. stays constant).

3.Benefits to society ie. less use of fossil fuels.

And risks

Hydrogen is flammable! :tongue:

Plus the hydrogen used to refill the fuel cells is often sourced from crude oil or other non 'carbon neutral' sources so it's often not as environmentally friendly as it first appears.

Reply 110

wasting-time

9

Original post by Alpha5

Good thing about Chem 5 is that the GB's are lower in comparison to Chem 4.

Yeah this. In june 2011 you had to get 87/100 to get full ums which is way nicer than chem 4.

Reply 111

Sheo

5

Hello!

Can someone explain to me why chelation leads to a more stable complex ion?

I understand it leads to a positive entropy change because there is more moles of products than reactants.
and this leads to increased disorder, but I don't get how this links to increased stability?

Reply 112

Original post by Sheo

Hello!

Can someone explain to me why chelation leads to a more stable complex ion?

I understand it leads to a positive entropy change because there is more moles of products than reactants.
and this leads to increased disorder, but I don't get how this links to increased stability?

If there's a positive entropy change then -TdeltaS is negative.

As you're essentially breaking and reforming the same bonds deltaH is close to zero.

This means for most if not all temperatures deltaG is negative which means the reaction is feasible. Because 'free energy' is being given out, the products are in a lower energy state and things in a lower energy state are more thermodynamically stable. :smile:

Reply 113

mrshinyshoes

13

Original post by wasting-time

1. The electrode reactions, so:
H2 + 2OH- --> H2O + 2e-

O2 + 2H2O + 4e- --> 4OH-

2. Understand why the cell does not need to be recharged (the reagants are constantly beign fed into the cell so their concentrations don't decrease over time thus p.d. stays constant).

3.Benefits to society ie. less use of fossil fuels.

Is there anyway to work the equations out or do we just have to learn them? Sorry for being a noob :smile:

Reply 114

wasting-time

9

Original post by mrshinyshoes

Is there anyway to work the equations out or do we just have to learn them? Sorry for being a noob :smile:

you can work them out if you remeber that the overall product is water, but its probobly easier to just learn them

Reply 115

Please can someone help me with these questions???

1. Ammonium iron (II) sulphate crystals have the following formula:
(NH4)2SO4.FeSO4.nH2O. In an experiment to determine n, 8.492g of the salt were dissolved and made up to 250 cm3 of solution with distilled water and dilute sulphuric acid. A 25 cm3 portion of the solution was further acidified and titrated against potassium manganate (VII) solution of concentration 0.0150 moldm-3. A volume of 22.5 cm3 was required. Determine n.

2. Calculate x in the formula FeSO4.xH2O from the following data: 24.4 g of iron (II)
sulphate crystals were made up to 1 dm3 of aqueous solution acidified with sulphuric acid. 25.0 cm3 of the solution required 16.6 cm3 of 0.022M K2Cr2O7 for complete reaction.

3. Draw the conventional representation of the electrochemical cells to show how you would use the following reactions to make electricity (they are all spontaneous). In each case indicate the polarity of the electrodes.

a) Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
b) 2H2(g) + O2(g)  2H2O(l)

(edited 11 years ago)

Reply 116

Original post by shuaib786

Please can someone help me with these questions???

1. Ammonium iron (II) sulphate crystals have the following formula:
(NH4)2SO4.FeSO4.nH2O. In an experiment to determine n, 8.492g of the salt were dissolved and made up to 250 cm3 of solution with distilled water and dilute sulphuric acid. A 25 cm3 portion of the solution was further acidified and titrated against potassium manganate (VII) solution of concentration 0.0150 moldm-3. A volume of 22.5 cm3 was required. Determine n.

2. Calculate x in the formula FeSO4.xH2O from the following data: 24.4 g of iron (II)
sulphate crystals were made up to 1 dm3 of aqueous solution acidified with sulphuric acid. 25.0 cm3 of the solution required 16.6 cm3 of 0.022M K2Cr2O7 for complete reaction.

3. Draw the conventional representation of the electrochemical cells to show how you would use the following reactions to make electricity (they are all spontaneous). In each case indicate the polarity of the electrodes.

a) Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
b) 2H2(g) + O2(g)  2H2O(l)

In the first one - if I gave you a clue with the half equation would that help?

Fe2+ ------> Fe3+ + e
MnO4- + 8H+ + 5e ------> Mn2+ + 4H2O

Reply 117

Original post by wibletg

In the first one - if I gave you a clue with the half equation would that help?

Fe2+ ------> Fe3+ + e
MnO4- + 8H+ + 5e ------> Mn2+ + 4H2O

Can you go throught the first one plz. I have got no clue.

(edited 11 years ago)

Reply 118

Original post by shuaib786

Can you go throught the first one plz. I have got no clue.

Potassium manganate is often used in redox titrations - you assume that it is reduced completely (to Mn2+) and that the other reactant (in this case Fe2+) is oxidised completely (to Fe3+)

It's a redox reaction, so you need to put the two half equations together.

The number of electrons produced by the oxidation of Fe2+ needs to equal the number of electrons required to reduce MnO4- to Mn2+

Fe2+ ------> Fe3+ + e
MnO4- + 8H+ + 5e ------> Mn2+ + 4H2O

So you need 5 moles of Fe2+ to produce 5 moles of electrons.

5Fe2+ + 8H+ + MnO4- -------> Mn2+ + 5Fe3+ + 4H2O

You know you're using 22.5cm cubed of 0.015M MnO4- so you should be able to work out the number of moles of MnO4- that has reacted and by extension the number of moles of Fe2+ that has reacted :smile:

(edited 11 years ago)

Reply 119