ALL CIE CHEMISTRY DOUBTS HERE !!

[Greed444]

i have a doubt on May/June 2007 paper 4
Question 1(b). need help and explanation to get the answer please

[Deadly_king]

i have a doubt on May/June 2007 paper 4
Question 1(b). need help and explanation to get the answer please

Conditions                                   Product at anode                          Product at cathode

ZnCl_2(l)                                        Cholrine                                      Zinc

ZnCl₂(concentrated aqueous)            Chlorine                                    Hydrogen

ZnCl₂(diltue aqueous)                       Oxygen                                    Hydrogen



NOTE : Positive ions get discharged at the cathode while negative ones are discharged at the anode.

i) Zinc Chloride in liquid form is composed of Zn²⁺ and Cl^- ions which are discharged at the electrodes forming the respective compounds.

ii) Concentrated aqueous Zinc Chloride contains the following ions : Zn²⁺, Cl^-, H⁺ and OH^-.
For cations, you should compare them from the reactivity series. The lower the cation is in the reactivity series the earlier it will be discharged in preference to the most reactive cation.
This is why hydrogen gets preferentially discharged at the expense of Zn²⁺

For anions, it's about the same thing but you should note that the second part is concentrated aqueous, which means that Cl^- is abundant while OH^- is present in very small amount.

Hence oxygen will be formed in both cases but it in so minimal amount that you won't even notice and Cl^- will immediately be discharged after that.

[moon]

plz help me in solving this question. thanx in advance.

Compounds containing manganese oxo-anions often disproportionate in aqueous solution.

(i) Use the following half-equation, and other data from the Data Booklet, to construct an
overall equation for the disproportionation of MnO₄^2- in strongly acidic solution.
Calculate the E _o cell for the process.
MnO₄^2- + 8H⁺ + 4e <=> ? Mn²⁺ + 4H₂O E^o = +1.74 V
 
(ii) When potassium manganate(VII) is reduced with aqueous sodium sulphite, the bright
blue salt K3MnO4 is produced. The salt readily disproportionates in acidic solution, giving
a brown precipitate of MnO2(s) and a purple solution.

Calculate the oxidation number of manganese in the blue salt, and construct a balanced
ionic equation for its disproportionation.

[moon]

can u help me in another Question plz.
 
June 08 Q2 a (ii) explain and draw the diagram. thank u very much for ur help.

[Deadly_king]

plz help me in solving this question. thanx in advance.

Compounds containing manganese oxo-anions often disproportionate in aqueous solution.

(i) Use the following half-equation, and other data from the Data Booklet, to construct an
overall equation for the disproportionation of MnO₄^2- in strongly acidic solution.
Calculate the E _o cell for the process.
MnO₄^2- + 8H⁺ + 4e <=> ? Mn²⁺ + 4H₂O E^o = +1.74 V
  
(ii) When potassium manganate(VII) is reduced with aqueous sodium sulphite, the bright
blue salt K3MnO4 is produced. The salt readily disproportionates in acidic solution, giving
a brown precipitate of MnO2(s) and a purple solution.

Calculate the oxidation number of manganese in the blue salt, and construct a balanced
ionic equation for its disproportionation.

i)Acidic conditions:

3MnO₄^2- + 4H⁺ –> MnO₂ + 2MnO₄^- + 2H₂O

This disproportionation can be though of as two half equations:
1. MnO₄^2- + 4H⁺ + 2e –> MnO₂ + 2H₂O ……….. Eº = +2.26V ( the Manganate ion is reduced)
2. MnO₄^2- –> MnO₄^- + 1e ……………………….Eº = +0.56V ( the Manganate ion is oxidised)

Calculating Eº = E(reduced state) – E(oxidised state) = +2.26V – +0.56V = + 1.70V
This is a large positive value (remember that any Eº value greater than 0.3v means that the reaction is spontaneous as shown) so the forward reaction proceeds.

Here is the reference.

ii) The blue salt is K₃MnO₄
Let the oxidation number of Mn in the salt be .
Oxidation number of O is -2 while that of K is +1. Overall charge is zero since it is an uncharged salt.

Hence 3(+1) + + 4(-2) = 0 ----> = +5

2MnO₄^3- + 4H⁺ ---> MnO₂ + 2H₂O + MnO₄^2-

[Deadly_king]

can u help me in another Question plz.
 
June 08 Q2 a (ii) explain and draw the diagram. thank u very much for ur help.

::O=:Cl.=O::

Valence electrons = 19

Chlorine attaches to the oxygens through double bonds, and and each oxygen has two lone pairs to complete its octet. The chlorine also one lone pair and one electron. Chlorine in this structure obviously does not obey octet rule but in chlorine there is 3d subshell in which electrons are promoted and hence chlorine can also have 3,5 and 7 bonds.

Since chlorine has one lone pair, the ClO₂ molecule will be a bent(v-shaped) one but not the same angle as H₂O since it has only one lone pair.

For more details click here.

[moon]

::O=:Cl.=O::

Valence electrons = 19

Chlorine in this structure obviously does not obey octet rule but in chlorine there is 3d subshell in which electrons are promoted and hence chlorine can also have 3,5 and 7 bonds.

How can we know the number of valence electrons??and how can we know that chlorine is not obeying octet rule???
how do we determine the number of bonds formed by the chlorine in this case?

For the question above
how to construct a balanced ionic equation for its disproportionation.
  Thanx in advance

[$H00t!N& $t@r]

can someone please help me with this:

Q) A flask containing a mixture of 0.200 mol of ethanoic acid and 0.11 mol of ethanol was maintained at 25 degrees Celsius until the following equilibrium had been established.
CH3C00H + C2H5OH <-> CH3COOC2H5 + H2O

The ethanoic present at equilibrium required 72.5 cm3 of a 1.50 mol dm-3 solution of sodium hydroxide for a complete reaction.
- Calculate the value of the equilibrium constant Kc for this reaction at 25 degrees Celsius
- The enthalpy change for this reaction is quite small. By reference to the number and type of bonds broken and made, explain how this might have been predicted.

Thanks in advance  
and sorry i dont have the answer to this question because i found it on a website 

[Deadly_king]

How can we know the number of valence electrons??and how can we know that chlorine is not obeying octet rule???
how do we determine the number of bonds formed by the chlorine in this case?

For the question above
how to construct a balanced ionic equation for its disproportionation.
  Thanx in advance

Valence electrons is the total number of electrons from the outermost shells of the atoms forming the molecule.

Number of electrons in the outermost shell of O : 6
Number of electrons in the outermost shell of Cl : 7

Hence valence electrons : 2(6) + 7 = 19

Hmm............in this case ClO₂ is a neutral molecule which means that oxygen has a stable electronic configuration. Otherwise there would have been a negative charge.

For oxygen to have a stable electronic configuration, it should form two covalent bonds or gain two electrons. Cl is a non-metal which does not give electrons. The only plausible solution is that it formed two bonds with each O atom.

Hence structure of ClO2 => ::O=:Cl.=O::

Moreover in the question itself, it is stated that the bonds between Cl and O is double bond.

I need to go now. I'll complete the answer asap

[moon]

Thank you very very much for ur help

[Hypernova]

Lets first find the number of moles of CH3C00H at equilibrium. It reacts with NaOH in 1:1 ration.

number of moles=concentration*volume
               =1.5 * (72.5/1000)
               =0.10875 moles of CH3C00H at equilibrium

When 1 mole of CH3C00H reacts with 1 mole of C2H5OH 1 mole of CH3COOC2H5 and H2O is produced.
So if 0.09125 moles of CH3C00H reacts, then 0.09125 of CH3COOC2H5 and H2O is formed.


                      CH3C00H      +     C2H5OH <->     CH3COOC2H5 +      H2O

Before equilibrium      0.2                0.11                     0              0
At equilibrium         0.10875           0.01875            0.09125        0.09125


  KC= 0.09125 * 0.09125
             0.10875 * 0.01875

    
    =4.08

    
Note: the concentration should be used to find Kc, but since they all in the same container, the volumes is the same and will cancel out in this case, so i didn't need to put in volumes.


For the second part, you need to find enthalpy change of reaction. enthalpy change = bonds broken - bonds made
but thats long.

[$H00t!N& $t@r]

Thanks alot for the detailed answer!
+rep 

[Greed444]

Conditions                                   Product at anode                          Product at cathode

ZnCl_2(l)                                        Cholrine                                      Zinc

ZnCl₂(concentrated aqueous)            Chlorine                                    Hydrogen

ZnCl₂(diltue aqueous)                       Oxygen                                    Hydrogen



NOTE : Positive ions get discharged at the cathode while negative ones are discharged at the anode.

i) Zinc Chloride in liquid form is composed of Zn²⁺ and Cl^- ions which are discharged at the electrodes forming the respective compounds.

ii) Concentrated aqueous Zinc Chloride contains the following ions : Zn²⁺, Cl^-, H⁺ and OH^-.
For cations, you should compare them from the reactivity series. The lower the cation is in the reactivity series the earlier it will be discharged in preference to the most reactive cation.
This is why hydrogen gets preferentially discharged at the expense of Zn²⁺

For anions, it's about the same thing but you should note that the second part is concentrated aqueous, which means that Cl^- is abundant while OH^- is present in very small amount.

Hence oxygen will be formed in both cases but it in so minimal amount that you won't even notice and Cl^- will immediately be discharged after that.

Thanks Bro! i Completely understand now

[Greed444]

does anyone know why some books say
1) E(std)cell = E(+ve terminal) - E(-ve terminal)
2) E(std)cell = E(right hand terminal) - E(left hand terminal)

in my class, we used the first one. but when i attempt June 2008 paper4 Q1, they used the 2nd one to get the answer. im confused :S

[moon]

does anyone know why some books say
1) E(std)cell = E(+ve terminal) - E(-ve terminal)
2) E(std)cell = E(right hand terminal) - E(left hand terminal)

in my class, we used the first one. but when i attempt June 2008 paper4 Q1, they used the 2nd one to get the answer. im confused :S

Yes I am confused too . I would really appreciate any help Thanks in advance