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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:
--- Quote from: Greed444 on November 04, 2010, 03:49:35 pm ---i have a doubt on May/June 2007 paper 4
Question 1(b). need help and explanation to get the answer please
--- End quote ---
Conditions Product at anode Product at cathode
ZnCl2(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 MnO42- in strongly acidic solution.
Calculate the E o cell for the process.
MnO42- + 8H+ + 4e <=> ? Mn2+ + 4H2O Eo = +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:
--- Quote from: moon on November 05, 2010, 08:50:13 pm ---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 MnO42- in strongly acidic solution.
Calculate the E o cell for the process.
MnO42- + 8H+ + 4e <=> ? Mn2+ + 4H2O Eo = +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.
--- End quote ---
i)Acidic conditions:
3MnO42- + 4H+ –> MnO2 + 2MnO4- + 2H2O
This disproportionation can be though of as two half equations:
1. MnO42- + 4H+ + 2e –> MnO2 + 2H2O ……….. Eº = +2.26V ( the Manganate ion is reduced)
2. MnO42- –> 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
2MnO43- + 4H+ ---> MnO2 + 2H2O + MnO42-
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