1) cal. the oxidation no. of Mn in KMnO4 and Cr in K2Cr2O7. 2) oxidation and reduction go side by side in redox reaction. justify it. 3) chlorine, bromine and iodine disproportionate in alkaline medium but fluorine does not. Why?
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3)becoz flouring is highly reactive.....
1) 1+x+4×(-2)=0 ,then
x will be 7 so ON of Mn is +7
same with K2Cr2O7
2+2x+7×(-2)=0
x will be 6 so ON of Cr will be +6
1) 1+x+4×(-2)=0 ,then
x will be 7 so ON of Mn is +7
same with K2Cr2O7
2+2x+7×(-2)=0
x will be 6 so ON of Cr will be +6
Answered by
7
Ans:- 1 (i) Mn in KMnO4 :-
O.N. of K =+1 ; O= -2
O.N. of Mn = 1+x+4×(-2)=0
=> x = 7
So,O.N. of Mn is +7
(ii) Cr in K2Cr2O7 :-
O.N. of K=+1 , O = -2
So, O.N. of Cr = 2+2x+7×(-2)=0
==> x = 6
So,O.N. of Cr will be +6
Ans :- 2 Oxidation is when an atom loses an electron (think LEO: Loose Electron Oxidation).
Reduction means an atom gains an electron (think GER: Gain Electron Reduction).
Look at the equation below.
2 Mg + O2 ----> 2 [Mg2+][O2-]
If we split the equation into its oxidation and reduction reactions we get:
Mg ----> Mg2+ + 2 e-
The magnesium atom loses two electrons in the above reaction, making this part of the reaction the oxidation part.
O2 + 4 e- ----> 2 O2-
Each oxygen molecule gains four electrons to form a pair of O2- ions, making this part of the reaction the reduction reaction.
Since matter is not created or destroyed, when electrons are lost by one atom they are gained by another atom, making oxidation and reduction linked.
Ans:- 3 In a disproportionation reaction, the same species is simultaneously oxidised as well as reduced. Therefore, for such a redox reaction to occur, the reacting species must contain an element which has adeast three oxidation states.
The element, in reacting species, is present in an intermediate state while lower and higher oxidation states are available for reduction and oxidation to occur (respectively). Fluorine is the strongest oxidising agent. It does not show positive oxidation state. That’s why fluorine does not show disproportionation reaction.
O.N. of K =+1 ; O= -2
O.N. of Mn = 1+x+4×(-2)=0
=> x = 7
So,O.N. of Mn is +7
(ii) Cr in K2Cr2O7 :-
O.N. of K=+1 , O = -2
So, O.N. of Cr = 2+2x+7×(-2)=0
==> x = 6
So,O.N. of Cr will be +6
Ans :- 2 Oxidation is when an atom loses an electron (think LEO: Loose Electron Oxidation).
Reduction means an atom gains an electron (think GER: Gain Electron Reduction).
Look at the equation below.
2 Mg + O2 ----> 2 [Mg2+][O2-]
If we split the equation into its oxidation and reduction reactions we get:
Mg ----> Mg2+ + 2 e-
The magnesium atom loses two electrons in the above reaction, making this part of the reaction the oxidation part.
O2 + 4 e- ----> 2 O2-
Each oxygen molecule gains four electrons to form a pair of O2- ions, making this part of the reaction the reduction reaction.
Since matter is not created or destroyed, when electrons are lost by one atom they are gained by another atom, making oxidation and reduction linked.
Ans:- 3 In a disproportionation reaction, the same species is simultaneously oxidised as well as reduced. Therefore, for such a redox reaction to occur, the reacting species must contain an element which has adeast three oxidation states.
The element, in reacting species, is present in an intermediate state while lower and higher oxidation states are available for reduction and oxidation to occur (respectively). Fluorine is the strongest oxidising agent. It does not show positive oxidation state. That’s why fluorine does not show disproportionation reaction.
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