Q. 21

Write balanced chemical equation for the following reactions:

(i) Permanganate ion (MnO4-) reacts with sulphur dioxide gas in acidic medium to produce Mn2+ and hydrogensulphate ion. (Balance by ion electron method)

(ii) Reaction of liquid hydrazine (N2H4) with chlorate ion (ClO-3) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method)

(iii) Dichlorine heptaoxide (Cl2O7) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO2-) and oxygen gas.

(Balance by ion electron method)

Answer :

Option (i) Permanganate ion (MnO4- ) reacts with sulphur dioxide gas in acidic medium to produce Mn2+ and hydrogensulphate ion. (Balance by ion electron method)

Equation balance :-


Step (i) Generate unbalanced equation so this reaction can be written as:



Here Mn undergoes reduction from +7 to +2 and sulphur ion undergoes oxidation from +4 to +6. By using ion electron or hal-reaction method,


Step (ii) The half reduction reaction is:


The half oxidation reaction is:


Step (iii) we balance the number of atoms other than O and H, as here they are already balanced we then balance the number of oxygen and hydrogen atoms.


Step (iv) We assume the reaction is undertaking under aqueous acidic conditions and thus balance by adding H2O on RHS of the equation to balance number of oxygen, and H+ on LHS to balance the hydrogen atoms.


Reduction:


Oxidation:


Step (v) in order to balance charges you need to add electrons on the unbalanced side, here reduction reaction has 1 electron less in LHS and 5 electron less in oxidation reaction in LHS(if needed you can multiply the equation by coefficients.)


Reduction:


Oxidation:


Step (vi) we add the Half-cell reactions to attain the overall reaction.



Step (vii) make electron gain and lost during oxidation and reduction equal, thus multiplying reduction reaction by 2 and oxidation reaction by 5.


Reduction:


Oxidation:


The overall reaction is met by adding thus half-reactions:



(ii) Reaction of liquid hydrazine (N2H4 ) with chlorate ion (ClO-3 ) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method)


Step (i) we generate the unbalanced skeleton:



Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.


Oxidation:


Reduction:


Step (iii) Balance the number of atoms of reduced and oxidized redox couples. Since its aready balanced, reactions are same as previous steps.


Oxidation:


Reduction:


Step (iv) For basic solutions, balance the charge by adding OH- on required side of the equation.


Oxidation:


Reduction:


Step (v) Balance the oxygen atoms. There are 6 oxygen atoms in LHS of oxidation reaction, so add water malecule to RHS of the equation to balance oxygen and 3 molecules to LHS of reduction reaction.


Oxidation:


Reduction:


Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. As they are same no need for any changes.


Oxidation:


Reduction:


Step (vii) Combine these two equation by adding them so that all reactants and products remain in the same side.



Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.



Check the charge on both sides whether is balanced or not for verification. As the charge is -1 on both sides reaction is balanced.


(iii) Dichlorine heptaoxide (Cl2O7 ) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO2- ) and oxygen gas.


(Balance by ion electron method)


Step (i) we generate the unbalanced skeleton:



Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.


Oxidation:


Reduction:


Step (iii) Balance the number of atoms of reduced and oxidized redox couples.


Oxidation:


Reduction:


Step (iv) For acidic solutions, balance the charge by adding H+ on required side of the equation.


Oxidation:


Reduction:


Step (v) Balance the oxygen atoms. There are 7 oxygen atoms in reduction reaction, so add needed water molecule to RHS of the equation to balance oxygen.


Oxidation:


Reduction:


Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. Multiply oxidation reaction by 3.


Oxidation:


Reduction:


Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.



Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.



The charge on both sides is 0, thus reaction is balanced.


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