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Hint: To solve this question you should have basic knowledge of oxidation state of an element, balancing redox reaction and some idea if you know the product of chemical reaction. In general, oxygen has -2 oxidation number in its compounds and 0 in free atomic state. Also, it has positive oxidation no. when it forms compounds with more electronegative atoms.
Complete step by step solution: Oxidation number or oxidation state is either the real charge or the virtual charge present on an element after cleavage (breaking) of all the bonds in a compound.
According to given question we have to find oxidation state of oxygen of ${{H}_{2}}{{O}_{2}}$ in the final products when it reacts with $ClO_{3}^{-}$
${{H}_{2}}{{O}_{2}}+ClO_{3}^{-}\to ClO_{2}^{-}+{{O}_{2}}$
Now, balance the redox reaction using the ion electron method.
1. Remove all the ions of which neither oxidation or reduction takes place.
2. Now break the equation in two half equations, one for oxidation and other for reduction.
3. Balance all the elements other than hydrogen and oxygen in the two-half equation.
4. Balance hydrogen and oxygen using ${{H}^{+}}$ and water in acidic medium and $O{{H}^{-}}$ water in basic medium.
5. Now, Balance the charge of two equations using the electrons.
6. Multiply the two-half equation with the appropriate number and add so that electrons get cancelled out in the final equation.
7. Now, convert the equation in the original format by adding the ions in which neither oxidation or reduction takes place.
Hence, balance equation is ${{H}_{2}}{{O}_{2}}+2ClO_{3}^{-}\to 2ClO_{2}^{-}+{{O}_{2}}+2{{H}^{+}}$
This is now balanced. The charges are the same on both sides and the number of atoms of each element is the same on both sides.
As here, we have ${{O}_{2}}$as the final product in this reaction, which is an element so this has an oxidation state to be 0.
Hence, option A is correct.
Note: Oxidation state of any element or compound is always 0.
In balancing of redox reaction, remember the most important step as it is common mistake done by students balance hydrogen and oxygen using ${{H}^{+}}$ and water in acidic medium and $O{{H}^{-}}$and water in basic medium.
Complete step by step solution: Oxidation number or oxidation state is either the real charge or the virtual charge present on an element after cleavage (breaking) of all the bonds in a compound.
According to given question we have to find oxidation state of oxygen of ${{H}_{2}}{{O}_{2}}$ in the final products when it reacts with $ClO_{3}^{-}$
${{H}_{2}}{{O}_{2}}+ClO_{3}^{-}\to ClO_{2}^{-}+{{O}_{2}}$
Now, balance the redox reaction using the ion electron method.
1. Remove all the ions of which neither oxidation or reduction takes place.
2. Now break the equation in two half equations, one for oxidation and other for reduction.
3. Balance all the elements other than hydrogen and oxygen in the two-half equation.
4. Balance hydrogen and oxygen using ${{H}^{+}}$ and water in acidic medium and $O{{H}^{-}}$ water in basic medium.
5. Now, Balance the charge of two equations using the electrons.
6. Multiply the two-half equation with the appropriate number and add so that electrons get cancelled out in the final equation.
7. Now, convert the equation in the original format by adding the ions in which neither oxidation or reduction takes place.
Hence, balance equation is ${{H}_{2}}{{O}_{2}}+2ClO_{3}^{-}\to 2ClO_{2}^{-}+{{O}_{2}}+2{{H}^{+}}$
This is now balanced. The charges are the same on both sides and the number of atoms of each element is the same on both sides.
As here, we have ${{O}_{2}}$as the final product in this reaction, which is an element so this has an oxidation state to be 0.
Hence, option A is correct.
Note: Oxidation state of any element or compound is always 0.
In balancing of redox reaction, remember the most important step as it is common mistake done by students balance hydrogen and oxygen using ${{H}^{+}}$ and water in acidic medium and $O{{H}^{-}}$and water in basic medium.
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