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Answer:- The balanced equation is, .
Solution:- Oxidation number of S is increasing from 0 to 4 and so it is oxidation. Oxidation number of N is decreasing from 5 to 1 and so it is reduction.
We write the oxidation and reduction half equations and balance them. The given reaction is taking place in an acidic medium.
First of all we balance all the atoms other than H and O. Then oxygen is balanced by adding and hydrogen is balanced by adding
. Charge is balanced by adding electrons.
To makes the electrons equal for both the half equations we multiply the equation/equations by appropriate numbers.
Oxidation half equation:
S is already balanced. To balance O, we need to add three water molecules to the left side:
For balancing hydrogen, we need to add 4 hydrogen ions to the right side:
Now to balance the charge we need to add 4 electrons to the right side:
Reduction half equation:
To balance N, we need to multiply left side by 2:
For balancing oxygen, we need to add 5 water molecules to the right side:
To balance hydrogen, we need to add 8 hydrogen ions to the left side:
Now, for charge balance, we need to add 8 electrons to the left side:
First half equation has 4 electrons and second half equation has 8 electrons.
To make the electrons equal, we need to multiply oxidation half equation by 2:
Now we add both of these two half equations and cancel out common species. What we get on doing this is: