Balance the following reactions in acidic and alkaline medium : `SO_(3)^(2-)(aq)toSO_(4)^(2-)(aq)`

Explanation

Acidic medium: `SO_(3)^(2-)aqtoSO_(4)^(2-)(aq)` In this reaction ,the left side of the equation is deficient in one O -atom. So the number of O - atoms on both sides is equalised by adding one `H_(2)O(l)` molecule to the left side .
`SO_(3)^(2-)aq+H_(2)O(l)toSO_(4)^(2-)(aq)`
For equalising the number of H -atoms one both sides `H^(+)`inos are added to the right side,So we get-
`So_(3)^(2)(aq)H_(2)O(l)toSO_(4)^(2-)(aq)+2H^(+)(aq)`
To balance the charge on both sides,2 electrons are added to the right side,we get-
`So_(3)^(2)(aq)H_(2)O(l)toSO_(4)^(2-)(aq)+2H^(+)(aq)`
Basic meduim : `SO_(3)^(2-)(aq)toSO_(4)^(2-)(aq)` In order to balance the number of O- atoms on the both side two `OH^(-)(aq)` ions and one `H_(2)O(l)`molecule are added to left side and right side respeectively . This gives-
`SO_(3)^(2-)(aq)+2OH^(-)(aq)toSO_(4)^(2-)(aq)+H_(2)O(l)`
To balance the change on both sides we add 2 electrons to the right side .Thus the balance equation is -
`SO_(3)^(2-)(aq)+2OH^(-)(aq)toSO_(4)^(2-)(aq)+H_(2)O(l)`