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(vii) S_(2)O_(3)^(2-)+I_(2)to S_(2)O_(6)^(2-)+I^(-). |
Answer» SOLUTION : . Eqalise the increase/ DECREASE in O.N by multiplying the S species by 1 and I species by 3. `S_(2)O_(3)^(-)+3I_(2)to S_(2)O_(6)^(2-)+3I^(-)`. Balance all other atoms EXCEPT O and H `S_(2)O_(3)^(2-)+3I_(2)to S_(2)O_(6)^(2-)+6I^(-)` Balance O ATOM by adding `H_2O` on the side falling short of Oxygen. `S_(2)O_(3)^(2-)+3I_(2)+3H_(2)Oto S_(2)O_(6)^(2-)+6I^(-)` Balance H atom by adding `H+` ion on the side falling short of hydrogen. `S_(2)O_(3)^(2-)+3I_(2)+3H_(2)Oto S_(2)O_(6)^(2-)+6I^(-)+6H^(+)`. Add EQUAL number of `OH^(-)` ion on the both side since hte medium is alkaline. `S_(2)O_(3)^(2-)+3I_(2)+3H_(2)O+6OH^(-)to S_(2)O_(6)^(2-)+6I^(-)+6H^(+)+60H^(-)`. `S_(2)O_(3)^(2-)+3I_(2)+3H_(2)O+6OH^(-)to S_(2)O_(6)^(2-)+6I^(-)+6H_(2)O`. |
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