Correct Answer - D
`overset(+3)(As_(2))overset(-2)(S_(3))+overset(+5)(N)overset(-2)(O_(3))+overset(+1)(H^(+))rarr overset(+5)(As)overset(-2)(O_(4)^(3-))+overset(0)(S)+overset(+2-2)(NO)+overset(+1)(H_(2))overset(-2)(O)`
The oxidation number of As increases from ` +3` to ` + 5`, of S increases from ` -2 ` to 0, whilen of N decreases from ` +5`, to `+2` . This implies that both As and S of ` As_2S_3` are oxidized while N of ` NO_3^(-)` is reduced. The change in oxidation numbers of As and S must be considered together as they must maintain their atomic ration fo ` 2: 3`.
Oxidation haf-reaction:
` As_2S_3 rarr AsO_4^(3-) + S`
Reduction half-reaction
` NO_3^(-) rarr NO`
Now we balance the two half -reactions
(a) Balance all atoms excpt H and O atoms
` As_2S_3 rarr 2 AsO_4^(3-) + 3S`
` NO_3^(-) rarr NO`
(b) Account for the change in oxidation number. Total inccrease in the oxidation numbern of As `( + 6 rarr + 10 )` is of four units, totla increase in the oxidation number fo ` S ( -6 rarr 0)` is of six units, and decrease in the oxidation number of ` N (+ 5 rarr +2)` is fo three units .
`As_2S_3 rarr 2AsO_4^(3-) + 3 S + 10e^(-)`
`NO_3^(-) + 3 e^(-) rarr NO`
(c ) Balance ionic charges by adding ` H^(+)` ions as the reaction is carried out in acidic medium
`As_(2)S_(3) rarr 2AsO_(4)^(3-) +3S+16H^(+) +10e^(-)`
`NO_(3)^(-)+4H^(+)+3e^(-) rarr NO`
(d) Balance H and O atoms by adding `H_(2)O` molecules:
`As_(2)S_(3)+8H_(2)O rarr 2AsO_(4)^(3-)+3S+16H^(+)+10e^(-)`
`NO_(3)^(-)+4H^(+)+3e^(-) rarr NO +2H_(2)O`
To equalize the number of electrons lost amnd gained, multiply oxidation half-reaction by 3 and resuction half-reaction by 10. Finally add them, cancelling electrons on both sides:
`3As_(2)S_(3)+10NO_(3)^(-)+4H_(2)O rarr 6AsO_(4)^(3-)+9S+8H^(+)`
