Balance the equation
`As_(2)(s)+NO_(3)^-)(aq)+H^(+)(aq)rarrAsO_+(4)^(3-)(aq)+S(s)NO(g)+H_(2)O(l)`

Explanation

To identify the atoms whose oxidation number have undergone a chage writing the oxidatin number of each atom above its symbol we have
Herer the oxidation number of As has increased fro +3 to 5 and that of S has increased form -2 to 0 while
that of N has decreased form +5 to +2 in other words both As and S have been oxidised when `NO_(3)^(-)` ahs been reduced since As nand S must maintain theri atoimic ration of 2:3 (as in `As_(2)S_(3)` therefore the change in oxidation numbers of these two atoms must be considered to gether keeping in view these points the above redox reaction can be split up unto the following two half reaction
Oxidation half equation : `As_(2)S_(3)^(-2)(s)rarrAsO_(4)^(3-)(aq)+S(s)`
Reduction half equation : `NO_(3)^(-)(aq) rarr NO(g)`
Step 2 To balace the oxidation half Eq (i)
(a) Balance all the atoms other than O Multiply As `O_(4)^(3-)` by 2 and S by 3 on R.H.S of Eq (i) We have
`overset(+3)As_(2)overset(-2)S_(3) rarr 2 AsO_(4)^(3-)+3S(s)`
(b)Balance the oxidain number by adding electron s since each as atom loses two electrons and there are two as atoms therefore due to the oxidation of As alone add `4 e^(-)` R.H.S of Eq (iii) further since each s atom loses two electron and there are three S atoms therfroe due to the oxiation of S alone add `6 e^(-)` to R.H.S of Eq (iii) Combing these two oxidation steps together and `10 e^(-)` to R.H.S of Eq (iii) we have
(c )Balance charge by adding `H^(+)` ions
The total charge on R.H.S of Eq (iv) is 16 nad zero on the L.H.S therefore add `16 H^(+)` to R.H.S of Eq (iv) we have
(d) Balance O atoms by adding `H_(2)O`
Step 3 To balance the reductionn half eq (ii)
(a) Balance oxidaton number byu adding electron oxidation number of N in `NO_(3)^(-)` is +5 on L.H.S while it is +2 in no on R.H.S therefore add 3 `e^(-)` L.H.S of Eq (ii) we have
(b) Balance charge by adding `H^(+)` ions
The total charge of L.H.S is -4 while it is zero on R.H.S therefore add 4 `H^(+)` to L.H.S of Eq (vii) We have
`NO_(3)^(-)(aq)+4h^(+)(aq)+3e^(-)rarrNO(g)`
step 4 To balance the electrons lost in Eq (vi) and gained in Eq (ix) multiply Eq (ix) by 10 Eq (vi) by 3 and add we have