Correct Answer - A
a. `Fe^(2+)rarrFe^(3+)+cancel(e^(-))]xx2`
`2H^(o+)+2cancel(e^(-))H_(2)O_(2)rarr2H_(2)O`
`2Fe^(2+)+H_(2)O_(2)+2H^(o+)rarr2Fe^(3+)+2H_(2)O`
b. `3H_(2)O+I^(o+)rarrIO_(3)^(ө)+4e^(-)+6H^(o+)`
`x=+1, x-6= -1`
`x=5`
Adding `Br^(ө)` to both sides.
`3H_(2)O+IBrrarrIO_(3)^(ө)+cancel(4e^(-))+6H^(o+)+br^(ө)]xx3`
`6H^(o+)+cancel(6e^(-))+BrO_(3)^(ө)rarrBr^(ө)+3H_(2)O]xx2`
`x-6=-1, x= -1`
`x=5`
`ulbar(3H_(2)O+3IBr+2BrO_(3)^(ө)rarr3IO_(3)^(ө)+5Br^(ө)+6H^(o+))`
c. `2I^(ө)rarrI_(2)+2e^(-)]xx5`
`2x=-2, 2x=0`
`12H^(o+)+10e^(-)+2IO_(3)^(ө)rarrI_(2)+6H_(2)O`
`2x-12= -2, 2x=0`
`2x=10`
`ulbar(10I^(ө)+2IO_(3)^(ө)+12H^(o+)rarr6I_(2)+6H_(2)O)`
d. Peroxide ion `(O_(2)^(ө))` undergoes reduction to give `overset(ө)OH` and `H_(2)O` undergoes oxidation to give `O_(2)`.
`2H^(o+)+3e^(-)+O_(2)^(ө)rarr2HO^(ө)]xx4`
`2x=-1, 2+2x= -2`
`2x= -4`
`2H_(2)OrarrO_(2)+cancel(4e^(-))+4H^(o+)]xx3`
`4+2x=0, 2x=0`
`2x= -4`
`ulbar(4O_(2)^(ө)+6H_(2)Orarr8HO^(ө)+3O_(2)+4H^(o+))`
e. `3H_(2)O+2HS^(ө)rarrS_(2)O_(3)^(2-)+8e^(-)+8H^(o+)`
`2+2x=-2, 2x-6=-2`
`2x= -4, 2x=4`
`4H^(o+)+4e^(-)+2HSO_(3)^(ө)rarrS_(2)O_(3)^(2-)+3H_(2)O]xx2`
`2+2x-12= -2, 2x=2`
`2x=8`
`ulbar(2HS^(ө)+4HSO_(3)^(ө)rarr3S_(2)O_(3)^(2-)+3H_(2)O)`
f. `[4H_(2)O+PrarrH_(2)PO_(4)^(ө)+5e^(-)+6H^(o+)]xx4`
`x=0, 2+x-8=-1`
`x=5`
or `16H_(2)O+P_(4)rarr4H_(2)PO_(4)^(ө)+cancel(20e^(-))+24H^(o+)]xx3`
`4H^(o+)+cancel(3e^(-))+NO_(3)^(ө)rarrNO+2H_(2)O]xx20`
`x-6=-1, x-2=0`
`x=5, x=2`
`ulbar(8H^(o+)+8H_(2)O+3P_(4)+20NO_(3)^(ө)rarr12H_(2)PO_(4)^(ө)+20NO)`
g. Here, `As_(2)S_(3)` is split into two parts `overset(+3xx2)As_(2)` and `overset(-2xx3)S_(3)` in which `As_(2)^(6+)` is oxididsed to `H_(3)AsO_(4)` and `S_(3)^(-6)` is also oxidised to `S`. Balance both the equations add them to get the final equation.
`8H_(2)O+As_(2)^(6+)rarr2H_(3)AsO_(4)+4e^(-)+10H^(o+)`
`2x=6, 6+2x-16=0`
`2x=10`
`S_(3)^(-6)rarr3S+6e^(-)`
`3x=-6,3x=0`
`cancel(8H_(2)O)+As_(2)S_(3)rarr2H_(3)AsO_(4)+3S+10H^(o+)+cancel(10e^(-))`...(i)
`8H^(o+)+cancel(5e^(-))+MnO_(4)^(ө)rarrMn^(2+)+4H_(2)O]xx2`...(ii)
`x-8=-1, x=2`
`x=7`
`ulbar(6H^(o+)+As_(2)S_(3)+2MnO_(4)rarr2Mn^(2+)+2H_(3)AsO_(4)+3S)`
h. Split `CrI_(3)` into parts `overset(+3)(Cr)and overset(-1xx3)(I_(3))` in which `Cr^(3+)` is oxidised to `CrO_(4)^(2-)` and `I_(3)^(3-)` is also oxidised to `IO_(4)^(ө)`. Balance both the equations in basic medium and add them to get the final equation.
`8overset(ө)OH+cancel(4H_(2)O)+Cr^(3+)rarrCrO_(4)^(2-)+3e^(-)+cancel(8H_(2)O)`
`x=3, x-8= -2`
`x=6`
`24overset(ө)OH+cancel(12H_(2)O)+I_(3)^(-3)rarr3I_(O)(4)^(ө)+24e^(-)+12H_(2)O`
`3x=-3, 3x-24= -3`
`3x=21`
`ulbar(32overset(ө)OH+CrI_(3)rarrCrO_(4)^(2-)+3IO_(4)^(ө)+16H_(2)O+cancel(27e^(-)]xx2)`
`2cancel(H_(2)O)+2cancel(e^(-))+H_(2)O_(2)rarr2H_(2)O+2overset(ө)OH]xx27`
`2+2x=0, 4+2x=0`
`2x= -2, 2x=-4`
`ulbar(10overset(ө)OH+2CrI_(3)+27H_(2)O_(2)rarr2CrO_(4)^(2-)+6IO_(4)^(ө)+32H_(2)O)`
i. Split `Fe_(2)S_(3)` into two parts `(overset(+3xx2)(Fe_(2))overset(-2xx3)(S_(3)))` in which `S_(3)^(6-)` undergoes oxidation to `S` and `O_(2)` undergoes reduction to `overset(ө)OH` in basic medium.
`S_(3)^(6-)rarr3S+cancel(6e^(-))]xx2`
`3x=-6, 3x=0`
`2H_(2)O+cancel(4e^(-))+O_(2)rarr2HO^(ө)+2HO^(ө)]xx3`
`2x=0, 2+2x=-2`
`2x=-4`
`ulbar(6H_(2)O+2S_(3)^(6-)+3O_(2)rarr6S+12HO^(ө))`
Add `4Fe^(3+)` to both sides.
`6H_(2)+2Fe_(2)S_(3)+3O_(2)rarr6S+4Fe(OH)_(3)`
j. `6H^(o+)+cancel(6e^(-))+IO_(3)^(ө)rarrI^(ө)+3H_(2)O`
`x-6=-1, x=-1`
`x=5`
`H_(2)O+HSO_(3)^(ө)rarrSO_(4)^(2-)+cancel(2e^(-))+3H^(o+)]xx3`
`1+x-6=-1, x-8=-2`
`x=4,x=6`
`ulbar(IO_(3)^(ө)+3HSO_(3)^(ө)rarrI^(ө)+3SO_(4)^(2-)+3H^(o+))`
k. `cancel(e-)+Fe^(3+)rarrFe^(2+)]xx4`
`2NH_(2)OHrarrN_(2)O+cancel(4e^(-))+H_(2)O+4H^(o+)`
`2x+4-4+2=0, 2x-2=0`
`2x=-2, 2x=2`
`ulbar(4Fe^(+3)+2NH_(2)OHrarr4Fe^(2+)+N_(2)O+H_(2)O+4H^(o+))`.
