Complete and balance the following equation
(a). `Bi(OH)_3+SnO_2^(2-)toBi+SnO_3^(2-)`
(b). `[Fe(CN)_6]^(3-)+Cr_2O_7^(2-)toFe^(3+)+Cr^(3+)+CO_2+NO_3^(ɵ)`
(c). `VtoPH_3+P_4H_2`
(d). `P_2H_4toPH_3+P_4H_2`
(e). `Ca_3(PO_4)_2+SiO_2+CtoCaSiO_3+P_4+CO`

Explanation

(a). In `Bi(OH)_3,Bi^(3+)` is reduced to Bi and `SnO_2^(2-)` (stannite) is oxidised to `NsNO_3` (Stannate)
`Bi+cancel3e^(-)toBi]xx2`
`underline(2overset(ɵ)(O)H+H_2O+SnO_2^(2-)toSn^(+4)O_2^(2-)+cancel(2e^(-))+2H_2Oxx3`
or `2Bi(OH)_3+3SnO_2^(2-)to2Bi+3SnO_3^(2-)+3H_2O`
(b). In `[Fe(CN)_6]^(3-),Fe^(3+)` does not change, and`CN^(ɵ)` is oxidised to `CO_2` and `NO_3^(ɵ)` [(refer inllustration 3.5 (a). ]
`30H_2O+[Fe(CN)_6]^(3-)to6CO_2+6NO_3^(ɵ)+cancel(60e^(-)+Fe^(3+))+60H^(o+)`.. (i)
`14H^(o+)+cancel(6e^(-))+Cr_2O_7^(2-)to2Cr^(3+)+7H_2O`.. (ii)
Multiply equation (ii) by 10 and add equation (i) and (ii) to get final redox equation.
`underline(80H^(o+)[Fe(CN)_6]^(3-)10Cr_2O_7^(2-)to6CO_2+6NO_3^(ɵ)+Fe^(3+)+20Cr^(3+)+40H_2O)`
(c). `33overset(ɵ)(O)H+17H_2O+6VtoHV_6O_(17)^(3-)`
ior `33overset(ɵ)(O)H+6VtoHV_6O_(17)^(3-)+16H_2O+cancel(30e^(-))` .(i)
`underline(3overset(ɵ)(O)H+14H_2O+6VtoHV_6O_(17)^(3-)+15H_2)`
(d). It is disproportionation reaction.
`cancel(2H^(o+))+cancel(2e^(-))+P_2H_4to2PH_3]xx3`
`2P_2H_4toP_4H_2+cancel(6e^(-))+6H^(o+)`
`underline(5P_2H_4to6PH_3+P_4H_2)`
(e). `PO_4^(3-)toP_4 and CtoCO`
`32H^(o+)+cancel20e^(-)+4PO_4^(3-)toP_4+16H_2O`.. (i)
`H_2O+CtoCO+cancal(2e^(-))+2H^(o+)]xx10` ..(ii)
`underline(12H^(o+)+10C+4PO_4^(3-)toP_4+10CO+6H_2O ...(iii))`
`SiO_2` to `SiO_3^(2-)` no change in oxidation state
`H_2O+SiO_2toSiO_3^(2-)+2H^(o+)`
`6H_2O+6SiO_2to6SiO_3^(2-)+12H^(o+)` .(iv)
Add equations (iii) and (iv) to get the final redox equation.
`+10C+4PO_4^(3-)+6SiO_2toP_4+10CO+6SiO_3^(2-)`
To balance other ions, add `6Ca^(2+)` to both sides.
`underline(10C+2Ca_3(PO_4)_2+6SiO_2toP_4+10CO+6CaSiO_3)`