Complete and balance the followig equations:
(a) `[Fe(CN)_(6)]^(4-)+H^(oplus)+MnO_(4)^ɵ rarr Fe^(3+)+CO_(2)+NO_(3)^ɵ+Mn^(2+)`
(b) `Cu_(3)P+Cr_(2)O_(7)^(2-) rarr Cu^(2+)+ H_(3)PO_(4)+Cr^(3+)`
(c) `P_(4)S_(6)+H^(oplus)+NO_(3)^ɵ rarr NO +H_(3)PO_(4)+SO_(2)+H_(2)O`
(d) `AuCl_(4)^ɵ+Zn rarr Au+Zn^(2+)+?`
(e) `Zn+oversetɵ(O)H rarrZn(OH)_(4)^(2-)+?`.

Explanation

`(CN)_(6)^(6-) rarr CO_(2)+NO_(3)^ɵ`
Proceed as in the above illustration
`underset(6x=-24)overset(-4 xx6)(C_(6)) underset(24-(-24)=48e^(-))(rarr) underset(underset(6x=24)(6x-24 =0))(6CO_(2)+48e^(-))` ....(i)
`underset(6x =18) overset(+3 xx 6)(N_(6)) underset(30-18 =12e^(-))(rarr) underset(underset(6x=30)(6x-36=-6))(6NO_(3)^ɵ+12e^(-))`....(ii)
`Fe^(2+) rarr Fe^(3+) + e^(-)`
`ulbar([Fe^(2+)overset(-6)((CN)_(6))]^(4-)rarr 6CO_(2)+6NO_(3)^ɵ +61 e^(-)+Fe^(3+) (iv))`
Balance equation (iv) in acidic medium
`30H_(2)O +[Fe(CN)_(6)]^(4-) rarr 6CO_(2)+6NO_(3)^ɵ+Fe^(3+) +61e^(-)+6 oversetɵ(O)H] xx5`....(v)
Balance the reduction reaction of `MnO_(4)^ɵ` to `Mn^(2+)`
`8H^(oplus)+5e^(-)+MnO_(4)^ɵ rarr Mn_(-)^(2+)+4H_(2)O]xx61`....(vi)
Add equation (v) and (vi) to get the final redox reaction.
`5[Fe(CN)_(6)]^(4-)+188H^(oplus)+61 MnO_(4)^ɵ rarr5Fe^(3+)`
`+30CO_(2)+30NO_(3)^ɵ +61Mn^(2+) +94H_(2)O`
(b) `overset(+1 xx3-3)((Cu_(3)P)) rarr 3Cu^(2+) +H_(2)PO_(4)`
`Cu_(3)^(3+) rarr 3Cu^(2+)+3e^(-)`
`underset(x=-3)overset(3x=3)(P^(3+)) rarr underset(underset(x=5)(3+x-8=0))overset(3x=6)(H_(3)PO_(4)+8e^(-))`
`ulbar(Cu_(3)P rarr 3Cu^(2+)+H_(3)PO_(4)+11e^(-))`
Balance the half oxidation reaction in acidic medium
`4H_(2)O+Cu_(3)P rarr 3Cu^(2+)+H_(3)PO_(4)+11e^(-)+5H^(oplus)`...(i)
Balance the half reduction reaction of `Cr_(2)O_(7)^(2-)` to `2Cr^(3+)`
`14 H^(oplus)+6e^(-)+Cr_(2)O_(7)^(2-) rarr 2Cr^(3+)+7H_(2)O`....(ii)
Multiply equation (i) by 6 and equation (ii) by 11 and add them to get the final redox reaction
`6Cu_(3)P+124H^(oplus)+11Cr_(2)O_(7)^(2-) rarr 18 Cu^(2+)+6H_(3)PO_(4)+22Cr^(3+)+53H_(2)O`
(c) `overset(+3xx 4-2 xx6)(P_(4)S_(6)) rarrH_(3)PO_(4)+SO_(2)`
`underset(4x=12) overset(+12)(P_(4)) rarr underset(underset(4x=20)(12+4x-32=0))(4H_(3)PO_(4)+8e^(-))`
`underset(6x=-12)overset(-12)(S_(6)) rarr underset(underset(6x=24)(6x-24=0))(6SO_(2)-36e^(-))`
`underline(P_(4)S_(6) rarr 4H_(3)PO_(4)+6SO_(2)+44e^(-))`....(i)
Balance equation (i) in acidic medium
`28H_(2)O+P_(4)S_(6) rarr 4H_(3)PO_(4) +6SO_(2)+44e^(-)+44H^(oplus)` ....(ii)
Balance the half reduction reaction of `NO_(3)^ɵ` to `NO`.
`4H^(oplus)+3e^(-)+ underset(underset(x-5)(x-6=-1))(NO_(3)^ɵ)rarr underset(underset(x=2)(x-2=0))(NO+2H_(2)O)`....(iii)
Multiply equation (ii) by 3 and equation (iii) by 44 and add them to get the final redox equation.
`3P_(4)S_(6)+44H^(oplus)+44NO_(3)^ɵ rarr 12H_(3)PO_(4)+44NO +18SO_(2) +4H_(2)O`
`(overset(+3-1xx4)(AuCl_(4)))^ɵ rarr Au`
`underset(ulbar(2Au^(3+)+3Zn rarr 2Au+3Zn^(2+)))([3e^(-)+ underset(Zn rarr Zn^(2+)+2e^(-))(Au^(3+)rarr Au)]_(xx3)^(xx2)`
To balance other and ions add `8Cl^ɵ` to both sides
`2AuCl_(4)^ɵ+3Zn rarr 2Au +3Zn^(2+)+8CL^ɵ`
Zn is oxidised to `Zn^(2+)`, so `H_(2)O` is reduced to `overset ɵ(O)H` and `1//2 H_(2)`
`{:(Zn rarr Zn^(2+)+cancel(2e^(-))),(cancele^(-)+H_(2)O rarr oversetɵ(O)H +(1)/(2)H_(2)"]"xx2),(ulbar(Zn +2H_(2)rarr Zn^(2+)+oversetɵ(O)H+H_(2))):}`
To balance other ions, add `2 oversetɵ(O)H` to both sides. So the net balanced redox equation is
`Zn+oversetɵ(2OH)+2H_(2)O rarr [Zn(OH)_(4)]^(2-)+H_(2)`.