(a) $\mathrm{P}_{4(s)} + \mathrm{OH}^{-}_{(aq)}$ → $\mathrm{PH}_{3(g)} + \mathrm{{H_2PO_2}^{-}_{(aq)}}$ Oxidation number method : Total increase in O.N. of P from $\mathrm{P_4}$ to $\mathrm{H_2PO_2}$ – = 1 × 4 = 4 Total decrease in O.N. of P from $\mathrm{P_4}$ to $\mathrm{PH_3}$ = 3 × 4 = 12 Therefore, to balance increase/decrease in O.N. multiply $\mathrm{PH_3}$ by 1 and $\mathrm{H_2PO_2^{-}}$ by 3, we have $\mathrm{P}_{4(s)}$ → $\mathrm{PH}_{3(g)} + 3\mathrm{H_2PO_2^{-}}_{(aq)}$ To balance O atoms, add $6\mathrm{OH^{-}}$ on the left side $\mathrm{P}_{4(s)} + 6\mathrm{OH^{-}_{(aq)}}$ → $\mathrm{PH_3}_{(g)} + 3\mathrm{H_2PO_2^{-}}_{(aq)}$ To balance H atoms, add $3\mathrm{H_2O}$ to L.H.S. and $3\mathrm{OH^{-}}$ to the R.H.S. $\mathrm{P}_{4(s)} + 6\mathrm{OH^{-}_{(aq)}} + 3\mathrm{H_2O_{(l)}}$ → $\mathrm{PH_3}_{(g)} + 3\mathrm{H_2PO_2^{-}}_{(aq)} + 3\mathrm{OH^{-}}_{(aq)}$ or, $\mathrm{P}_{4(s)} + 3\mathrm{OH^{-}_{(aq)}} + 3\mathrm{H_2O_{(l)}}$ → $\mathrm{PH}_{3(g)} + 3\mathrm{H_2PO_2^{-}}_{(aq)}$ Ion electron method : The two half reactions are : Oxidation half reaction : $\mathrm{P}_{4(s)} + 8\mathrm{OH^{-}_{(aq)}}$ → $4\mathrm{H_2PO_2^{-}}_{(aq)} + 4e^{-}$ ... (i) Reduction half reaction : $\mathrm{P}_{4(s)} + 12\mathrm{H_2O_{(l)}} + 12e^{-}$ → $4\mathrm{PH}_{3(g)} + 12\mathrm{OH^{-}_{(aq)}}$ ... (ii) Multiply eq. (i) by 3 and add it to eq. (ii), we get $4\mathrm{P}_{4(s)} + 24\mathrm{OH^{-}_{(aq)}} + 12\mathrm{H_2O_{(l)}}$ → $4\mathrm{PH}_{3(g)} + 12\mathrm{H_2PO_2^{-}}_{(aq)} + 12\mathrm{OH^{-}}_{(aq)}$ or, $\mathrm{P}_{4(s)} + 3\mathrm{OH^{-}_{(aq)}} + 3\mathrm{H_2O_{(l)}}$ → $\mathrm{PH}_{3(g)} + 3\mathrm{H_2PO_2^{-}}_{(aq)}$ Reductant - phosphorus; oxidant-phosphorus (b) $\mathrm{N_2H_{4(l)}} + \mathrm{ClO_3^{-}{}_{(aq)}}$ → $\mathrm{NO_{(g)}} + \mathrm{Cl^{-}_{(aq)}}$ Oxidation number method : Net reaction is $6\mathrm{N_2H_4} + 8\mathrm{ClO^{-3}}$ → $12\mathrm{NO} + 8\mathrm{Cl^{-}} + 12\mathrm{H_2O}$ $3\mathrm{N_2H_4} + 4\mathrm{ClO^{-3}}$ → $6\mathrm{NO} + 4\mathrm{Cl^{-}} + 6\mathrm{H_2O}$ Ion-electron method : Oxidation half-reaction : [$\mathrm{N_2H_4} + 8\mathrm{OH^{-}}$ → $2\mathrm{NO} + 8e^{-} + 6\mathrm{H_2O}$] × 6 Reduction half-reaction : [$\mathrm{ClO_3^{-}} + 6e^{-} + 3\mathrm{H_2O}$ → $\mathrm{Cl^{-}} + 6\mathrm{OH^{-}}$] × 8 Net reaction is $6\mathrm{N_2H_4} + 8\mathrm{ClO_3^{-}}$ → $12\mathrm{NO} + 8\mathrm{Cl^{-}} + 12\mathrm{H_2O}$ $3\mathrm{N_2H_4} + 4\mathrm{ClO_3^{-}}$ → $6\mathrm{NO} + 4\mathrm{Cl^{-}} + 6\mathrm{H_2O}$ Reductant : $\mathrm{N_2H_4}$ ; Oxidant : $\mathrm{ClO_3^{-}}$ (c) $\mathrm{Cl_2O_{7(g)}} + \mathrm{H_2O_{2(aq)}}$ → $\mathrm{ClO_2^{-}}_{(aq)} + \mathrm{O_{2(g)}} + \mathrm{H^{+}_{(aq)}}$ Oxidation number method : Net reaction is $\mathrm{Cl_2O_7} + 4\mathrm{H_2O_2} + 2\mathrm{OH^{-}}$ → $2\mathrm{ClO_2^{-}} + 5\mathrm{H_2O} + 4\mathrm{O_2}$ Ion-electron method : Oxidation half-reaction : [$\mathrm{H_2O_2} + 2\mathrm{OH^{-}}$ → $\mathrm{O_2} + 2e^{-} + 2\mathrm{H_2O}$] × 4 Reduction half-reaction : $\mathrm{Cl_2O_7} + 8e^{-} + 3\mathrm{H_2O}$ → $2\mathrm{ClO_2^{-}} + 6\mathrm{OH^{-}}$ Net reaction is $\mathrm{Cl_2O_7} + 4\mathrm{H_2O_2} + 2\mathrm{OH^{-}}$ → $2\mathrm{ClO_2^{-}} + 5\mathrm{H_2O} + 4\mathrm{O_2}$ Reductant : $\mathrm{H_2O_2}$; Oxidant : $\mathrm{Cl_2O_7}$