Balance the following equations in basic medium by the ion-electron method and identify the oxidising agent and the reducing agent.
$(a)$ $P_{4(s)} + OH^{-}_{(aq)} \rightarrow PH_{3(g)} + H_{2}P{O_{2}}^{-}_{(aq)}$
$(b)$ $N_{2}H_{4(l)} + Cl{O_{3}}^{-}_{(aq)} \rightarrow NO_{(g)} + Cl^{-}_{(g)}$
$(c)$ $Cl_{2}O_{7(g)} + H_{2}O_{2(aq)} \rightarrow Cl{O_{2}}^{-}_{(aq)} + O_{2(g)} + H^{+}_{(aq)}$

(A) For $(a)$: $P_{4(s)} + 3OH^{-}_{(aq)} + 3H_{2}O_{(l)} \rightarrow PH_{3(g)} + 3H_{2}P{O_{2}}^{-}_{(aq)}$.
Oxidation half: $P_{4(s)} + 8OH^{-}_{(aq)} \rightarrow 4H_{2}P{O_{2}}^{-}_{(aq)} + 4e^{-}$.
Reduction half: $P_{4(s)} + 12H_{2}O_{(l)} + 12e^{-} \rightarrow 4PH_{3(g)} + 12OH^{-}_{(aq)}$.
Multiplying oxidation half by $3$ and adding to reduction half gives the balanced equation.
$P_{4}$ acts as both the oxidising and reducing agent (disproportionation reaction).