Oxidation number method: The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence, \\\\(\\displaystyle P_4 \\\\) is oxidizing as well as reducing agent.During reduction, the total decrease in the oxidation number for 4 P atoms is 12.During oxidation, total increase in the oxidation number for 4 P atoms is 4.The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying \\\\(\\displaystyle H_2PO_2^-\\\\) with 3. \\\\(\\displaystyle P_4(s) +OH^-(aq) \\rightarrow PH_3(g) + 3 H_2PO_2^-(aq) \\\\)To balance O atoms, multiply \\\\(\\displaystyle OH^- \\\\) ions by 6.\\\\(\\displaystyle P_4(s) +6OH^-(aq) \\rightarrow PH_3(g) + 3 H_2PO_2^-(aq) \\\\)To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.\\\\(\\displaystyle P_4(s) +6OH^-(aq) +3H_2O(l)\\rightarrow PH_3(g) + 3 H_2PO_2^-(aq) +3OH^-(aq)\\\\)Subtract 3 hydroxide ions from both sides.\\\\(\\displaystyle P_4(s) +3OH^-(aq) +3H_2O(l)\\rightarrow PH_3(g) + 3 H_2PO_2^-(aq) \\\\)Ion electron method:The oxidation half reaction is \\\\(\\displaystyle P_4(s) \\rightarrow H_2PO_2^-(aq) \\\\).The P atom is balanced.\\\\(\\displaystyle P_4(s) \\rightarrow 4H_2PO_2^-(aq) \\\\)The oxidation number is balanced by adding 4 electrons on RHS.\\\\(\\displaystyle P_4(s) \\rightarrow 4H_2PO_2^-(aq) +4e^-\\\\)The charge is balanced by adding 8 hydroxide ions on LHS.\\\\(\\displaystyle P_4(s) +8OH^-(aq)\\rightarrow 4H_2PO_2^-(aq) \\\\)The O and H atoms are balanced.The reduction half reaction is \\\\(\\displaystyle P_4(s) \\rightarrow PH_3(g) \\\\).The oxidation number is balanced by adding 12 eelctrons on LHS.\\\\(\\displaystyle P_4(s) +12e^- \\rightarrow PH_3(g) \\\\)The charge is balanced by adding 12 hydroxide ions on RHS.\\\\(\\displaystyle P_4(s) +12e^- \\rightarrow PH_3(g)+12OH^- \\\\)The oxidation half reaction is multiplied by 3 and the reduction half reaction is multiplied by 2. The half reactions are then added to obtain balanced chemical equation.\\\\(\\displaystyle P_4(s) + 3OH^-(aq) +3H_2O (l) \\rightarrow PH_3(g)+3H_2PO_2^- (aq) \\\\)