$$\begin{gathered} Balance the followingwith ion electron method: \\ {P}_{4_{\left(s\right)}}+O{H^{-}}_{\left(aq\right)}\to P{H}_3\left(g\right)+H_{2}P{O}_2^{-}\left(aq\right) \end{gathered}$$

Step 1: Write the two half reactions

Oxidation half reaction : P₄ à PH3

Reduction half reaction: P₄ à H₂PO₂ ^-

Step:2

Balance the atoms other than O and H

Oxidation half reaction : P₄ à 4PH3

Reduction half reaction: P₄ à 4H₂PO₂ ^-

Step:3

Balance the O and H atoms using H₂O and H⁺.

Oxidation half reaction : P₄ +12H⁺ à 4PH3

Reduction half reaction: P₄+8H₂O à 4H₂PO₂ ^-  +8H⁺

Step:3

Balance the charges using electrons

Oxidation half reaction : P₄ +12H⁺ + 12e à 4PH₃

Reduction half reaction: P₄+8H₂O à 4H₂PO₂ ^- +8H⁺  +4e

Step: 4

Multiply the reduction half reaction with 3 and add to the oxidation half reaction in order to get the net reaction after canceling the electrons.

Oxidation half reaction : P₄ +12H⁺ + 12e à 4PH₃

Reduction half reaction: P₄+8H₂O à 4H₂PO₂ ^- +8H⁺  +4e ] ×3

Net reaction:

4 P₄ +12H⁺ +24H₂O à 4PH₃ +12H₂PO₂ ^- +24H⁺

Or

4 P₄ +24H₂O à 4PH₃ +12H₂PO₂ ^- +12H⁺

Step:5

As the reaction takes place in basic medium add 12OH^- ions on both sides in order to neutralize 12H⁺.

4 P₄ +24H₂O+12OH^- à 4PH₃ +12H₂PO₂ ^- +12H⁺+12OH^-

Or

4 P₄ +24H₂O+12OH^- à 4PH₃ +12H₂PO₂ ^- +12H₂O

Balanced chemical equation is:

  4P₄ + 12OH^- + 12H₂O  à 4PH₃ + 12H₂PO₂ ^-