Balance equation with oxidation number of method.
$P_{4} + {O H}^{-} \to {P H}_{3} + {H P O}_{2}^{-}$

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Solution

${0 P}_{4} + - 2 O^{-} ⇌ - 3 P {+ 1 H}_{3} + {+ 1 H}_{2} + 1 P {- 2 O}_{2}^{-}$
$\Rightarrow$ $P_{4}^{0} ⟶ 4 {+ 1 H}_{2} + 1 P {- 2 O}_{2} + 4 e^{-}$ (oxidation)
$\Rightarrow$ ${0 P}_{4} + 12 e^{-} ⟶ 4 - 3 P {+ 1 H}_{3}$ (Reduction).
Balance the oxygen atoms in oxidation reaction.
${0 P}_{4} + 8 {O H}^{-} ⟶ 4 {+ 1 H}_{2} + 1 P {2 O}_{2} + 4 e^{-}$
$3 {0 P}_{4} + 24 {O H}^{-} ⟶ 12 {+ 1 H}_{2} + 1 P {+ 2 O}_{2} + 12 e^{-} (O)$
${0 P}_{4} + 12 e^{-} + 12 H_{2} O ⟶ 4 - 3 P {+ 1 H}_{3} + 12 {O H}^{-} (R) - ----------------------------------------------------- -$
$4 {0 P}_{4} + 24 {O H}^{-} + 12 e^{-} + 12 H_{2} O ⟶ 12 {+ 1 H}_{2} + 1 P {- 2 O}_{2} + 4 - 3 P {+ 1 H}_{3} + 12 e^{-} + 12 {O H}^{-}$
Simplity the above equation.
${0 P}_{4} + 3 {O H}^{-} + 3 H_{2} O ⟶ 3 {+ 1 H}_{2} + 1 P {- 2 O}_{2} + - 3 P {+ 1 H}_{3}$
$P_{4} + 3 {O H}^{-} + 3 H_{2} O ⇌ {P H}_{3} + 3 H_{2} {P O}_{2}$

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Balance equation with oxidation number of method.P4+OH−→PH3+HPO−2

Balance equation with oxidation number of method.
$P_{4} + {O H}^{-} \to {P H}_{3} + {H P O}_{2}^{-}$