How does elemental phosphorus react to give $P H_{3}$ $PH_3$ and $H P O_{2}^{-}$ $HPO_2^(-)$ under basic conditions?

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How does elemental phosphorus react to give $P H_{3}$ $PH_3$ and $H P O_{2}^{-}$ $HPO_2^(-)$ under basic conditions?

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Answer 1 · 2016-12-19T15:29:49

This is a disproportionation reaction. Elemental phosphorus is reduced and oxidized.

Explanation:

$\frac{1}{4} P_{4} + 3 H^{+} + 3 e^{-} \to P H_{3}$ $1/4P_4+3H^+ + 3e^(-) rarrPH_3$ $(i) reduction$ $"(i) reduction"$

$\frac{1}{4} P_{4} + 2 H O^{-} \to H P O_{2}^{-} + H^{+} + 2 e^{-}$ $1/4P_4 + 2HO^(-) rarr HPO_2^(-) +H^(+) + 2e^(-)$ $(ii) oxidation$ $" (ii) oxidation"$

Overall: $2 \times (i) + 3 \times (i i)$ $2xx(i) + 3xx(ii)$

$\frac{5}{4} P_{4} + 6 H_{2} O \to 2 P H_{3} + 3 H P O_{2}^{-} + 3 H^{+}$ $5/4P_4+6H_2O rarr2PH_3 +3HPO_2^(-) +3H^(+)$

I will add $3 \times H O^{-}$ $3xxHO^-$ to each side of the equation to represent that the reaction is performed in basic conditions:

$\frac{5}{4} P_{4} + 3 H O^{-} + 3 H_{2} O \to 2 P H_{3} + 3 H P O_{2}^{-}$ $5/4P_4+3HO^(-) + 3H_2O rarr2PH_3 +3HPO_2^(-)$

Is this balanced with respect to mass and charge?

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How does elemental phosphorus react to give $P H_{3}$ $PH_3$ and $H P O_{2}^{-}$ $HPO_2^(-)$ under basic conditions?

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How does elemental phosphorus react to give $P H_{3}$ $PH_3$ and $H P O_{2}^{-}$ $HPO_2^(-)$ under basic conditions?