Question

balance the following equation
(in basic medium)
$\bold{P _{4} \to \: H _{2} P{ O_{2}}^{ - } + P H_{3}}$
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Answer 1

Explanation:

P₄ → H₂PO₂⁻ + PH₃

Oxidation no. of:

P₄: 0

H₂PO₂⁻: 2

PH₃: -3

Now,

Oxidation half:

P₄ + 8H₂O → 4H₂PO₂⁻ + 12H⁺

[ P₄ + 8H₂O → 4H₂PO₂⁻ + 12H⁺ + 8e⁻ ] × 3

.... (i)

Reduction half:

P₄ + 12H⁺ → 4PH₃

[ P₄ + 12H⁺ + 12e⁻ → 4PH₃ ] × 2 ..... (ii)

From (i) and (ii), we get:

3P₄ + 24H₂O → 12H₂PO₂⁻ + 36H⁺ + 24e⁻

2P₄ + 24H⁺ + 24e⁻ → 8PH₃

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3P₄ + 24H₂O + 2P₄ + 24H⁺ + 24e⁻ → 12H₂PO₂⁻ + 36H⁺ + 24e⁻ + 8PH₃

5P₄ + 24H₂O → 12H₂PO₂⁻ + 12H⁺ + 8PH₃

Now,

Adding 12OH⁻ both sides we get:

5P₄ + 24H₂O + 12OH⁻ → 12H₂PO₂⁻ + 12H⁺ + 8PH₃ + 12OH⁻

5P₄ + 24H₂O + 12OH⁻ → 12H₂PO₂⁻ + 12H₂O + 8PH₃

5P₄ + 12H₂O + 12OH⁻ → 12H₂PO₂⁻ + 8PH₃

(Basic medium)