Balance the following redox reaction:
$P_{2} H_{4} ⟶ P H_{3} + P_{4} H_{2}$

5 P_{2} H_{4} ⟶ 6 P H_{3} + P_{4} H_{2}

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3 P_{2} H_{4} ⟶ 6 P H_{3} + P_{4} H_{2}

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3 P_{2} H_{4} ⟶ 4 P H_{3} + P_{4} H_{2}

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5 P_{2} H_{4} ⟶ 4 P H_{3} + P_{4} H_{2}

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Solution

The correct option is B $5 P_{2} H_{4} ⟶ 6 P H_{3} + P_{4} H_{2}$
The unbalanced redox equation is as follows:
$P_{2} H_{4} ⟶ P H_{3} + P_{4} H_{2}$
Balance all atoms other than H and O.
$5 P_{2} H_{4} ⟶ 2 P H_{3} + 2 P_{4} H_{2}$
The oxidation number of P changes from -2 to -3. The change in the oxidation number per P atom is 1.
Total change in the oxidation number for 2 P atoms is 2.
The oxidation number of P changes from -2 to -0.5. The change in the oxidation number per O atom is 1.5.
Total change in the oxidation number for 8 P atoms is 12.
The increase in the oxidation number is balanced with decrease in the oxidation number by changing the coefficient of $P_{2} H_{4}$ to 6 and the coefficient of $P H_{3}$ to 1.
$5 P_{2} H_{4} ⟶ 6 P H_{3} + P_{4} H_{2}$
This is the balanced chemical equation.

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Similar questions

P_{2} H_{4} \to P H_{3} + P_{4} H_{2}

P_{2} H_{4} \to {PH}_{3} + P_{4} H_{2}

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P H_{3} + \oplus N H_{4} ⟶ \oplus P H_{4} + N H_{3}

N H_{3} + P {\oplus H}_{4} ⟶ N {\oplus H}_{4} + P H_{3}

(C H_{3})_{3} P + \oplus N H_{4} ⟶ (C H_{3})_{3} \oplus P H + N H_{3}

(C H_{3})_{3} N + P H_{4}^{\oplus} ⟶ (C H_{3})_{3} \oplus N H + P H_{3}

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(M .wt N H_{4} OH - 35; pK,N H_{4} OH - 4 .75)

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