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Class-12-science»Chemistry

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P_{2} H_{4} \to {PH}_{3} + P_{4} H_{2} This equation need to be balanced in order to calculate n - factor Oxidation : P_{2} H_{4} \to P_{4} H_{2} Reduction : P_{2} H_{4} \to {PH}_{3} Step 1 : Balance the P atom in both the reactions Oxidation : 2 P_{2} H_{4} \to P_{4} H_{2} Reduction : P_{2} H_{4} \to 2 {PH}_{3} Step 2 : H is balanced by adding H^{+} to deficient side Oxidation : 2 P_{2} H_{4} \to P_{4} H_{2} + 6 H^{+} Reduction : P_{2} H_{4} + 2 H^{+} \to 2 {PH}_{3} Step 3 : Charge is balanced by adding electrons to the side with excess positive charge Oxidation : 2 P_{2} H_{4} \to P_{4} H_{2} + 6 H^{+} + 6 e^{-} . . (i) Reduction : P_{2} H_{4} + 2 H^{+} + 2 e^{-} \to 2 {PH}_{3} . . (ii) Both the reactions are now added and multiplied (ii) by 3 Therefore, 5 P_{2} H_{4} \to 6 {PH}_{3} + P_{4} H_{2} Now, 5 mol of P_{2} H_{4} involes 6 e^{-} SO, 1 mol of P_{2} H_{4} involes \frac{6}{5} e^{-} Hence, n = \frac{6}{5} and . Equivalent mass = \frac{Molar mass}{n} = \frac{M}{\frac{6}{5}} = \frac{5 M}{6} Hence, the correct option is (2)

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