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In the electrolysis of H(2)SO(4), 9.72 l...

In the electrolysis of `H_(2)SO_(4)`, 9.72 litres and 2.35 litres of `H_(2)` and `O_(2)` were liberated. Number of equivalent of persulphuric acid `(H_(2)S_(2)O_(8))` produced is

A

0.448

B

0.224

C

0.868

D

0.42

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The correct Answer is:
To solve the problem regarding the electrolysis of \( H_2SO_4 \) and the production of persulphuric acid \( H_2S_2O_8 \), we can follow these steps: ### Step 1: Understand the Electrolysis Process During the electrolysis of sulfuric acid, hydrogen gas \( H_2 \) is produced at the cathode, and oxygen gas \( O_2 \) is produced at the anode. The overall reactions can be summarized as follows: - At the cathode: \( 2H^+ + 2e^- \rightarrow H_2 \) - At the anode: \( 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \) ### Step 2: Calculate the Equivalent of Hydrogen Gas The volume of hydrogen gas liberated is given as 9.72 liters. The equivalent of hydrogen can be calculated using the molar volume of a gas at standard conditions, which is 22.4 liters per mole. The number of equivalents of hydrogen can be calculated using the formula: \[ \text{Equivalent of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar Volume}} = \frac{9.72 \, \text{L}}{22.4 \, \text{L/mol}} = 0.433 \, \text{mol} \] ### Step 3: Calculate the Equivalent of Oxygen Gas The volume of oxygen gas liberated is given as 2.35 liters. Similarly, we can calculate the equivalent of oxygen: \[ \text{Equivalent of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar Volume}} = \frac{2.35 \, \text{L}}{22.4 \, \text{L/mol}} = 0.105 \, \text{mol} \] ### Step 4: Relate the Equivalents to Persulphuric Acid In the electrolysis process, the equivalents of \( H_2S_2O_8 \) produced will be equal to the difference between the equivalents of hydrogen and oxygen, since each mole of \( H_2S_2O_8 \) corresponds to the consumption of 2 moles of hydrogen and 1 mole of oxygen: \[ \text{Equivalent of } H_2S_2O_8 = \text{Equivalent of } H_2 - \text{Equivalent of } O_2 \] Substituting the values we calculated: \[ \text{Equivalent of } H_2S_2O_8 = 0.433 - 0.105 = 0.328 \] ### Step 5: Final Calculation The number of equivalents of persulphuric acid produced is: \[ \text{Number of equivalents of } H_2S_2O_8 = 0.328 \, \text{mol} \] ### Conclusion Thus, the number of equivalents of persulphuric acid produced is approximately **0.328 equivalents**.

To solve the problem regarding the electrolysis of \( H_2SO_4 \) and the production of persulphuric acid \( H_2S_2O_8 \), we can follow these steps: ### Step 1: Understand the Electrolysis Process During the electrolysis of sulfuric acid, hydrogen gas \( H_2 \) is produced at the cathode, and oxygen gas \( O_2 \) is produced at the anode. The overall reactions can be summarized as follows: - At the cathode: \( 2H^+ + 2e^- \rightarrow H_2 \) - At the anode: \( 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \) ### Step 2: Calculate the Equivalent of Hydrogen Gas ...
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Peroxy disulphuric acid (H_(2)S_(2)O_(8)) can be prepared by electrolytic oxidation of H_(2)SO_(4) as : 2H_(2)SO_(4) rarr H_(2)S_(2)O_(8)+2H^(+)+2e^(-) Oxygen and hydrogen are by products. In such an electrolysis 9.72 litre of H_(2) and 2.35 litre of O_(2) were generated at NTP. What is the mass of peroxy disulphuric acid formed ?

During the electrolysis of conc H_(2)SO_(4) , it was found that H_(2)S_(2)O_(8) and O_(2) liberated in a molar ratio of 3:1 . How many moles of H_(2) were found of moles of H_(2)S_(2)O_(8) ? ( Express your answer as :3xx mol e s of H_(2), integer answer is between 0 and 5 0

Statement-1: During electrolysis of H_(2)SO_(4),H_(2) and O_(2) are liberated under normal concentrations and H_(2)S_(2)O_(8) at higher concentrations and lower temperature. Statement-2: Liberation of H_(2) and O_(2) at elecrode requires some potential drop known as overvoltage.

RC MUKHERJEE-ELECTROLYSIS AND ELECTROLYTIC CONDUCTANCE-Objective Problems
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  2. The time required (approx) to remove electrolytically one half from 0....

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  3. In the electrolysis of H(2)SO(4), 9.72 litres and 2.35 litres of H(2) ...

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  4. In the electrolysis of H(2)O, 11.2 litres of H(2) was liberated at cat...

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  9. In which of the following aqueous solutions H(2) and O(2) are not libe...

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  10. The aqueous solutions of the following substances were electrolysed. I...

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  11. Electrolytic conductance is due to movement of

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  14. Which of the following solutions of KCl has the lowest value of equiva...

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