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Standard X

Chemistry

Electrolysis and Electrolytes

During an ele...

Question

During an electrolysis of conc. $H_{2} S O_{4}$ , perdisulphuric acid $(H_{2} S_{2} O_{8})$ and $O_{2}$ form in equimolar amount. The amount of $H_{2}$ that will form simultaneously will be: $(2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-})$

Thrice that of O

_{2}

in moles

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Twice that of O

_{2}

in moles

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Equal to that of O

_{2}

in moles

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Half of that of O

_{2}

in moles

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Solution

The correct option is A Thrice that of O $_{2}$ in moles
Reaction of process:
$A n o d e {\begin{matrix} 2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-} 2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-} \end{matrix}}$

$C a t h o d e {2 H_{2} O \to H_{2} + 2 O H^{-} - 2 e^{-}} \times 3$
---------------------------------------------------------------------------------------------------------
Net: $2 H_{2} S O_{4} + 8 H_{2} O \to H_{2} S_{2} O_{8} + O_{2} + 3 H_{2} + 6 H^{+} + 6 O H^{-}$
Hence ratio of $n_{O_{2}}$ and $n_{H_{2}}$ is 1 : 3.

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Similar questions

Q. Marshall's acid or perdisulphuric acid

(H_{2} S_{2} O_{8})

can be prepared by electrolytic oxidation of

H_{2} S O_{4}

as :

2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{\oplus} + 2 e^{-}

O_{2}

and

H_{2}

are byproducts. In such an electrolysis 9.72 L of

H_{2}

and 2.35 L of

O_{2}

were produced at S.T.P. What is the weight of

H_{2} S_{2} O_{8}

formed ?

Q. Peroxydisulfuric acid

(H_{2} S_{2} O_{8})

can be prepared by electrolytic oxidation of

H_{2} S O_{4}

as:

2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e

O_{2}

and

H_{2}

are by products. In such on electrolysis

0.87

g of

H_{2}

and

3.36

O_{2}

were generated at STP. Report the mass of

H_{2} S_{2} O_{8}

formed in decagram (write nearest integer).

Marshalls acid is prepared by the electrolytic oxidation of $H_{2} S O_{4}$ as $2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{\oplus} + 2 e^{-}$

Oxygen and hydrogen are byproducts. In such Electrolysis 2.24 L of $H_{2}$ amd 0.56 L $O_{2}$ were produced at STP. The weight of $H_{2} S_{2} O_{8}$ formed is:

Q. Peroxy disulphuric acid

(H_{2} S_{2} O_{8})

can be prepared by electrolytic oxidation of

H_{2} S O_{4}

as:

2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-}

Oxygen and hydrogen are by products. In such an electrolysis

9.72 l i t r e

H_{2}

and

2.35

litre of

O_{2}

were generated at NTP. What is the mass of peroxy disulphuric acid formed?

Q. One of the methods of preparation of per disulphuric acid,

H_{2} S_{2} O_{8}

involve electrolytic oxidation of

H_{2} S O_{4}

at anode

(2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-})

with oxygen and hydrogen as by-products. In such an electrolysis,

10.08 L of H_{2}

and 2.24 L of

O_{2}

were generated at STP. What is the weight of

H_{2} S_{2} O_{8}

formed?

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During an electrolysis of conc. H2SO4, perdisulphuric acid (H2S2O8) and O2 form in equimolar amount. The amount of H2 that will form simultaneously will be: (2H2SO4→H2S2O8+2H++2e−)

During an electrolysis of conc. $H_{2} S O_{4}$ , perdisulphuric acid $(H_{2} S_{2} O_{8})$ and $O_{2}$ form in equimolar amount. The amount of $H_{2}$ that will form simultaneously will be: $(2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-})$