Marshall-s acid or perdisulphuric acid H2S2O8 can be prepared by electrolytic oxidation of H2SO4 as - 2H2SO4 - H2S2O8 - 2H- - 2e- O2 and H2 are byproducts- In such an electrolysis 9-72 L of H2 and 2-35 L of O2 were produced at S-T-P- What is the weight of H2S2O8 formed -

H_2SO_4 → 2H^⊕ + SO^2 _4 Or H_2SO_4 → H^⊕ + HSO^ _4 Two reactions are competing at anode : 2HSO^⊖_4 → H_2S_2O_8 + 2e^ and H_2O → 2 ...

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Byju's Answer

Standard XII

Chemistry

Faraday's Second Law

Marshall's ac...

Question

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 ?

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Solution

$H_{2} S O_{4} \to 2 H^{\oplus} + S O_{4}^{2 -}$
Or $H_{2} S O_{4} \to H^{\oplus} + H S O_{4}^{-}$
Two reactions are competing at anode :
$2 H S O_{4}^{⊖} \to H_{2} S_{2} O_{8} + 2 e^{-}$
and $H_{2} O \to 2 H^{\oplus} + \frac{1}{2} O_{2} + 2 e^{-}$
At cathode : $2 H^{\oplus} + 2 e^{-} \to H_{2} (g)$
Since one reaction at cathode and two reactions at anode are taking place, therefore, equivalent of $H_{2} (g)$ produced at cathode should be equal to the equivalent of $O_{2}$ produced and equivalent of $H_{2} S_{2} O_{8}$ formed at anode.
$∴$ Eq of $H_{2} =$ Eq of $O_{2} +$ Eq of $H_{2} S_{2} O_{8}$
$\frac{9.72}{22.4 / 2} = \frac{2.35}{22.4 / 4} + \frac{W_{(H_{2} S_{2} O_{8})}}{194 / 2}$
$∴ W_{H_{2} S_{2} O_{8}} = 43.456 g$
Note :-
1 Eq of $H_{2} = \frac{22.4 L}{2} 11.2 L$ at STP (n factor = 2)
1 Eq of $O_{2} = \frac{22.4 L}{4} = 5.6 L$ at STP (n factor = 4)
Ew of $H_{2} S_{2} O_{8} = \frac{M w}{2}$ (n factor = 2)
Mw of $H_{2} S_{2} O_{8} = 194$ g $m o l^{- 1}$

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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

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

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0.87

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3.36

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were generated at STP. Report the mass of

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

formed in decagram (write nearest integer).

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}

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were generated at STP. What is the weight of

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

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 node

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

with oxygen and hydrogen as by -products. In such an electrolysis, 9.722 L of

H_{2}

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O_{2}

were generated at STP. What is the weight of

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

formed in gms? (write the value to the nearest integer).

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Marshall's acid or perdisulphuric acid (H2S2O8) can be prepared by electrolytic oxidation of H2SO4 as :2H2SO4→H2S2O8+2H⊕+2e−O2 and H2 are byproducts. In such an electrolysis 9.72 L of H2 and 2.35 L of O2 were produced at S.T.P. What is the weight of H2S2O8 formed ?