H2O2 can be prepared by the hydrolysis of peroxydisulphuric acid H2S2O8- H2S2O8 can be prepared by anodic oxidation of H2SO4 as follows - 2H2SO4 - H2S2O8 - 2H- - 2e- H2S2O8 - 2H2O - 2H2SO4 - H2O2For the production of 340 g of H2O2 in solution- the current strengthin A required- if it was passed for 10 hours is-

H_2S_2O_8 +, 2H_2O → 2H_2SO_4 + H_2O_2 2H_2SO_4 → H_2S_2O_8 + 2H^⊕ + 2e^ Moles of H_2O_2 = 340/34 = 10 = Moles of H_2S_2O_8 2F ...

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Byju's Answer

Standard XII

Chemistry

Decarboxylation Reaction of Alkanes

H2O2 can be p...

Question

$H_{2} O_{2}$ can be prepared by the hydrolysis of peroxydisulphuric acid $(H_{2} S_{2} O_{8})$ . $H_{2} S_{2} O_{8}$ can be prepared by anodic oxidation of $H_{2} S O_{4}$ as follows :
$2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{\oplus} + 2 e^{-}$
$H_{2} S_{2} O_{8} + 2 H_{2} O \to 2 H_{2} S O_{4} + H_{2} O_{2}$
For the production of 340 g of $H_{2} O_{2}$ (in solution), the current strength(in A) required, if it was passed for 10 hours is___________.

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Solution

$H_{2} S_{2} O_{8} +, 2 H_{2} O \to 2 H_{2} S O_{4} + H_{2} O_{2}$
$2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{\oplus} + 2 e^{-}$
Moles of $H_{2} O_{2} = \frac{340}{34} = 10 = M o l e s o f H_{2} S_{2} O_{8}$
2F = 1 mol $H_{2} S_{2} O_{8}$
$\Rightarrow 10 m o l H_{2} S_{2} O_{8} = 20 F$
$\Rightarrow \frac{I \times 10 \times 3600}{96500} = 20 \Rightarrow I = 53.6 A$

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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^{-}

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and

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are byproducts. In such an electrolysis 9.72 L of

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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

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0.87

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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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and

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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:

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

involve electrolytic oxidation of

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H2S2O8+,2H2O→2H2SO4+H2O22H2SO4→H2S2O8+2H⊕+2e−Moles of H2O2=34034=10=MolesofH2S2O82F = 1 mol H2S2O8⇒10molH2S2O8=20F⇒I×10×360096500=20⇒I=53.6A