One of the methods of preparation of per disulphuric acid- H2S2O8 involve electrolytic oxidation of H2SO4 at anode 2H2SO4 - H2S2O8 - 2H- - 2e- with oxygen and hydrogen as by-products- In such an electrolysis- 10-08 L of H2 and 2-24 L of O2 were generated at STP- What is the weight of H2 S2O8 formed-

At anode 2H_2SO_4 → H_2S_2O_8 +2H^+ +2e^ 2H_2O → 4H^+ + O_2(g) +4e^ At Cathode, 2H_2O +2e^ → 2OH^ +H_2(g) The number of moles of hydrogen liberated nbsp; ...

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Standard XII

Chemistry

Percentage Composition

One of the me...

Question

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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Solution

At anode
$2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-}$
$2 H_{2} O \to 4 H^{+} + O_{2} (g) + 4 e^{-}$
At Cathode,
$2 H_{2} O + 2 e^{-} \to 2 O H^{-} + H_{2} (g)$
The number of moles of hydrogen liberated

$= \frac{10.08}{22.4}$

$= 0.45 mol$
Number of moles of electrons required for liberation of hydrogen $= 2 \times 0.45$ mole

Number of moles of electron required for liberation of hydrogen = $2 \times 0.45 = 0.90 mol e^{-}$
The number of moles of oxygen liberated $\frac{2.24}{22.4} = 0.10 mol$
The number of moles of oxygen electron released during liberation of oxygen = $4 \times 0.1 = 0.4 m o l e^{-}$
The number of moles of electrons released during formation of $H_{2} S_{2} O_{8} = 0.9 - 0.4$
$= 0.5 mol$
Moles of $H_{2} S_{2} O_{8}$ formed $= \frac{0.5}{2}$
Mass of $H_{2} S_{2} O_{8}$ formed $= 0.25 \times 194 g$
$= 48.5 g$

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Similar questions

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}

and 2.35 L of

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).

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: