Anodic oxidation of ammonium hydrogen sulphate produces ammonium persulphate- NH4HSO4- NH4SO4- - H- 2NH4SO4- NH42S2O8 - 2e- Anodic oxidation 2H- - 2e- H2 Cathodic reductionHydrolysis of ammonium persulphate forms H2O2- NH42 S2O8 - 2H2O - 2NH4HSO4 - H2O2Current efficiency inn electrolytic process is 60

Given, 228 g(NH_4)_2S_2O_8 + H_2O→ 2NH_4HSO_4 + 34 gH_2O_2 ∵ 34 g H_2O_2 is produced by 228 g(NH_4)_2 S_2O_8 ∴ 85 g H_2O_2 will be produced ...

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Byju's Answer

Standard XII

Chemistry

Faraday's First Law

Anodic oxidat...

Question

Anodic oxidation of ammonium hydrogen sulphate produces ammonium persulphate.
$N H_{4} H S O_{4} \to N H_{4} S O_{4}^{-} + H^{+}$
$2 N H_{4} S O_{4}^{-} \to (N H_{4})_{2} S_{2} O_{8} + 2 e^{-}$ (Anodic oxidation)
$2 H^{+} + 2 e^{-} \to H_{2}$ (Cathodic reduction)
Hydrolysis of ammonium persulphate forms $H_{2} O_{2}$ .
$(N H_{4})_{2} S_{2} O_{8} + 2 H_{2} O \to 2 N H_{4} H S O_{4} + H_{2} O_{2}$
Current efficiency inn electrolytic process is $60$ %. Calculate the amount of current required to produce $85 g$ of $H_{2} O_{2}$ per hour.
Hydrolysis reaction shows $100$ % yield.

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Solution

Given,
$(N H_{4})_{2} S_{2} O_{8} 228 g + H_{2} O \to 2 N H_{4} H S O_{4} + H_{2} O_{2} 34 g$
$∵ 34 g H_{2} O_{2}$ is produced by $228 g (N H_{4})_{2} S_{2} O_{8}$
$∴ 85 g H_{2} O_{2}$ will be produced by $\frac{228}{34} \times 85 g (N_{4})_{2} S_{2} O_{8}$
$= 570 g$
Equivalent mass of $(N H_{3})_{2} S_{2} O_{8}$ may be calculated using the following reaction:
$2 N H_{3} S O_{4}^{-} \to (N H_{4})_{2} S_{2} O_{8} + 2 e^{-}$
Equivalent mass of $(N H_{4})_{2} S_{2} O_{8} = \frac{M o l . m a s s}{2} = \frac{228}{2} = 114$
From first law of electrolysis,
$W = \frac{I t E}{96500}$
$570 = \frac{I \times 3600 \times 114}{96500}$
$I = 134.0277 a m p e r e$
Give that, current efficiency is $60$ %, the actual amount of current
$= \frac{100}{60} \times 134.0277 = 223.379 a m p e r e$ .

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Similar questions

Q. Hydrogen peroxide can be prepared by successive reactions :

2 N H_{4} H S O_{4} \to H_{2} + (N H_{4})_{2} S_{2} O_{8}

(N H_{4})_{2} S_{2} O_{8} + 2 H_{2} O \to 2 N H_{4} H S O_{4} + H_{2} O_{2}

The first reaction is an electrolytic reaction the second is steam distillation. What amount of current would have to be used in the first reaction to produce enough intermediate to yield 100 g pure

H_{2} O_{2}

per hour? Assume 50% anode current efficiency.

H_{2} O_{2}

can be prepared by successive reactions:

2 N H_{4} H S O_{4} \to H_{2} + (N H_{4})_{2} S_{2} O_{8}, (N H_{4})_{2} S_{2} O_{8} + 2 H_{2} O \to 2 N H_{4} H S_{4} + H_{2} O_{2}

The first reaction is an electrolytic reaction and second is steam distillation. The amount of current in ampere would have to be used in first reaction to produce enough intermediate to yield 200 g pure

H_{2} O_{2}

per hr is: Assume current efficiency 50%.

In a fuel cell, hydrogen and oxygen react to produce electricity. In the process, hydrogen gas is oxidized at anode and oxygen at cathode. If 67.2 L of $H_{2}$ at STP reacts in 15 min, the average current (in A) produced is :

Anode reaction : $H_{2} + 2 ⊖ O H \to 2 H_{2} O + 2 e^{-}$

Cathode reaction :

O_{2} + 2 H_{2} O + 2 e^{-} \to 4 ⊖ O H

(write your answer to nearest integer)

Q. In which of the following reactions,

H_{2} O_{2}

acts as reducing agent?
(A)

H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O

(B)

H_{2} O_{2} - 2 e^{-} \to O_{2} + 2 H^{+}

(C)

H_{2} O_{2} + 2 e^{-} \to 2 O H^{-}

(D)

H_{2} O_{2} + 2 O H^{-} 2 e^{-} \to O_{2} + 2 H_{2} O

Q. In which of the following reactions

H_{2} O_{2}

acts as a reducing agent?
(a)

H_{2} O_{2} + 2 H^{+} + 2 e \to 2 H_{2} O

(b)

H_{2} O_{2} - 2 e \to O_{2} + 2 H^{+}

(c)

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

(d)

H_{2} O_{2} + 2 O H^{-} - 2 e \to O_{2} + 2 H_{2} O

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