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Anodic oxidation of ammonium hydrogen su...

Anodic oxidation of ammonium hydrogen sulphate produces ammonium persulphate.
`NH_(4) HSO_(4) rarr NH_(4)SO_(4)^(-) + H^(+)`
`2NH_(4) SO_(4)^(-) rarr (NH_(4))_(2) S_(2) O_(8) + 2e^(-)` (Anodic oxidation)
`2H^(+) + 2e^(-) rarr H_(2)` (Cathodic reduction)
Hydrolysis of ammonium persulphate forms `H_(2)O_(2)`
`(NH_(2))_(2) S_(2)O_(8) + 2H_(2)O rarr 2NH_(4) HSO_(4) + H_(2)O_(2)`
Current efficiency in 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.

Text Solution

Verified by Experts

Given,
`underset(228g) ((NH_(4))_(2)) S_(2)O_(8) + H_(2)O rarr 2NH_(4) HSO_(4) + underset(34g)(H_(2)O_(2))`
`:' 34 g H_(2)O_(2)` is produced by 228 g `(NH_(4))_(2) S_(2)O_(8)`
`:. 85 g H_(2)O_(2)` will be produced by `(228)/(34) xx 85 g (NH_(4))_(2) S_(2)O_(8) = 570g`
Equivalent mass of `(NH_(4))_(2)S_(2)O_(8)` may be calculated using the following reaction:
`2NH_(4)SO_(4)^(-) rarr (NH_(4))_(2) S_(2) O_(8) + 2e^(-)`
Equivalent mass of `(NH_(4))_(2) S_(2)O_(8) = ("Mol. mass")/(2) = (228)/(2) = 114`
From first law of electrolysis,
`W = (I tE)/(96500)`
`570 = (I xx 3600 xx 114)/(96500)`
`I = 134.0277` ampere
Given that, current efficiency is 60%, the actual amount of current
`=(100)/(60) xx 134.0277 = 223.379` ampere
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