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Electrolysis of a solution of MnSO(4) in...

Electrolysis of a solution of `MnSO_(4)` in aqueous sulphuric acid is a method for the preparation of `MnO_(2)` as per reaction,
`Mn_((aq.))^(2+)+2H_(2)O rarr MnO_(2(s))+2H_((aq))^(+)+H_(2(g))`
Passing a current of 27 ampere for 24 hour gives one of `MnO_(2)`. Waht is the value of current efficiency ? Write the reaction taking place at the cathode the anode.

Text Solution

Verified by Experts

`W=(Eit)/(96500)`
`rArr 1000=(87 xx I 24 xx 60 xx 60)/(2 xx 96500) ("Eq. weight of "MnO_(2)=(87)/(2))`
i=25.6 ampere
Current efficiency `=(25.6)/(27)xx 100=94.8%`
Reaction `Mn^(2+) to Mn^(4+) ("anode")`
`2H^(+) +2e to H_(2) ("cathode")`
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