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When electrolysis of KCl is doen in alka...

When electrolysis of `KCl` is doen in alkaline medium, `10g` of `KClO_(3)` is produced as follows `:`
`Cl^(c-)+6overset(c-)(O)H rarr ClO_(3) +3H_(2)O+6e^(-)`
A current of `2A` is passed for `10.941 ` hours. Calculate the `((Percent a g e current efficiency)/(10))` used in the process.
`(Mw` of `KClO_(3)=122.5)`

Text Solution

Verified by Experts

The correct Answer is:
6
Let `x%` is the current efficiency of `KClO_(3)=` Number of Faradays.
`(10g)/(122.5//6)=(2xx x xx10.941xx3600)/(100xx96500)`
`:. x=60%`
`:. ("Percentage current efficiency ")/(10)=(60)/(10)=6`
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