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 `kg` of `MnO_(2)`. What is the value of current efficiency ? Write the reaction taking place at the cathode and at the anode.

Explanation

We know , `w = (E.i.t)/(96500)`
`1000 = (87 xx i xx 24 xx 60 xx 60)/(2 xx 96500)`
`i = 25.6` ampere
`:.` Current efficiency `= (25.6)/(27) xx 100 = 94.8%`
Reactions
Anode: `Mn^(2+) rarr Mn^(4+)+2e`
Cathode: `2H^(+)+2e rarr H_(2)`