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.

Explanation

We know, `w=("E.i.t")/(96500)`
`therefore 1000=(87xxixx24xx60xx60)/(2xx96500)`
`i=25.6` ampere
`therefore` current efficiency `=(25.6)/(27)xx100=96%`
Reactions
Anode : `Mn^(2+)rarr Mn^(4+)+2e^(-)`
Cathode : `2H^(+)+2e^(-)rarr H_(2)`