Electrolysis of a solution of `MnSO_(4)` in aqueous sulphuric acid is a method for the preparation of `MnO_(2)` as the per reaction,
`Mn^(2+) (aq.)+2H_(2)O rarr MnP_(2) +2H^(+) (aq.) +H_(2)`
Passing a current of 27 A for 24 hours gives one kg of `MnO_(2)`. what is the value of current efficiency ? Write the reactions taking place at the cathode and at the anode.

Explanation

Apply `w=(Exxixxt)/(96500)`
`1000=(87xxixx24xx60xx60)/(2xx96500)`
`i=25.6` ampere
Current efficiency `=25.6/27.0xx100=94.8 %`
Reaction :
`{:("Anode:",Mn^(2+) rarr Mn^(4+)+2e^(-)),("Cathode :",2H^(+) +2e^(-) rarr H_(2)):}`