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.A. ` 100%`B. ` 95.185%`C. ` 80%`D. ` 82. 951%`

Electrolysis of a solution of `MnSO_(4)` in aqueous sulphuric acid is

1.	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.A. ` 100%`B. ` 95.185%`C. ` 80%`D. ` 82. 951%`
Answer» Correct Answer - B ` (1000 xx2)/((55+ 32)) = ( 27 xx 24xx 3600 xx ita)/(96500) or eta=0.951 = 95. 1%`.

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