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Electrolysis of a solution of MnSO(4) i...

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.

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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.

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.

Electrolysis of a solution of MnSO_(4) in aqueous sulphuric acid is a method for the preparation of MnO_(2) . Passig a curret of 27 A for 24 hours gives 1 kg of MnO_(2) . The current efficiency in this process is:

Electrolysis of a solution of MnSO_(4) in aqueous sulphuric acid is a method for the preparation of MnO_(2) . Passing a current of 27 A for 24 hours gives 1 kg of MnO_(2) . The current efficiency in this process is:

The eq. mass of KMnO_4 in the reaction, MnO_4^(-) +Mn^(2+) +H_(2)O rarr MnO_2 +H^(+) (unbalanced) is