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:

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) 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.

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.

Disproportionation products of one mole of MnO_(4)^(-2) in aqueous acidic medium are

In the electrolysis of 7.2L aqueous solution of CuSO_(4) , a current of 9.65A passed for 2 hours. Find the weight of Cu deposited at cathode.