[" Electrolysis of a solution of "MnSO_(4)" in aqueous sulphuric acid is a method for the preparation of MnO "_(2).],[" Passing a current of "27A" for "24" hours gives "1kg" of "MnO_(2)" .The current efficiency in this process is: "],[[" (1) "100%," (2) "95.185%," (3) "80%," (4) "82.951%]]

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