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[" Electrolysis of a solution of MnSO,in aqueous sulphuric acid is a method for the preparation of Mud "],[" Passing a current of "27A" for "24" hours gives "1kg" of MnOz.The current efficiencey in this process is: "],[[" (A) "100%," (B) "95.185%," (C) "80%," (D) "82.951%]]

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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:

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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 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.