Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of MnO2. Passing a current of 27 A for 24 hours gives 1 kg of MnO2 The current efficiency in this process is:

Question

Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of MnO2. Passing a current of 27 A for 24 hours gives 1 kg of MnO2 The current efficiency in this process is: (A) 100 % (B) 95.185 % (C) 80 % (D) 82.951 %

Tardigrade Admin · Accepted Answer

(1000×2/(55+32))=(27×24×3600×η/96500)or η=0.921=95.1 %

Q. Electrolysis of a solution of $M n S O_{4}$ in aqueous sulphuric acid is a method for the preparation of $M n O_{2}$ . Passing a current of $27 A$ for $24$ hours gives $1 k g$ of $M n O_{2}$ The current efficiency in this process is:

$100%$

$95.185%$

$80%$

$82.951%$

$\frac{1000 \times 2}{( 55 + 32 )} = \frac{27 \times 24 \times 3600 \times η}{96500} orη = 0.921 = 95.1%$

Q. Electrolysis of a solution of MnSO4​ in aqueous sulphuric acid is a method for the preparation of MnO2​. Passing a current of 27A for 24 hours gives 1kg of MnO2​ The current efficiency in this process is:

100%

95.185%

80%

82.951%