ElectrochemistryHard
Question
Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of MnO2 as per the reaction: Mn2+ (aq) + 2H2O → MnO2(s) + 2H+ (aq) + H2(g).
Passing a current of 19.3 A for 2 h gives only 52.2 g of MnO2. The current efficiency is (Mn = 55)
Options
A.8.33%
B.83.33%
C.41.67%
D.100%
Solution
$\frac{w}{E} = \frac{Q}{F} \Rightarrow \frac{52.2}{87} \times 2 = \frac{19.3 \times 2 \times 3600 \times \eta}{96500} \Rightarrow \eta = 0.8333\text{ or 83.33\%}$
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
A hydrogen electrode placed in a buffer solution of CH3COONa and CH3COOH in the ratio’s x:y and y:x has electrode potent...Any redox reaction would occur spontaneously, if :...A solution containing Na+, NO3-, Cl-, and SO42- ions, all at unit concentrations, is electrolyzed between nickel anode a...Consider the following reduction processes :$${{Al}^{3 +} + 3e^{-} \rightarrow Al(s),E^{\circ} = - 1.66\text{ }V }{{Fe}^...Which is correct about fuel cells?...