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
The following electrochemical cell has been set up: Pt(s)|Fe3+, Fe2+ (a = 1)||Ce4+, Ce3+ (a = 1)|Pt(s); E°(Fe3+|Fe2+) = ...How does the electrical conductivity of 20 ml of 0.2 M – MgSO4 change when 0.5 M – Ba(OH)2 solution is gradually added i...During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.5 to 1.1 g/ml. Sulphuric acid ...The ionization constant of a weak electrolyte is 25 × 10-6 while the equivalent conductance of its 0.01 M solution ...Adding powdered Pb and Fe to a solution containing 1.0 M is each of Pb2+ and Fe2+ ions would result into the formation o...