ElectrochemistryMedium

Question

The current required to produce oxygen at the rate of 2.8 ml (0°C, 1 atm) per second during electrolysis of acidulated water is

Options

A.48.25 A/s
B.24.12 A/s
C.96.5 A/s
D.0.0048 A/s

Solution

neq = n × n-factor = $\frac{Q}{F}$

Or, $\frac{2.8}{22400} \times 4 = \frac{i \times 1}{96500} \Rightarrow i = 48.25\text{ A/sec}$

Create a free account to view solution

View Solution Free
Topic: Electrochemistry·Practice all Electrochemistry questions

More Electrochemistry Questions

For Zn2+ / Zn, Eo = - 0.76 V, for Ag+/Ag Eo = 0.799 V. The correct statement is -...Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 S m-1. The ...The cell potential for the voltaic cell depicted below is 0.109V under standard conditions ; Pb(s)| Pb2+(aq.)||Ni2+(aq.)...The quantity of electricity required for the reduction of 1 mole of Fe2O3 to Fe is...The EMF of the cell: Zn|Zn2+ (0.01 M)||Fe2+ (0.001 M)|Fe at 298 K is 0.2905 V, then the value of equilibrium constant fo...