ElectrochemistryHard
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 FreeMore Electrochemistry Questions
A metal having negative reduction potential, when dipped in the solution of its own ions, has a tendency to...A certain current liberated 0.50 g of hydrogen in 2 h. How many grams of copper can be liberated by the same current flo...The electrode through which electrons enter the electrolytic solution is...Two weak acid solutions HA1 and HA2 each with the same concentration and having pKa values 3 and 5 are placed in contact...Given below are the half-cell reactions Mn2+ + 2e- → Mn ; Eo = - 1.18 V (Mn3+ + e- → Mn2+) ; Eo = + 1.51 V T...