ElectrochemistryMedium
Question
Passage of 96,500 coulomb of electricity liberates ________ L of O2 at 273°C and 2 atm during electrolysis.
Options
A.5.6
B.16.8
C.22.4
D.11.2
Solution
neq = n × n-factor = $\frac{Q}{F}$
Or, $\frac{1 \times V}{0.0821 \times 546} \times 4 = \frac{96500}{96500} \Rightarrow V = 5.6\text{ L}$
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