Question

The STP volume of oxygen liberated by 2 ampere of current when passed through acidulated water for 3 minutes and 13 seconds, is:

• A. 120 cc
• B. 22.4 cc
• C. 11.2 cc
• D. 44.8 cc

Answer 1

The correct option is B 22.4 cc

Total time $=3\times 60+13=193$ seconds

$Current=2A$

$\therefore$ Moles of electron $=\frac{charge}{\text{total charge on one mole}}$

$=\frac{193\times 2}{96500}=0.004$

$4O{H}^{-}\to O_2+2H_{2}O+4e^{-}$

4 moles of $e^{-}$ gives 1 mole of $O_2$.

$\therefore$ 0.004 mole of $e^{-}$ give 0.001 mole of $O_2$.

Now,

$PV=RTn$

$1\times V=0.0821\times 273\times 0.001$.

$V=22.4\times {10}^{-3}l=22.4cc$.