# Question #39e4b

Oct 3, 2016

$c \to \text{Current strength} = 2.5 A$

$t \to \text{Time for current} = 1 h r 20 \min$
$= \left(3600 + 1200\right) s = 4800 s$

$\text{Quantity of eletricity passed}$

$= c \times t = 2.5 \times 4800 C = \frac{12000}{96500} F$

The reaction at anode

$2 O {H}^{-} - 2 e \to {H}_{2} O + \frac{1}{2} {O}_{2}$

This equation reveals that 2F of electricity produces 0.5 moles ${O}_{2}$

So $\frac{12000}{96500} F$ electricity will produce

$\frac{1}{2} \times 0.5 \times \frac{12000}{96500} \text{ moles} {O}_{2}$

Given molar volume at r.t.p is $24 {\mathrm{dm}}^{3}$.
The volume of oxygen produced at anode will be

$= \frac{1}{2} \times 0.5 \times \frac{12000}{96500} \times 24 {\mathrm{dm}}^{3}$

$= 0.746 {\mathrm{dm}}^{3}$