Question

When electric current is passed through acidified water for $1930s,1120mL$ of $H_2$ gas is collected (at STP) at the cathode. What is the current passed in amperes?

• A. 0.05
• B. 0.50
• C. 5.0
• D. 50

Answer 1

Answer: (C)

Electrolysis of water takes place as follows.
$H_{2}O\rightleftharpoons \underset{\text{ cathode }}{H^{+}}+\underset{\text{ anode }}{O{H}^{-}}$
At anode
$O{H}^{-}\stackrel{\text{ Oxidation }}{\to }OH+e^{-}$ $4OH⟶2H_{2}O+O_2$
At cathode
$2H^{+}+2e^{-}\stackrel{\text{ Reduction }}{\to }{H}_2$
Given, time, $t=1930s$
Number of moles of hydrogen collected
$=\frac{1120\times 10^{-3}}{22.4}$ moles
$=0.05$ moles
$\because 1$ mole of hydrogen is deposited by $=2$ moles of
electrons
$\therefore 0.05$ moles of hydrogen will be deposited by
$=2\times 0.05$
$=0.10$ mole of electrons
Charge, $Q=nF$
$=0.1\times 96500$
Charge, $Q=$ it
$0.1\times 96500=i\times 1930$
$i=\frac{0.1\times 96500}{1930}$
$i=5.0A$