In the electrolysis of acidulated water,it is | vedclass

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(B) The reaction for the evolution of hydrogen at the cathode is: $2H^+ + 2e^- \rightarrow H_2(g)$.From the stoichiometry,$1 \ mole$ of $H_2$ requires

Vedclass · Accepted Answer

$9.65$

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(B) The reaction for the evolution of hydrogen at the cathode is: $2H^+ + 2e^- ightarrow H_2(g)$.From the stoichiometry,$1 \ mole$ of $H_2$ requires $2 \ moles$ of electrons $(2 \ F)$.Volume of $H_2$ produced per second $= 1.12 \ cc = 1.12 \ mL$.At $STP$,$22400 \ mL$ of $H_2$ corresponds to $1 \ mole$.Therefore,moles of $H_2$ produced per second $= \frac{1.12}{22400} = 5 	imes 10^{-5} \ mol$.Number of moles of electrons required per second $= 2 	imes (5 	imes 10^{-5}) = 10^{-4} \ mol$.Since $1 \ mole$ of electrons $= 96500 \ C$,the charge required per second is $Q = 10^{-4} 	imes 96500 = 9.65 \ C$.Since $Current (I) = \frac{Q}{t}$ and $t = 1 \ s$,the current is $9.65 \ A$.

In the electrolysis of acidulated water,it is desired to obtain $1.12 \ cc$ of hydrogen gas per second at $STP$ conditions. The current to be passed is: (in $A$)

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