Question

A current of 1.26 A is passed through an electrolytic cell containing an aqueous solution of sulphuric acid for 7.44 h. Write the half-cell reactions and calculate the volume of gases generated at STP?

Answer 1

Answer:

Determine half cell reactions

oxidation:$2H_{2} O(I)$→$O_{2} (g)+4H^{+} (aq)+4e^{-}$

Reduction:$2H^{+} +2e^{-}$→$1H_{2} (g)$

find the volume of O2 gas and then convert current to charge

1.26A*7.44hr*(3600s/1hr)=$3.37*10^{4} C$

convert charge to moles of electrons

$3.37*10^{4} C*(1mole/96500C)=0.349mole$

use mole ratio of oxygen and electrons

$2H_{2} O(I)$→$O_{2} (g)+4H^{+} (aq)+4e^{-}$

for every mole of oxygen formed 4moles of electrons are transferred.

0.349mole*(1mol O2/4mole)=0.0873mol O2

PV=nRT

V=nRT/P

V=(0.0873)(0.0821)(273)/(1atm)

=1.96L

similarly for H2 gas we get

2$H^{+} +2e^{-}$→$1H_{2} (g)$

$3.37*10^{4} C*(1mole e^{-} /96500C)*(1mol H_{2} /2 mol e^{-} )$=0.175 mol H2

V=nRT/P

=(0.175mol)(0.821)(273)/(1atm)

=3.92L