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Q.
The volume of oxygen gas liberated at NTP by passing a current of $9650$ coulombs through acidified water is
Electrochemistry
Solution:
Oxidation reaction at anode, upon electrolysis of water:
$2 H _{2} O (l) \longrightarrow O _{2}(g)+4 H ^{+}( aq )+4 e ^{-} ; E_{c e l l}^{\circ}=+1.23 V$
Thus, 1 mole of oxygen is liberated by 4 moles of electrons. $4 \times 96500$ coulombs electricity liberates $=22.4 \,L _{2} \,O _{2}$ gas
$9650$ coulombs electricity liberates $=\frac{22.4}{4 \times 96500} \times 9650$ $=0.56 L . O _{2} $ gas