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Q.
Calculate the volume of $O_{2}$ liberated at $STP$ by passing $5A$ current for $193 \, sec$ through acidified water:
NTA AbhyasNTA Abhyas 2020Electrochemistry
Solution:
As per the first law of electrolysis, The mass deposited or volume liberated at the electrode is directly proportional to the charge passing between the electrode.
$m=ZQ$
Where $Z=\frac{E}{F}$ (electrovalent equivalent)
$Q=i\times t$ .
mass of $O_{2}$ $=\frac{E \times i \times t}{F}$ (Where $F=96500C$ )
$=\frac{8 \times 5 \times 193}{96500}g$
$=0.08g$
$\because $ At $STP$ , volume of $32gO_{2}=22400mL$
$\therefore $ The volume of $0.08gO_{2}$ $=\frac{22400 \times 0 \text{.} 08}{32}$