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Q.
The weight of silver (at. $w t .=108$ ) displaced by a quantity of electricity which displaces $5600\, mL$ of $O _{2}$ at $STP$ will be
AIPMTAIPMT 2014Electrochemistry
Solution:
At STP, $1$ mole of oxygen occupies $22400\, mL$.
Hence, the number of moles of oxygen corresponding to $5600\, ml$ is
$n _{ O _{2}}=\frac{5600}{22400}=\frac{1}{4}=\frac{ W _{ O _{2}}}{ M _{ O _{2}}}$.
$1 $ mole of oxygen produces $4$ moles of lectrons and $1$ mole of silver requires 1 mole of electrons.
$\frac{W_{ Ag }}{ M _{ Ag }} \times 1=\frac{ W _{ O _{2}}}{ M _{ O _{2}}} \times 4$
$\frac{ W _{ Ag }}{ M _{ Ag }} \times 1=\frac{1}{4} \times 4$
$\frac{ W _{ Ag }}{108}=\frac{1}{4} \times 4$
$W _{ Ag }=108\, g$
Thus, the weight of silver displaced by a quantity of electricity which displaces $5600\, ml _{2} O _{2}$ at STP will be $108.0 \, g$.