Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper

Question

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of $\mathrm{Cu}: 63 \mathrm{~g} \mathrm{~mol}^{-1}, 1 \mathrm{~F}=96487 \mathrm{C}$ )

Professor PC Thomas Classes · Answer

(b) : Given, Current $(I)=9.6487 \mathrm{~A}$, Time $(t)=100$ seconds $ 1 \mathrm{~F}=96487 \mathrm{C} \ $ Reaction : $\mathrm{Cu}_{(a q)}^{2+}+2 e^{-} ightarrow \mathrm{Cu}_{(s)}$ \ Here, 2 electrons are exchanged, so valency factor $(n)=2$ Applying Faraday's first law, $ W=Z I t$

$\left(\because Z=\frac{	ext { Equivalent weight }}{F}=\frac{63}{2 	imes 96487}ight)$
So, $W=\frac{63 	imes 9.6487 	imes 100}{2 	imes 96487}=0.315 \mathrm{~g}$