6.5 , g of Mg is deposited during the electro | vedclass

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(B) According to Faraday's law of electrolysis,the number of equivalents of $Mg$ deposited is equal to the number of equivalents of $Cl_2$ gas evolved

Vedclass · Accepted Answer

$6.06$

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(B) According to Faraday's law of electrolysis,the number of equivalents of $Mg$ deposited is equal to the number of equivalents of $Cl_2$ gas evolved.$Eq. 	ext{ of } Mg = \frac{	ext{mass}}{	ext{equivalent mass}} = \frac{6.5}{12} \approx 0.5417$Since $Eq. 	ext{ of } Cl_2 = Eq. 	ext{ of } Mg$,we have $Eq. 	ext{ of } Cl_2 = 0.5417$.Number of moles of $Cl_2 = \frac{Eq.}{n	ext{-factor}} = \frac{0.5417}{2} = 0.27085 	ext{ mol}$.Volume of $Cl_2$ at $STP = 	ext{moles} 	imes 22.4 \, L/mol = 0.27085 	imes 22.4 = 6.067 \, L$.

$6.5 \, g$ of $Mg$ is deposited during the electrolysis of fused $MgCl_2$. What will be the volume of chlorine gas (in $L$) discharged at the electrode at $STP$?

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