The ratio of weights of hydrogen gas liberate | vedclass

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(B) According to Faraday's second law of electrolysis,when the same amount of electricity is passed through different electrolytes,the masses of subst

Vedclass · Accepted Answer

$1:12$

Vedclass · Answer

(B) According to Faraday's second law of electrolysis,when the same amount of electricity is passed through different electrolytes,the masses of substances deposited are proportional to their equivalent weights.The reaction for hydrogen liberation is: $2H^+ + 2e^- ightarrow H_2$.The equivalent weight of $H_2$ is $E_{H_2} = \frac{	ext{Molar mass}}{	ext{n-factor}} = \frac{2}{2} = 1$.The reaction for magnesium deposition is: $Mg^{2+} + 2e^- ightarrow Mg$.The equivalent weight of $Mg$ is $E_{Mg} = \frac{	ext{Atomic mass}}{	ext{n-factor}} = \frac{24}{2} = 12$.The ratio of weights is equal to the ratio of their equivalent weights:$\frac{W_{H_2}}{W_{Mg}} = \frac{E_{H_2}}{E_{Mg}} = \frac{1}{12}$.Therefore,the ratio is $1:12$.

The ratio of weights of hydrogen gas liberated and magnesium deposited by passing the same amount of electricity from aqueous $H_2SO_4$ and fused $MgSO_4$ are :

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