Calculate the volume of O_2 gas liberated at | vedclass

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(A) $1$. The reaction at the anode for the electrolysis of aqueous $Na_2SO_4$ is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. $2$. Total charge passed $Q =

Vedclass · Accepted Answer

$573.93$

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(A) $1$. The reaction at the anode for the electrolysis of aqueous $Na_2SO_4$ is: $2H_2O ightarrow O_2 + 4H^+ + 4e^-$. $2$. Total charge passed $Q = I 	imes t = 2.5 \ A 	imes 3600 \ s = 9000 \ C$. $3$. Number of moles of electrons $n(e^-) = Q / F = 9000 / 96500 \approx 0.09326 \ mol$. $4$. From the stoichiometry,$4 \ moles$ of $e^-$ produce $1 \ mole$ of $O_2$. $5$. Moles of $O_2 = 0.09326 / 4 = 0.023315 \ mol$. $6$. Volume of $O_2 = n 	imes V_m = 0.023315 	imes 24.6 \ L \approx 0.5735 \ L = 573.5 \ mL$.

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