A dilute solution of sulphuric acid is electr | vedclass

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(A) The reactions at the electrodes are:At anode: $2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$At cathode: $4H^+ + 4e^- \rightarrow 2H_2(g)$Total charge pa

Vedclass · Accepted Answer

$127$

Vedclass · Answer

(A) The reactions at the electrodes are:At anode: $2H_2O ightarrow O_2(g) + 4H^+ + 4e^-$At cathode: $4H^+ + 4e^- ightarrow 2H_2(g)$Total charge passed $Q = i 	imes t = 0.10 \ A 	imes 2 	imes 3600 \ s = 720 \ C$.Number of moles of electrons $n_e = \frac{Q}{F} = \frac{720}{96500} \approx 0.00746 \ mol$.From the stoichiometry,$4 \ mol$ of electrons produce $1 \ mol$ of $O_2$ and $2 \ mol$ of $H_2$ (total $3 \ mol$ of gas).Total moles of gas produced $n_{total} = \frac{n_e}{4} + \frac{n_e}{2} = \frac{3}{4} n_e = 0.75 	imes 0.00746 = 0.005595 \ mol$.Volume at $STP = n_{total} 	imes 22.7 \ L \ mol^{-1} = 0.005595 	imes 22.7 \approx 0.127 \ L = 127 \ cm^3$.

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