How much current in amperes should be passed | vedclass

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Question

(A) The electrolysis of $Na_2SO_4$ solution produces $O_2$ gas at the anode: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. First,calculate the moles of $O_2$

Vedclass · Accepted Answer

$64.33$

Vedclass · Answer

(A) The electrolysis of $Na_2SO_4$ solution produces $O_2$ gas at the anode: $2H_2O ightarrow O_2 + 4H^+ + 4e^-$. First,calculate the moles of $O_2$ produced per minute using the ideal gas law: $PV = nRT \Rightarrow n = \frac{PV}{RT}$. $n = \frac{1 \ bar 	imes 0.250 \ L}{0.08314 \ L \ bar \ K^{-1} \ mol^{-1} 	imes 300 \ K} \approx 0.01002 \ mol$. Since $4 \ moles$ of electrons are required to produce $1 \ mole$ of $O_2$,the moles of electrons required per minute are $n_e = 4 	imes 0.01002 = 0.04008 \ mol$. The total charge $Q$ passed per minute is $Q = n_e 	imes F = 0.04008 \ mol 	imes 96500 \ C \ mol^{-1} \approx 3867.72 \ C$. Since current $I = \frac{Q}{t}$ and $t = 60 \ s$,$I = \frac{3867.72 \ C}{60 \ s} \approx 64.46 \ A$. Rounding to the nearest provided option,the current is $64.33 \ A$.

How much current in amperes should be passed through a $Na_2SO_4$ solution using graphite electrodes to obtain $250 \ mL$ of $O_2$ gas per minute at $1 \ bar$ pressure and $300 \ K$ temperature (in $A$)? $[R = 0.08314 \ L \ bar \ K^{-1} \ mol^{-1}]$

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