In a fuel cell,H_2 and O_2 react to produce e | vedclass

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Question

(C) The oxidation reaction at the anode is: $H_2(g) \rightarrow 2H^+(aq) + 2e^-$. Number of moles of $H_2 = \frac{67.2 \ L}{22.4 \ L/mol} = 3 \ mol$.

Vedclass · Accepted Answer

$643.33$

Vedclass · Answer

(C) The oxidation reaction at the anode is: $H_2(g) ightarrow 2H^+(aq) + 2e^-$. Number of moles of $H_2 = \frac{67.2 \ L}{22.4 \ L/mol} = 3 \ mol$. Since $2 \ mol$ of electrons are produced per mole of $H_2$,total moles of electrons $(n)$ = $3 	imes 2 = 6 \ mol$. Total charge $(Q)$ = $n 	imes F = 6 	imes 96500 \ C = 579000 \ C$. Time $(t)$ = $15 \ min = 15 	imes 60 \ s = 900 \ s$. Current $(I)$ = $\frac{Q}{t} = \frac{579000 \ C}{900 \ s} = 643.33 \ A$.

In a fuel cell,$H_2$ and $O_2$ react to produce electricity. In this process,$H_2$ gas is oxidized at the anode and $O_2$ gas is reduced at the cathode. If $67.2 \ L$ of $H_2$ gas at $\text{STP}$ is consumed in $15 \ min$,what is the current produced (in $A$)?

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