A fuel cell uses CH_4 gas and it forms CO_3^{ | vedclass

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(A) The anode reaction for the oxidation of methane in a fuel cell is: $CH_4 + 3H_2O 	o CO_3^{2-} + 10H^+ + 8e^-$. From the stoichiometry,$1 \ mol$ o

Vedclass · Accepted Answer

$28$

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(A) The anode reaction for the oxidation of methane in a fuel cell is: $CH_4 + 3H_2O 	o CO_3^{2-} + 10H^+ + 8e^-$. From the stoichiometry,$1 \ mol$ of $CH_4$ releases $8 \ mol$ of electrons. Calculate the total moles of electrons transferred: $n_{e^-} = \frac{I 	imes t}{F} = \frac{96.5 	imes 10000}{96500} = 10 \ mol$. Calculate the moles of $CH_4$ required: $n_{CH_4} = \frac{1}{8} 	imes n_{e^-} = \frac{10}{8} = 1.25 \ mol$. At $STP$ ($1 \ atm$ and $273 \ K$),$1 \ mol$ of gas occupies $22.4 \ L$. Therefore,the volume of $CH_4$ required is $1.25 	imes 22.4 \ L = 28 \ L$.

$A$ fuel cell uses $CH_4$ gas and it forms $CO_3^{2-}$ at the anode. It is used to power a car with $96.5 \ A$ for $10000 \ s$. The volume of $CH_4$ gas required at $1 \ atm$ and $273 \ K$ is $............ \ L$.

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