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A fuell cell uses CH(4)(g) and forms CO(...

A fuell cell uses `CH_(4)(g)` and forms `CO_(3)^(2-)` at the anode. It is used to power a car with 80 amp, for 0.96 hr. how many litres of `CH_(4)(g)` (at 1 atm, 273 K) would be required? (V_(m) = 22.4 L/mol) (F = 96500). Assume `100%` efficiency.

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The correct Answer is:
8
`CH_4+10OH^(-)toCO_3^(2-)+7H_2O+8e^(-)`
No . of Faradays required `=(80xx3600xx0.96)/96500`
Hence mol. of `CH_4` required =`1/8xx(80xx3600xx0.96)/96500`
`V_(CH_4)=1/8xx(80xx3600xx0.96)/96500xx22.4 L`
=8.356x0.96=8.02 L
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