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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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Verified by Experts

The correct Answer is:
`8`
`CH_(4)+10OH^(-)toCO_(3)^(2-)+7H_(2)O+8e^(-)`
No. of faradays required`=(80xx3600xx0.96)/(96500)`
`V_(CH_(4))=(1)/(8)xx(80xx3600xx0.96)/(96500)xx22.4L=8356xx0.96=8.02L`
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