A fuel cell is supplied 1 mole of `H_(2)` gas and 10 moles `O_(2)`gas. If the fuel cell is operated at 96.5 mA currecnt, how long will it deliver power?

Explanation

In `H_(2)-O_(2)` fuel cell, the net reaction is :
` underset((2" mol"))(2H_(2)(g))+underset((1 " mol"))(O_(2)(g)) to 2H_(2)O(l)`
In the present case, the cell is supplied with 1 mole of `H_(2)`gas and 10 moles of `O_(2)`gas . This means that `H_(2)`(g) is the limiting reactant.
Thus, 1 mole of `H_(2)`reacts with 0.5 mole of `O_(2)`.
`H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l),n=2e^(-)`
Now, `Q=nF=2 molxx(96500 C mol^(-1))=2xx(96500 AS)`
Also `Q=Ixxt=96.5" mA "=(96.5xx10^(-3)" A")xxt`
upon equating , `t=( 2xx(96500" AS"))/((96.5xx10^(-3)" A"))=2xx10^(6)s`