In fuel cell `H_(2) and O_(2)` react to produce electricity. In the process, `H_(2)` gas is oxidised at the anode and `O_(2)` at cathode. If 44.8 litre of `H_(2)` at `25^(@)C` and also pressure reacts in 10 minutes, what is average current produced? If the entire current is used for electro deposition of Cu from `Cu^(2+)`, how many grams of Cu deposited?

Explanation

Oxidation at anode :
`2H_(2(g))+4OH_((aq))^(-) rarr 4H_(2)O(l)+4e^(-)`
1 mole of hydrogen gas produces 2 moles of electrons at `25^(@)C` and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres
`therefore` no. of moles of hydrogen gas produced
`=(1" mole")/(22.4" litres") times 44.8" litres"`
= 2 moles of hydrogen
`therefore` 2 of moles of hydrogen produces 4 moles of electro i.e., 4F charge.
We know that Q = It
`I=Q/t`
`" "=(4F)/(10" mins")`
`" "=(4 times 96500C)/(10 times 60s)`
I = 643.33 A
Electro deposition of copper
`Cu^(2+)""_((aq))+2e^(-) rarr Cu_((s))`
2F charge is required to deposit
1 mole of copper i.e., 63.5 g
if the entire current produced in the fuel cell i.e., 4 F is utilised for electrolysis, then
`2 times 63.5` i.e., 127.0 g copper will be deposited at cathode.