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In a fuel cell H(2) and O(2) react to pr...

In a 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)` is reduced at the cathode. If 6.72 litre of `H_(2)` at NTP reacts in 15 minute, what is the average current produced ? If the entire current is used for electro-deposition of Cu from `Cu^(2+)`, how many g of Cu are deposited ?

Text Solution

Verified by Experts

`O_(2)+2H_(2)O+4e to 4OH^(-) ("cathode")`
`H_(2)+2OH^(-) to 2H_(2)O +2e ("at anode")`
Moles of `H_(2)` reacting `=(67.2)/(22.4) =3`
Eq. of `H_(2)` used =`3 xx 2=6`
`W=("Eit")/(96500) rArr W/E=("it")/(96500)`
`rArr 6=(I xx 15 xx 60)/(96500) therefore i=643.33"ampere"`
Eq. of `H_(2)="eq. of Cu formed"=6`
`W_(cu)=6 xx (63.5)/(2)=190.5g`
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