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Question
In fuel cell H2 and O2 react to produce electricity. In the process, H2 gas is oxidised at the anode and O2 at cathode. If 44.8 litre of H2 at 25°C and 1 atm pressure reacts in 10 minutes, what is average current produced? If the entire current is used for electro deposition of Cu from Cu2+, how many grams of deposited?
Solution
Oxidation at anode:
\[\ce{2H2_{(g)} + 4OH^-_{( aq)} -> 4H2O_{(l)} + 4e^-}\]
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
∴ No. of moles of hydrogen gas produced = `(1 "mole")/(22.4 "litres") xx 44.8 "litres"`
= 2 moles of hydrogen
∴ 2 of moles of hydrogen produces 4 moles of electron i.e., 4F charge.
We know that Q = It
I = `"Q"/"t"`
= `(4"F")/(10 "mins")`
= `(4 xx 96500 "C")/(10 xx 60 "s")`
I = 643.33 A
Electro deposition of copper
\[\ce{Cu^{2+}_{( aq)} + 2e^- -> Cu_{(s)}}\]
2F charge is required to deposit
1 mole of copper i.e., 63.5 g
If the entire current produced in the fuel cell ie., 4F is utilised for electrolysis, then 2 × 63.5 i.e., 127.0 g copper will be deposited at cathode.
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