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Question

In a hydrogen oxygen fuel cell, electricity is produced. In this process H2(g) is oxidised at anode and O2(g) reduced at cathode. Given : Cathode: O2(g)+2H2O(l)+4e4OH(aq) Anode : H2(g)+2OH(aq)2H2O(l)+2e 4.48 L of H2 at 1 atm and 273 K is oxidised in 9650 sec.
The mass of water produced is (in grams):

A
7.2 g
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B
1.8 g
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C
3.6 g
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D
0.9 g
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Solution

The correct option is C 3.6 g
Using Faraday’s Law:
mH2O=MMn×96500×i×t
Now, from the previous question:
i= 4 A
Putting the values:
mH2O= 18×4×96502×96500
m= 3.6 g

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