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I came across this question:

A quantity of electrical charge that brings about the deposition of 4.5 g $\ce{Al}$ from $\ce{Al^{+3}}$ at cathode will also produce the following volume (STP) of $\ce{H2}$ (g) from $\ce{H+}$ at cathode:

A)22.4

B)11.2

C)5.6

D)2.8

If we use faradays first law: For $\ce{Al^{+3}}$

4.5-=(27*c)/ (96500*3)

c=96500/2

For $\ce{H2}$:

m=c/(96500*2)

m=0.25

Hence V=2.8

If we use 2nd law, we get mass of $\ce{H2}$ produced as 0.5 Hence, V=5.6

So then what was the mistake (Please reply if I have made calculation mistake, I generally do that)

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  • $\begingroup$ Your data are rather disagreeable to read, simply because you never mention the units. Is the volume in liters ? in cubic meters ? $\endgroup$
    – Maurice
    Jun 5 at 14:46
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I am not sure what your answer was (2.8 or 5.6dm3) and therfore am not sure what the mistake is. I suggest an alternative application of Faraday's law which strikes me as simpler. The law states that the same electrical charge will liberate the same mumber of moles of different atoms divided by the charge on their ions. In this case the No. of moles of aluminium = 4.5/27 = 1/6 therefore No. of moles of H atoms = 3*1/6 (because Al has three times the charge of H+ ions) therfore No. of moles of H2 molecules = 3/2*1/6 = 1/4. therefore volume of H2 at STP =22.4/4 = 5.6 dm3. which is one of the answers that you quoted.

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