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Question: 4.5g of aluminium (at. Mass 27 amu) is deposited at cathode from \(A{{l}^{3+}}\) solution by a certa...

4.5g of aluminium (at. Mass 27 amu) is deposited at cathode from Al3+A{{l}^{3+}} solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H+{{H}^{+}} ions in solution by the same quantity of electric charge will be:
A. 44.8 L
B. 11.2 L
C. 22.4 L
D. 5.6 L

Explanation

Solution

To solve this question we have to apply the formula given by Faraday's second law of electrolysis. According to it the amount of substances deposited through the passage of the same amount of electric current is directly proportional to their equivalent weight. It is given as W1W2=E1E2\dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{{{E}_{1}}}{{{E}_{2}}}.

Complete step by step solution:
From your chemistry lessons you have learned about the Faraday's second law of electrolysis. Faraday's second law of electrolysis states that the amount or masses of substance deposited at the electrode when the same amount of electricity is passed through the different electrolytes are directly proportional to their equivalent weight. So, we can say that the weight or mass of any element deposited is directly proportional to their equivalent weight. Let us take the amount of deposited substance as W1,W2{{W}_{1}},{{W}_{2}} and E1,E2{{E}_{1}},{{E}_{2}} will be their respective equivalent weight and it is given as:
W1W2=E1E2\dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{{{E}_{1}}}{{{E}_{2}}}
AS we know that the formula to find the equivalent weight is:
Eq. weight=molecular weightn−factorEq.\,weight=\dfrac{molecular\,weight}{n-factor}
In the question the mass of aluminium is given as 4.5 g and the equivalent weight of H will be 1. So, to find the equivalent weight of aluminium will be:
Eq. weight=273=9Eq.\,weight=\dfrac{27}{3}=9
Here we have to find the mass of H. So, the formula will be:
mAlmH=EALEH\dfrac{{{m}_{Al}}}{{{m}_{H}}}=\dfrac{{{E}_{AL}}}{{{E}_{H}}}
Now, put all the values in the formula we will get:
4.5mH=91\dfrac{4.5}{{{m}_{H}}}=\dfrac{9}{1}
∴mH=0.5g\therefore {{m}_{H}}=0.5g
Now, the 2g of H2{{H}_{2}} H2{{H}_{2}} volume at STP = 22.42=11.2L\dfrac{22.4}{2}=11.2L
Therefore, 0.5 g of H2{{H}_{2}} volume at STP = 11.2×0.5=5.6L11.2\times 0.5=5.6L

Thus the correct option will be (D).

Note: According to faraday law 1 Faraday charge liberates 1 eq of substance. Faraday is defined as the amount of electricity flowing by an electrolyte which liberates 1 gram equivalent of any substance at any electrode. And the value of Faraday constant (F) is 96500 C.mol−1C.mo{{l}^{-1}}. Faraday's second law is very helpful in determining the equivalent weight of different electrolyte