
4.5 g of aluminium (at. mass 27amu) is deposited at cathode from \[A{l^{3 + }}\] solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from \[{H^ + }\] ions in solution by the same quantity of electric charge will be:
A.44.8L
B.11.2L
C.22.4L
D.5.6L
Answer
509.4k+ views
Hint: We have to calculate moles of aluminium and are used to find the moles of hydrogen produced. Then using 1 mole of gas at STP is equal to 22.4L.
Complete step by step answer:
Let’s start with writing the chemical equation that will be happening at cathode and anode separately, the equations are given below\[\]
\[
{\text{2Al }} \to {\text{ 2A}}{{\text{l}}^{{\text{3 + }}}}{\text{ + 6}}{{\text{e}}^{\text{ - }}} \\
{\text{6}}{{\text{H}}^{\text{ + }}}{\text{ + 6}}{{\text{e}}^{\text{ - }}}{\text{ }} \to {\text{ 3}}{{\text{H}}_{\text{2}}} \\
\]
Also, it is to be noted that the no. of moles of aluminium deposited are
Mass given/Molecular mass of the aluminium = \[4.5g/27g\] = 1/6 moles of aluminium.
It is to be notice that the same quantity of charge which is required to change the \[A{l^{3 + }}\]to Al is used to produce \[{H_2}\] from \[{H^ + }\]ions. \[{H_2}\]is having 2 atoms of H mean 2+ charge is being removed and in case of Al +3 charge is being removed. So 2 moles of Aluminium is being deposited and 3 mole of hydrogen is being produced.
So 2 mole of aluminium -> 3 mole of Hydrogen.
We know only 1/6 mole of Aluminium is being deposited hence,
1/6 mole of aluminium $ \to $ \[3/2{\text{ }} \times {\text{ }}1/6\] moles of Hydrogen, which gives
¼ moles of hydrogen is being produced.
Further it is to be noticed that 1 mole of gas at STP has a volume of 22.4L, So ¼ mole will have the volume of 5.6L (22.4/4).
$\therefore $The answer to this question is D. 5.6L.
Note:
We must know that the STP stands for standard temperature and pressure is widely used in the field of chemistry. All around the world it has different values. The most widely used is the one given by IUPAC which is 273 K and 1 bar pressure. Many gas companies in South America, Europe and Australia take 283K and 1atm pressure as STP.
Complete step by step answer:
Let’s start with writing the chemical equation that will be happening at cathode and anode separately, the equations are given below\[\]
\[
{\text{2Al }} \to {\text{ 2A}}{{\text{l}}^{{\text{3 + }}}}{\text{ + 6}}{{\text{e}}^{\text{ - }}} \\
{\text{6}}{{\text{H}}^{\text{ + }}}{\text{ + 6}}{{\text{e}}^{\text{ - }}}{\text{ }} \to {\text{ 3}}{{\text{H}}_{\text{2}}} \\
\]
Also, it is to be noted that the no. of moles of aluminium deposited are
Mass given/Molecular mass of the aluminium = \[4.5g/27g\] = 1/6 moles of aluminium.
It is to be notice that the same quantity of charge which is required to change the \[A{l^{3 + }}\]to Al is used to produce \[{H_2}\] from \[{H^ + }\]ions. \[{H_2}\]is having 2 atoms of H mean 2+ charge is being removed and in case of Al +3 charge is being removed. So 2 moles of Aluminium is being deposited and 3 mole of hydrogen is being produced.
So 2 mole of aluminium -> 3 mole of Hydrogen.
We know only 1/6 mole of Aluminium is being deposited hence,
1/6 mole of aluminium $ \to $ \[3/2{\text{ }} \times {\text{ }}1/6\] moles of Hydrogen, which gives
¼ moles of hydrogen is being produced.
Further it is to be noticed that 1 mole of gas at STP has a volume of 22.4L, So ¼ mole will have the volume of 5.6L (22.4/4).
$\therefore $The answer to this question is D. 5.6L.
Note:
We must know that the STP stands for standard temperature and pressure is widely used in the field of chemistry. All around the world it has different values. The most widely used is the one given by IUPAC which is 273 K and 1 bar pressure. Many gas companies in South America, Europe and Australia take 283K and 1atm pressure as STP.
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