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On passing 1.5 F charge, the number of moles of aluminium deposited at the cathode is:
[molar mass of Al = 27 g mol$^{ -1 }$]
A.1.0
B.13.5
C.0.50
D.0.75

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Last updated date: 25th Feb 2024
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Hint: To answer this question you first need to write the chemical equation, from there you can just find the number of moles of electrons involved, and then just relate it with the number of moles of Aluminium deposited.

Complete step by step answer:
We can write the balanced chemical equation for the deposition of aluminium as:
$Al^{ 3+ }+3e^{ - }\quad \rightarrow\quad Al$
Faraday's first law says the mass of the substance generated by electrolysis is proportional to the amount of electricity used.
Q = I x t
Where,
Q = quantity of electricity in coulombs
I = current in amperes
T= time in seconds
1 mole of electrons carry charge = 96500C = 1 Faraday(F)
Here, we can see that one mole of Al3+ ions requires 3 moles of electrons to neutralize and deposit in the form of aluminium.
So, on passing 3 moles of electrons (3F charge), the number of moles of aluminium deposited at the cathode is 1 mole.
Hence, on passing 1.5F charge (1.5 moles of electrons), the number of moles of aluminium deposited at the cathode will be,

$=\quad \dfrac { 1\quad \times \quad 1.5 }{ 3 } \quad =\quad 0.5mole$
Therefore, we can conclude that the correct answer to this question is option C.

Note: We should also know that the coulomb (symbolized C) is the standard unit of electric charge in the International System of Units (SI). This is a dimensionless quantity.