Question

$\text{1 mole}$ of aluminium is deposited by X coulomb of electricity passing through the aluminium nitrate solution. Calculate the number of moles of silver deposited by X coulombs electricity from silver nitrate solution.
A.3
B.4
C.2
D.1

Answer 1

One faraday is defined in the terms of electricity as 96500 coulombs. This is the charge of 1 mole of electrons, where 1 mole is equal to$6.023\times {{10}^{\text{23}}}$. We can calculate the number of moles of silver deposited from the reduction half-cell equation of aluminium.

Complete answer:
The electrolysis of aluminium nitrate leads to the dissociation of the salt into the cation $\text{A}{{\text{l}}^{\text{+3}}}$and anion$\text{N}{{\text{O}}_{\text{3}}}^{\text{-}}$.
The cation proceeds towards to the cathode to accept the electrons and gets reduced as the metal atoms by the following equation:
$\text{A}{{\text{l}}^{\text{+3}}}+3{{\text{e}}^{\text{-}}}\to \text{Al}$
1 mole of aluminium atoms is liberated by 3 faradays of electricity.
The electrolysis of silver nitrate liberates the silver cations and the nitrate anions.
The reaction of the cathode is as follows: $\text{A}{{\text{g}}^{+}}+{{\text{e}}^{\text{-}}}\to \text{Ag}$
So for the liberation of one mole of silver atoms, one faraday of electricity is required.
Hence 3 faradays of electricity can liberate 3 moles of silver atoms at the cathode.

So, the correct answer is option A.

Note:
The electrolytic cell consists of two electrodes: the cathode and the anode. The positive electrode of a cell is the anode and here the anions or the negative anions are liberated after they give up the electrons to the cathode. The negative electrode of the cell is called the cathode and here the cations or the positive anions are liberated at the liberated here after they accept electrons from the cathode.