Question

When $3.86\ A$ current is passed through an electrolyte for $50\ mins$, $2.4\ g$ of a divalent metal is deposited. The gram atomic weight of the metal (in grams) is:
A) $24$
B) $12$
C) $64$
D) $40$

Answer 1

Hint: Here, we have to apply the concept of Faraday’s law of electrolysis, which states that the weight of a chemical substance deposited after electrolysis is equal to the product of the equivalent weight of the substance, the current passing through the electrodes in amperes, and the time, divided by $96500$ coulombs (the value of $1$ Faraday).

Complete step by step answer:
The weight of metal deposited after electrolysis is given by Faraday’s law of electrolysis in the form of the given equation:

$Weight\ in\ grams\ = \dfrac{Equivalent\ weight\times Current\ in\ amperes\times Time\ in\ seconds}{96500\ C}$

Where, $Equivalent\ Weight\ = \dfrac{Atomic\ weight}{Valency\ factor}$

So, Faraday’s law of electrolysis can also be represented as:

$Weight\ in\ grams\ = \dfrac{Atomic\ weight\times Current\ in\ amperes\times Time\ in\ seconds}{Valency\ Factor\times 96500\ C}$

Given,
Current passed through the electrolyte $ = 3.86\ A$
Time for which the current is passed $ = 50\ min$
Weight of the metal deposited $ = 2.4\ g$
Valency factor of the deposited metal $ = 2$
The gram atomic weight of the metal $ = ?$
According to Faraday’s equation, the time must be in seconds. So, we have to convert $50\ minutes$ into $seconds$.
Now, we know that $1\ minute = 60\ seconds$
So, to convert a minute into seconds, we shall multiply the time by $60$.

Therefore, $50\ min = 50\times 60\ sec$

Substituting all the given values in Faraday’s equation, we get;

$2.4\ g=\dfrac{Atomic\; weight\times 3.86\ A\times (50\times 60)\ sec}{2\times 96500\ C}$

$\Rightarrow Atomic\; weight = \dfrac{2.4\ g\times 2\times 96500\ C}{3.86\ A\times (50\times 60)\ sec}$

$\Rightarrow Atomic\ weight = 40\ g$

So, the correct answer is Option D .

Note: While doing the calculations, the unit of each value must be considered.
Since $1\ coulomb=1\ ampere\times 1\ second$, the units $coulomb$ and $ampere\times sec$ cancel each other in the equation.
Faraday’s law of electrolysis is generally written as:

$Weight\ in\ grams\ = Z\times\ current\ in\ amperes(i)\times time\ in\ second(t)$

Where, $Z=\dfrac{Equivalent\ weight}{96500\ C}$

Valency is the number of electrons lost, gained, or shared by an element during a reaction. For metal, it is the number of electrons lost during a reaction. In the question, the metal is said to be “divalent”. The prefix before the valency of metal tells us about the number of electrons lost by it. The prefix “di” means two, which tells us that the metal has a valency of two.