Question

A current of $9.65{{ }}amp$ flowing for $10{{ }}minute$ deposits $3.0g$ of a metal. The equivalent wt. of the metal is:
A. $10$
B. $30$
C. $50$
D. $96.5$

Answer 1

This particular type of question can be solved using Faraday’s first law of electrolysis. First we will calculate the quantity of electricity passed and then using the law we will calculate the equivalent weight of the metal.
Formula used:
1. $q = I \times t$
where, $q$ is the quantity of charge passed
$I$ is the current; and $t$ is the time for which the current is passed or flowed.
2. $Eq.{{ }}wt.{{ }}of{{ }}metal\;\; = (Weight\;of\;metal/q) \times F$
Where, $q$ is again the amount of charge passed
And $F$ is the Faraday constant.

Complete step by step answer:
Firstly we will calculate the amount of charge passed for the given current flowed in the given time
We know that $q = I \times t$
And it is given that $I = 9.65A$ and $t = 10\min = 10 \times 60\sec = 600\sec$
Putting these values in the formula we have, $q = 9.65 \times 600 = 5790C$
Now we will calculate the equivalent weight:
The weight of the metal deposited is given in the question to be $3.0g$
Now as we already know, the formula for calculating the equivalent weight is:
$Eq.{{ }}wt.{{ }}of{{ }}metal\;\; = (Weight\;of\;metal/q) \times F$
Also, the value of Faraday’s constant is $96500$
Now, we will put all the values in the formula:
Equivalent weight of the metal $= (3/5970) \times 96500$
$\Rightarrow$ Equivalent weight of the metal $= 48.49g \sim 50g$
Hence the option C is correct.

Additional Information: Faraday’s first law of electrolysis states that the amount of chemical change which is produced by current at the boundary of an electrode and electrolyte is directly proportional to the amount of electricity that was used.

Note:
The amount of electricity that can bring about the change of one equivalent weight unit has been designated as a Faraday and it is approximately equal to $96500Coulomb$ of electricity.