How you will show that 96500 coulomb of charge is required to liberate 1Gram equivalent of a metal?

Answer 1

In fact it's actually 96483 C (96500 is sometimes used to make approximate calculations)

The idea is to know how many electrons we need to have a total charge of 96483 C.

For this we divide the 96483 C by the charge of one single electron (i.e. the elementary charge=1.60217646 × 10-19 C).

So we have:

96483 C/1.60217646 × 10-19 C = 6.022 × 1023

The number to the right of the equals sign is nothing but Avogadro's number, which means that we need exactly one mole of electrons to get 96483 coulombs of charge.

According to IUPAC, 1 gram equivalent is the mass of a given substance which will:

* supply or react with one mole of hydrogen ions (H+) in an acid-base reaction; or

* supply or react with one mole of electrons in a redox reaction.

So, using the results of the calculation we made above, we can replace in the IUPAC definition (the second part) "one mole of electrons" by "96583 C of charge", and then the definition becomes the answer to your question.

Answer 2

To show that 96500 coulombs of charge are needed to liberate 1 gram equivalent of a metal, you can use Faraday's law of electrolysis. Since 1 gram equivalent of a metal is equivalent to the number of grams equal to its equivalent weight, you can calculate the total charge required by multiplying the equivalent weight (in grams) by the Faraday constant (96500 C/equivalent). This calculation shows the relationship between the amount of substance being liberated and the charge required.