Question

Silver is monoatomic and has an atomic mass of$108$. Copper is divalent and has an atomic mass of$63.6$. The same electric current is passed, for the same length of time through a silver coulometer and a copper coulometer. If $27.0$g of silver is deposited, then the corresponding amount of copper deposited is--- (in gm).

Answer 1

Hint: We should know the Faraday’s second law to determine the answer of this question. We will determine the equivalent weight of silver and copper. Then by using Faraday’s second law of electrolysis we will determine the amount of copper deposited. According to which the amount deposited or liberated on the electrode is directly proportional to its equivalent weight.

Formula used:
 $\dfrac{{{{\text{w}}_{\text{1}}}}}{{{{\text{w}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{1}}}}}{{{{\text{E}}_{\text{2}}}}}$
${\text{equivalent}}\,{\text{weight = }}\dfrac{{{\text{Atomic}}\,{\text{weight}}\,{\text{(M)}}}}{{{\text{valency}}\,{\text{(n)}}}}$

 Complete step by step answer:
According to Faraday’s second law of electrolysis when a certain amount of charge is passed through a cell, the amount deposited or liberated on the electrode is directly proportional to its equivalent weight.
The mathematical expression of Faraday’s second law of electrolysis is shown as follows:
$\dfrac{{{{\text{w}}_{\text{1}}}}}{{{{\text{w}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{1}}}}}{{{{\text{E}}_{\text{2}}}}}$
Where,
${{\text{w}}_{\text{1}}}$and ${{\text{w}}_{\text{2}}}$ are the weight deposited of different elements on the electrodes.
${{\text{E}}_{\text{1}}}$and ${{\text{E}}_{\text{2}}}$ are the equivalent weight of elements deposited on the electrodes.
For silver and copper metal cell the equation can be written as follows:
\[\dfrac{{{{\text{w}}_{{\text{Ag}}}}}}{{{{\text{w}}_{{\text{Cu}}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{{\text{Ag}}}}}}{{{{\text{E}}_{{\text{Cu}}}}}}\]
The formula to determine the equivalent weight is as follows:
${\text{equivalent}}\,{\text{weight = }}\dfrac{{{\text{Atomic}}\,{\text{weight}}\,{\text{(M)}}}}{{{\text{valency}}\,{\text{(n)}}}}$
Silver is monoatomic so the valency factor for silver is one.
${\text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{Ag}}\,{\text{ = }}\dfrac{{108}}{1}$
${\text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{Ag}}\,{\text{ = }}\,{\text{108}}$
Copper is divalent so, the valency factor for silver is two.
${\text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{Cu}}\,{\text{ = }}\dfrac{{63.6}}{2}$
${\text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{Cu}}\,{\text{ = }}\,{\text{31}}{\text{.8}}$

On substituting $27.0$g for mass of silver metal deposited, $108$ for equivalent weight of silver metal and $31.8$ for equivalent weight of copper metal.

\[\dfrac{{27.0}}{{{{\text{w}}_{{\text{Cu}}}}}}{\text{ = }}\dfrac{{108}}{{{\text{31}}{\text{.8}}}}\]
${{\text{w}}_{{\text{Cu}}}}{\text{ = }}\dfrac{{{\text{27}}{\text{.0}} \times \,31.8}}{{108}}$
${{\text{w}}_{{\text{Cu}}}}{\text{ = }}\,{\text{7}}{\text{.95}}\,{\text{g}}$
So, the corresponding amount of copper deposited is${\text{7}}{\text{.95}}\,{\text{g}}$.

Therefore ${\text{7}}{\text{.95}}\,{\text{g}}$ is the correct answer.

Note:The equivalent weight is determined by dividing the atomic weight by valency. Valency is the charge or oxidation number of the atom. In the case of acids, the valency is determined as the number of protons donated. In the case of oxidation number one, the equivalent weight will be equal to atomic weight. The equivalent weight of deposited metals is compared in the faraday second law, not the Equivalent weight of salts.