Question

Salts of metals X, Y and Z are electrolyzed under identical conditions using the same quantity of electricity. It was observed that $4.2\,{\text{g}}$ of X, $5.4\,{\text{g}}$ of Y and $19.2\,{\text{g}}$ of Z were deposited at the respective cathode, if the atomic weights of X, Y and Z are$7$, $27$and $64$respectively, then their ratio valencies is,
A. $1:2:3\,$
B.$1:3:2$
C.$2:3:1$
D.$3:2:2$

Answer 1

Hint: We can determine the valency by using Faraday’s second law of electrolysis. According to which the amount deposited or liberated on the electrode is directly proportional to its equivalent weight so, we will substitute molar mass and valency in place of equivalent

Formula used: $\dfrac{{{{\text{w}}_{\text{1}}}}}{{{{\text{w}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{1}}}}}{{{{\text{E}}_{\text{2}}}}}$

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.
The above relation can be rearranged as follows:
$\dfrac{{{{\text{w}}_{\text{1}}}}}{{{{\text{E}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{w}}_2}}}{{{{\text{E}}_{\text{2}}}}}$
The above relation can be written for salts of metals X, Y and Z that are electrolyzed as follows:
$\dfrac{{{{\text{w}}_{\text{X}}}}}{{{{\text{E}}_{\text{X}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{Y}}}}}{{{{\text{E}}_{\text{Y}}}}} = \dfrac{{{{\text{w}}_{\text{Z}}}}}{{{{\text{E}}_{\text{Z}}}}}$.....$(1)$
The formula of equivalent weight is as follows:
${\text{equivalent}}\,{\text{weight = }}\dfrac{{{\text{Atomic}}\,{\text{weight}}\,{\text{(M)}}}}{{{\text{valency}}\,{\text{(n)}}}}$
So, the ratio of metals X, Y and Z is as follows:
$\dfrac{{{{\text{w}}_{\text{X}}}}}{{{{\text{M}}_{\text{X}}}/{{\text{n}}_{\text{X}}}}}{\text{:}}\dfrac{{{{\text{w}}_{\text{Y}}}}}{{{{\text{M}}_{\text{Y}}}/{{\text{n}}_{\text{Y}}}}}:\dfrac{{{{\text{w}}_{\text{Z}}}}}{{{{\text{M}}_{\text{Z}}}/{{\text{n}}_{\text{Z}}}}}$
On substituting $4.2\,{\text{g}}$for equivalent weight of X, $5.4\,{\text{g}}$for Y and $19.2\,{\text{g}}$for Z and, are$7$ for atomic mass of X,, $27$ for Y and $64$ for Z.
$\dfrac{{{\text{4}}{\text{.2}}\,{\text{g}}}}{{{\text{7}}/{{\text{n}}_{\text{X}}}}}{\text{:}}\dfrac{{{\text{5}}{\text{.4}}\,{\text{g}}}}{{{\text{27}}/{{\text{n}}_{\text{Y}}}}}:\dfrac{{{\text{19}}{\text{.2}}\,{\text{g}}}}{{{\text{64}}/{{\text{n}}_{\text{Z}}}}}$
$0.6\,{{\text{n}}_{\text{x}}}{\text{:}}\,0.2\,{{\text{n}}_{\text{Y}}}:0.3\,{{\text{n}}_{\text{Z}}}$
We will take the simplest ratio as follows:
Divide each with $0.6$.
$\dfrac{{0.6\,{{\text{n}}_{\text{x}}}}}{{0.6}}{\text{:}}\,\dfrac{{0.2\,{{\text{n}}_{\text{Y}}}}}{{0.6}}:\dfrac{{0.3\,{{\text{n}}_{\text{Z}}}}}{{0.6}}$
${{\text{n}}_{\text{x}}}\,{\text{:}}\,\dfrac{{1\,{{\text{n}}_{\text{Y}}}}}{{3}}:\dfrac{{1\,{{\text{n}}_{\text{Z}}}}}{{2}}$
So,$1\,\,{{\text{n}}_{\text{x}}}{\text{:}}\,3\,{{\text{n}}_{\text{Y}}}:2\,{{\text{n}}_{\text{Z}}}$
So, the ratio of valencies is $1\,{\text{:}}\,3\,:2$.
Therefore, option (B) $1\,{\text{:}}\,3\,:2$, is correct.

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. Ratio of equivalent weight of substance deposited gives the ratio of amount of substance deposited on an electrode.