Question

Same quantity of electricity being used to liberate iodine (at anode) and a metal (at cathode). The mass of metal liberated at cathode is $0.617$g and the liberated iodine completely reduced by $46.3\,{\text{ml}}$ of $0.124\,{\text{M}}$ sodium thiosulphate solution. What is the equivalent weight of metal?
A.$23\,{\text{g}}$
B.${\text{127}}\,{\text{g}}$
C.$108\,{\text{g}}$
D.$46\,{\text{g}}$

Answer 1

Hint:We can determine the equivalent weight of metal of aluminium 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. First we will determine the grams of thiosulfate by using molarity and mole formula then the equivalent weight of metal.

Formula used: $\dfrac{{{{\text{w}}_{\text{1}}}}}{{{{\text{w}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{1}}}}}{{{{\text{E}}_{\text{2}}}}}$
${\text{equivalent}}\,{\text{weight = }}\dfrac{{{\text{Atomic}}\,{\text{weight}}\,{\text{(M)}}}}{{{\text{valency}}\,{\text{(n)}}}}$
${\text{Molarity}}\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Litter}}\,{\text{of}}\,{\text{solution}}}}$
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$

Complete step-by-step solution:First, we will convert the volume of sodium thiosulphate solution from millilitre to litre as follows:
$1000\,{\text{ml}}\,{\text{ = }}\,\,{\text{1L}}$
$46.3\,{\text{ml}}\,{\text{ = }}\,\,0.0463\,{\text{L}}$
Now we will use the molarity formula to determine the moles of sodium thiosulphate as follows:
${\text{Molarity}}\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Litter}}\,{\text{of}}\,{\text{solution}}}}$
On substituting $0.0463\,{\text{L}}$ for volume of solution and $0.124\,{\text{M}}$ for molarity of sodium thiosulphate solution,
${\text{0}}{\text{.124}}\,{\text{M}}\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{0.0463\,{\text{L}}}}$
$\Rightarrow{\text{Moles}}\, = \,{\text{0}}{\text{.124}}\,{\text{M}}\,\, \times 0.0463\,{\text{L}}$
$\therefore {\text{Moles}}\, = \,{\text{0}}{\text{.0057}}$
We will use the mole formula to determine the mass of sodium thiosulphate.

${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
The molar mass of sodium thiosulphate is $248.17$g/mol.
On substituting $248.17$g/mol for molar mass and ${\text{0}}{\text{.0057}}$for moles,
$0.0057\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{248.17}}$
$\Rightarrow{\text{mass}}\,{\text{ = }}\,{\text{0}}{\text{.0057}}\,\,{\text{ \times }}\,{\text{248}}{\text{.17}}\,$
$\therefore {\text{mass}}\,{\text{ = }}\,1.414\,{\text{g}}$
So, the mass of sodium thiosulphate is$1.414\,{\text{g}}$.
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 metal and sodium thiosulphate electrolytic cell the equation can be written as follows:
\[\dfrac{{{{\text{w}}_{\text{M}}}}}{{{{\text{w}}_{{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{OM}}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{M}}}}}{{{{\text{E}}_{{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}}}}}\]
The formula to determine the equivalent weight is as follows:
${\text{equivalent}}\,{\text{weight = }}\dfrac{{{\text{Atomic}}\,{\text{weight}}\,{\text{(M)}}}}{{{\text{valancy}}\,{\text{(n)}}}}$
So, the valency factor for \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\] is $1$.
${\text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\,{\text{ = }}\dfrac{{{\text{248}}{\text{.17}}}}{1}$
${\text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\,{\text{ = }}\,248.17$
On substituting $0.617$g for mass of metal liberated, $1.414\,{\text{g}}$for the mass of sodium thiosulphate,$248.17$ for equivalent weight of sodium thiosulphate ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$.
$\dfrac{{0.617}}{{1.414\,{\text{g}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{M}}}}}{{248.17}}$
$\Rightarrow {{\text{E}}_{\text{M}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.617}} \times \,248.17}}{{1.414}}$
$\therefore {{\text{E}}_{\text{M}}}{\text{ = }}108\,{\text{g}}$
So, the equivalent weight of metal is $108\,{\text{g}}$.

Therefore option (C) $108\,{\text{g}}$ 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. 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.