Question

Two electrolytic cells are connected in series containing
${\text{CuS}}{{\text{O}}_{\text{4}}}$ solution and molten ${\text{AlC}}{{\text{l}}_{\text{3}}}$. If the electrolysis is ${\text{0}}{\text{.4}}$ moles of Cu are deposited on the cathode of the first cell. The number of moles of Al deposited on cathode of the second cell is:
A.$0.6$moles
B.$0.27$moles
C.$0.18$moles
D.$0.4$moles

Answer 1

Hint: We can determine the mole 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.

Complete step by step solution: 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${\text{CuS}}{{\text{O}}_{\text{4}}}$ and ${\text{AlC}}{{\text{l}}_{\text{3}}}$ electrolytic cell the equation can be written as follows:
$\dfrac{{{{\text{w}}_{{\text{Al}}}}}}{{{{\text{w}}_{{\text{Cu}}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{{\text{Al}}}}}}{{{{\text{E}}_{{\text{Cu}}}}}}$
Determine the mass of Cu in ${\text{0}}{\text{.4}}$ moles of Cu as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Molar mass of Cu is $63.55\,{\text{g/mol}}$
Substitute $63.55\,{\text{g/mol}}$ for molar mass and ${\text{0}}{\text{.4}}$ mol for mole of Cu.
${\text{0}}{\text{.4}}\,{\text{mol}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{63.55\,{\text{g/mol}}}}$
${\text{Mass}}\,{\text{of}}\,{\text{Cu = }}\,{\text{0}}{\text{.4}}\,{\text{mol}}\,\, \times \,{\text{63}}{\text{.55}}\,{\text{g/mol}}$
${\text{Mass}}\,{\text{of}}\,{\text{Cu = }}\,25.42\,{\text{g}}$
So, the mass of Cu is gram.
The formula to determine the equivalent weight is as follows:
${\text{equivalent}}\,{\text{weight = }}\dfrac{{{\text{Atomic}}\,{\text{weight}}\,{\text{(M)}}}}{{{\text{valency}}\,{\text{(n)}}}}$
The dissociation of the copper sulphate and aluminium chloride is as follows;
\[{\text{CuS}}{{\text{O}}_{\text{4}}}\, \to {\text{C}}{{\text{u}}^{2 + }} + \,{\text{SO}}_4^{2 - }\]
So, the valency factor for Cu is $2$.
\[{\text{AlC}}{{\text{l}}_3}\, \to {\text{A}}{{\text{l}}^{3 + }} + \,{\text{3}}\,{\text{C}}{{\text{l}}^ - }\]
So, the valency factor for Al is $3$.
${\text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{Cu}}\,{\text{ = }}\dfrac{{{\text{63}}{\text{.55}}}}{{\text{2}}}$
${\Rightarrow \text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{Cu}}\,{\text{ = 31}}{\text{.77}}$
${\Rightarrow \text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{Al}}\,{\text{ = }}\dfrac{{26.98}}{3}$
${\Rightarrow \text{equivalent}}\,{\text{weight}}\,{\text{of}}\,{\text{Al}}\,{\text{ = 8}}{\text{.99}}$
Substitute $31.77$ for equivalent weight of Cu, $8.99$ for Al and $25.42$ gram for mass of Cu in Farraday second law formula.
$\dfrac{{{{\text{w}}_{{\text{Al}}}}}}{{{\text{25}}{\text{.42}}}}{\text{ = }}\dfrac{{8.99}}{{31.77}}$
${{\text{w}}_{{\text{Al}}}}{\text{ = }}\dfrac{{{\text{25}}{\text{.42}} \times \,{\text{8}}{\text{.99}}}}{{31.77}}$
${{\text{w}}_{{\text{Al}}}}{\text{ = }}7.19\,{\text{g}}$
Determine the mole of Al in $7.19\,$ gram of Al as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Molar mass of Al is $26.98\,{\text{g/mol}}$
Substitute $26.98\,{\text{g/mol}}$ for molar mass and $7.19\,$ gram for mass of Al.
${\text{Mole}}\,{\text{of}}\,{\text{Al}}\,{\text{ = }}\,\dfrac{{{\text{7}}{\text{.19}}\,{\text{g}}}}{{26.98\,{\text{g/mol}}}}$
${\Rightarrow \text{Mole}}\,{\text{of}}\,{\text{Al}}\,{\text{ = }}\,0.27\,{\text{mol}}$
So, the mole of Al deposited is $0.27$.

Therefore option (B) $0.27$ mole 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.