Answer
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Hint: We must understand that there is no net loss in mass during the electrolysis. Therefore, the product will be \[Sn\]and \[S{n^{4 + }}\]respectively at cathode and anode. And the mass ratio is used according to the stoichiometry of the reaction.
Complete step by step answer:
We can write the chemical reaction during electrolysis as
$
2SnC{l_2}{\text{ }} \to {\text{ }}SnC{l_4}{\text{ }} + {\text{ }}Sn \\
2{\text{ }}mol{\text{ 1 mol 1 mol}} \\
{\text{2 x 190 g 1x261 g 1x119 g}} \\
$
No loss in mass of solution so mass loss due to deposition of \[Sn\] will be equal to mass gain due to the production of \[SnC{l_4}\].
Chlorine gas will not release on the anode as no loss in mass given in the question.
Here, we comparing the reacted amounts of the reactants
$
2SnC{l_2}{\text{ }} \to {\text{ }}SnC{l_4}{\text{ }} + {\text{ }}Sn \\
2{\text{ }}mol{\text{ 1 mol 1 mol}} \\
{\text{2 x 190 g 1x261 g 1x119 g}} \\
\dfrac{{mass{\text{ of Sn deposited}}}}{{mass{\text{ of SnC}}{{\text{l}}_4}{\text{ produced }}}} = \dfrac{{119}}{{261}} \\
$$
\dfrac{{mass{\text{ }} of {\text{ }} Sn {\text{ }}deposited}}{{mass{\text{ }}of{\text{ }} SnC{l_4}{\text{ }}produced{\text{ }}}} = \dfrac{{119}}{{261}} \\
\dfrac{{0.119}}{x} = \dfrac{{119}}{{261}} \\
x = {\text{ }}0.261{\text{ }}g \\
$
Here x is the mass of \[SnC{l_4}\] produced at the anode.
Mass of \[SnC{l_2}\]reacted = $\dfrac{{380}}{{119}}x0.119{\text{ }}g = 0.380{\text{ }}g$
Remaining unreacted mass of \[SnC{l_2}\] = $19 - {\text{ }}0.380 = 18.62{\text{ }}g$
So, mass ratio of \[SnC{l_2}\] and \[SnC{l_4}\] = $\dfrac{{18.62}}{{0.261}} = \dfrac{{71.34}}{1}$
Mass of \[SnC{l_2}\] : mass of \[SnC{l_4}\] = \[71.34:1\].
Note:
We can define a redox reaction as a chemical reaction in which electrons are transferred between two reactant species participating in the reaction. Redox reactions are based on electron transfers and electron density around any atom will change during redox reactions. Electron density will decrease in case of oxidation and increase in case of reduction.
Also, we must understand that the electron acceptor species is known as an oxidizing agent while the electron donor species is known as the reducing agent. The oxidizing agent undergoes a reduction of itself while the reducing agent undergoes oxidation of itself. The oxidation process is known as de-electronation while the reduction process is known as the electronation process.
Complete step by step answer:
We can write the chemical reaction during electrolysis as
$
2SnC{l_2}{\text{ }} \to {\text{ }}SnC{l_4}{\text{ }} + {\text{ }}Sn \\
2{\text{ }}mol{\text{ 1 mol 1 mol}} \\
{\text{2 x 190 g 1x261 g 1x119 g}} \\
$
No loss in mass of solution so mass loss due to deposition of \[Sn\] will be equal to mass gain due to the production of \[SnC{l_4}\].
Chlorine gas will not release on the anode as no loss in mass given in the question.
Here, we comparing the reacted amounts of the reactants
$
2SnC{l_2}{\text{ }} \to {\text{ }}SnC{l_4}{\text{ }} + {\text{ }}Sn \\
2{\text{ }}mol{\text{ 1 mol 1 mol}} \\
{\text{2 x 190 g 1x261 g 1x119 g}} \\
\dfrac{{mass{\text{ of Sn deposited}}}}{{mass{\text{ of SnC}}{{\text{l}}_4}{\text{ produced }}}} = \dfrac{{119}}{{261}} \\
$$
\dfrac{{mass{\text{ }} of {\text{ }} Sn {\text{ }}deposited}}{{mass{\text{ }}of{\text{ }} SnC{l_4}{\text{ }}produced{\text{ }}}} = \dfrac{{119}}{{261}} \\
\dfrac{{0.119}}{x} = \dfrac{{119}}{{261}} \\
x = {\text{ }}0.261{\text{ }}g \\
$
Here x is the mass of \[SnC{l_4}\] produced at the anode.
Mass of \[SnC{l_2}\]reacted = $\dfrac{{380}}{{119}}x0.119{\text{ }}g = 0.380{\text{ }}g$
Remaining unreacted mass of \[SnC{l_2}\] = $19 - {\text{ }}0.380 = 18.62{\text{ }}g$
So, mass ratio of \[SnC{l_2}\] and \[SnC{l_4}\] = $\dfrac{{18.62}}{{0.261}} = \dfrac{{71.34}}{1}$
Mass of \[SnC{l_2}\] : mass of \[SnC{l_4}\] = \[71.34:1\].
Note:
We can define a redox reaction as a chemical reaction in which electrons are transferred between two reactant species participating in the reaction. Redox reactions are based on electron transfers and electron density around any atom will change during redox reactions. Electron density will decrease in case of oxidation and increase in case of reduction.
Also, we must understand that the electron acceptor species is known as an oxidizing agent while the electron donor species is known as the reducing agent. The oxidizing agent undergoes a reduction of itself while the reducing agent undergoes oxidation of itself. The oxidation process is known as de-electronation while the reduction process is known as the electronation process.
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