Answer
438k+ views
Hint: When the same amount of current is passed from two electrolytes, the masses liberated from each electrolyte will be directly proportional to the equivalent weight of the substance liberated.
Complete answer:
Faraday’s First Law of Electrolysis- The mass of the substance (m) deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge (Q) passed.
Faraday’s Second Law of Electrolysis- the amount of substances deposited by the passage of the same amount of electric current will be proportional to their equivalent weights. When, the electric current (It or Q) remains the same, mass is directly proportional to equivalent weight.
Given, 5L of $CuS{{O}_{4}}$ of 0.1M is electrolysed by current of 2.0A . By Faraday’s Second Law of Electrolysis, the weight of $C{{u}^{+2}}$ can be evaluated.
\[\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{E}_{1}}}{{{E}_{2}}}\]
Where, ${{m}_{1}}$ is the mass of copper,
${{m}_{2}}$ is the mass of iron , given mass of iron is 9.70g
${{E}_{1}}$ is the equivalent weight of copper, equivalent weight of copper is 31.75g/eq
${{E}_{2}}$ is the equivalent weight of iron,
\[\begin{align}
& {{m}_{1}}={{m}_{2}}\times \dfrac{{{E}_{1}}}{{{E}_{2}}} \\
& {{m}_{1}}=9.70\times \dfrac{31.75}{27.9} \\
& {{m}_{1}}=11.03g \\
\end{align}\]
The amount of $C{{u}^{+2}}$ is 11.03g. Now applying the Faraday’s First Law of Electrolysis, we get,
\[\begin{align}
& {{m}_{1}}=z\times i\times t \\
& z=\dfrac{{{E}_{1}}}{96500} \\
& \Rightarrow t=\dfrac{{{m}_{1}}}{z\times i} \\
& \Rightarrow t=\dfrac{{{m}_{1}}\times 96500}{{{E}_{1}}\times i} \\
& \Rightarrow t=\dfrac{11.03\times 96500}{31.75\times 2.0} \\
& \Rightarrow t=16762.1s \\
\end{align}\]
Converting, seconds into hours by dividing the obtained value with 60x60,
\[\begin{align}
& t=\dfrac{16762.1}{60\times 60} \\
& \Rightarrow t=4.6hr \\
\end{align}\]
Therefore, the time required for the whole $C{{u}^{+2}}$ to be deposited is 4.6 hours.
Note: The time used in the formula of Faraday’s First Law of Electrolysis is in seconds, we have to change the time to hours or minutes according to the requirement of the question. Equivalent weight of a substance can be evaluated by dividing the molecular weight of the substance with its valency.
Complete answer:
Faraday’s First Law of Electrolysis- The mass of the substance (m) deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge (Q) passed.
Faraday’s Second Law of Electrolysis- the amount of substances deposited by the passage of the same amount of electric current will be proportional to their equivalent weights. When, the electric current (It or Q) remains the same, mass is directly proportional to equivalent weight.
Given, 5L of $CuS{{O}_{4}}$ of 0.1M is electrolysed by current of 2.0A . By Faraday’s Second Law of Electrolysis, the weight of $C{{u}^{+2}}$ can be evaluated.
\[\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{E}_{1}}}{{{E}_{2}}}\]
Where, ${{m}_{1}}$ is the mass of copper,
${{m}_{2}}$ is the mass of iron , given mass of iron is 9.70g
${{E}_{1}}$ is the equivalent weight of copper, equivalent weight of copper is 31.75g/eq
${{E}_{2}}$ is the equivalent weight of iron,
\[\begin{align}
& {{m}_{1}}={{m}_{2}}\times \dfrac{{{E}_{1}}}{{{E}_{2}}} \\
& {{m}_{1}}=9.70\times \dfrac{31.75}{27.9} \\
& {{m}_{1}}=11.03g \\
\end{align}\]
The amount of $C{{u}^{+2}}$ is 11.03g. Now applying the Faraday’s First Law of Electrolysis, we get,
\[\begin{align}
& {{m}_{1}}=z\times i\times t \\
& z=\dfrac{{{E}_{1}}}{96500} \\
& \Rightarrow t=\dfrac{{{m}_{1}}}{z\times i} \\
& \Rightarrow t=\dfrac{{{m}_{1}}\times 96500}{{{E}_{1}}\times i} \\
& \Rightarrow t=\dfrac{11.03\times 96500}{31.75\times 2.0} \\
& \Rightarrow t=16762.1s \\
\end{align}\]
Converting, seconds into hours by dividing the obtained value with 60x60,
\[\begin{align}
& t=\dfrac{16762.1}{60\times 60} \\
& \Rightarrow t=4.6hr \\
\end{align}\]
Therefore, the time required for the whole $C{{u}^{+2}}$ to be deposited is 4.6 hours.
Note: The time used in the formula of Faraday’s First Law of Electrolysis is in seconds, we have to change the time to hours or minutes according to the requirement of the question. Equivalent weight of a substance can be evaluated by dividing the molecular weight of the substance with its valency.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)