Answer
64.8k+ views
Hint:In electrochemistry we can’t always measure the amount of flow of charges. Instead the concept of current is used. Current is the amount of electrical charge that flows per unit time. It is measured in Amperes (A).
Complete step by step solution:
> For this we use Faraday's Constant, F. Faraday's Constant is given by the symbol F and is defined as the charge in coulombs (C) of 1 mole of electrons. Faraday's constant is approximately 96485 C mol-1. You can calculate F by multiplying the charge on one electron (1.602 x 10-19) by Avogadro's number (6.022 x 1023),
We have been given in the problem that the:
Mass of \[Mn{{O}_{2}}\] is 26.08g.
Atomic mass of Mn is 55 amu.
Current = 0.5 A
\[M{{n}^{4+}}+4{{e}^{-}}\to Mn\]
We can see from the above that 4 mol of electrons give 1 mol of Magnesium, Mn.
> Now we will put the numbers in. 1 mol of electrons is 1 faraday.
4 x 96500 coulombs give 55 g of copper.
Number of coulombs = \[=\dfrac{26.08\times 2\times 96485}{55}=94502.86\]C
Now we know how many coulombs we need, and the current was already given in the problem. Finally, we can calculate the time required to oxidize \[Mn{{O}_{2}}\].
Number of coulombs = current in amps x time in seconds
\[\begin{align}
& 94502.86=0.5\times \,t \\
& t=18901.72\,\sec \\
\end{align}\]
To convert into hours, we will divide by 60.
\[\dfrac{18901.72}{60}=315\,hours\]
So, the total time taken will be 315 hours.
Note: The same question can be given with time and asking to calculate the amount of \[Mn{{O}_{2}}\] produced after oxidation. In that case the procedure should be:
\[\text{Current}\,\text{and time}\to \text{Charge(C)}\to \text{faradays}\to \text{moles}\to \text{grams}\]
Complete step by step solution:
> For this we use Faraday's Constant, F. Faraday's Constant is given by the symbol F and is defined as the charge in coulombs (C) of 1 mole of electrons. Faraday's constant is approximately 96485 C mol-1. You can calculate F by multiplying the charge on one electron (1.602 x 10-19) by Avogadro's number (6.022 x 1023),
We have been given in the problem that the:
Mass of \[Mn{{O}_{2}}\] is 26.08g.
Atomic mass of Mn is 55 amu.
Current = 0.5 A
\[M{{n}^{4+}}+4{{e}^{-}}\to Mn\]
We can see from the above that 4 mol of electrons give 1 mol of Magnesium, Mn.
> Now we will put the numbers in. 1 mol of electrons is 1 faraday.
4 x 96500 coulombs give 55 g of copper.
Number of coulombs = \[=\dfrac{26.08\times 2\times 96485}{55}=94502.86\]C
Now we know how many coulombs we need, and the current was already given in the problem. Finally, we can calculate the time required to oxidize \[Mn{{O}_{2}}\].
Number of coulombs = current in amps x time in seconds
\[\begin{align}
& 94502.86=0.5\times \,t \\
& t=18901.72\,\sec \\
\end{align}\]
To convert into hours, we will divide by 60.
\[\dfrac{18901.72}{60}=315\,hours\]
So, the total time taken will be 315 hours.
Note: The same question can be given with time and asking to calculate the amount of \[Mn{{O}_{2}}\] produced after oxidation. In that case the procedure should be:
\[\text{Current}\,\text{and time}\to \text{Charge(C)}\to \text{faradays}\to \text{moles}\to \text{grams}\]
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)