
Silver could be plated out on a serving tray by electrolysis of a solution containing silver in $ + 1 $ oxidation state for a period of $ 8.0 $ hour at a current of $ 8.46 $ ampere. What is the area of the tray if the thickness of the silver plating is $ 0.00254\;{\text{cm}} $ ?
(The density of silver is $ 10.5{\text{g/c}}{{\text{m}}^{\text{3}}} $ )
(A) $ {\text{a}} = 601.95\;{\text{c}}{{\text{m}}^2} $
(B) $ {\text{a}} = 651.95\;{\text{c}}{{\text{m}}^2} $
(C) $ {\text{a}} = 701.95\;{\text{c}}{{\text{m}}^2} $
(D) $ {\text{a}} = 10204.72{\text{c}}{{\text{m}}^2} $
Answer
556.5k+ views
Hint: In this question, we will first calculate the weight of silver that is plated on the tray. After calculating the weight of the silver, we have to calculate the area of the tray. We can do it easily since the thickness of the plating and density of silver is provided to us.
Formula Used
We will use the following formulas to find out the answer to this question:
$ {W_{Ag}} = \dfrac{{E \times i \times t}}{{96500}} $
Where
$ {W_{Ag}} $ is the weight of the silver
$ E $ is the equivalent weight of silver
$ i $ is the current
$ t $ is the time in seconds
And
$ a = \dfrac{v}{T} $
Where
$ a $ is the area of the tray
$ v $ is the volume of tray
$ T $ is the thickness of the plating.
Complete step by step solution
According to the question, the following information is provided to us
Thickness of the silver plating, $ T = 0.00254\;{\text{cm}} $
Time of electrolysis, $ t = 8 {\text{hours}} = 8 \times 60 \times 60 = 28800 \sec $
Current, $ i = 8.46 {\text{A}} $
The density of silver is $ 10.5{\text{g/c}}{{\text{m}}^{\text{3}}} $
The equivalent weight of silver is given by $ \dfrac{{107.8{\text{g}}}}{1} = 107.8{\text{g}} $
$ {W_{Ag}} = \dfrac{{E \times i \times t}}{{96500}} $
Now, let us substitute the values into this equation to get the weight of Silver
$ {W_{Ag}} = \dfrac{{107.8 \times 8.46 \times 28800}}{{96500}} $
Upon solving, we get
$ {W_{Ag}} = 272.18{\text{g}} $
Now, we have to find the area of the tray. So, let us use our formula
$ a = \dfrac{v}{T} $
In the above formula, thickness is known to us. But we need to find out the volume in order to calculate the area.
We know that
$ {\text{volume = }}\dfrac{{{\text{mass}}}}{{{\text{density}}}} $
We have just now calculated the mass of silver. And the density of silver is already provided to us in the question. Now, let us easily substitute to find out the volume
$ {\text{volume}} = \dfrac{{272.18}}{{10.5}} $
Upon solving, we get
$ {\text{volume}} = 25.92{\text{ ml}} $
Now, we will substitute the value of volume to get the area of the tray
$ a = \dfrac{v}{T} $
$ a = \dfrac{{25.92}}{{0.00254}} $
Upon solving, we get
$ \therefore a = {\text{10204}}{\text{.72c}}{{\text{m}}^{\text{2}}} = 1.02 \times {10^4}{\text{c}}{{\text{m}}^{\text{2}}} $
Hence, the correct option is (D.)
Note
Electrolysis is defined by passing a direct electric current through the compound in a fluid form as a process of decomposing ionic compounds into their elements. At the cathode, the cations are decreased and anions at the anode are oxidized.
Formula Used
We will use the following formulas to find out the answer to this question:
$ {W_{Ag}} = \dfrac{{E \times i \times t}}{{96500}} $
Where
$ {W_{Ag}} $ is the weight of the silver
$ E $ is the equivalent weight of silver
$ i $ is the current
$ t $ is the time in seconds
And
$ a = \dfrac{v}{T} $
Where
$ a $ is the area of the tray
$ v $ is the volume of tray
$ T $ is the thickness of the plating.
Complete step by step solution
According to the question, the following information is provided to us
Thickness of the silver plating, $ T = 0.00254\;{\text{cm}} $
Time of electrolysis, $ t = 8 {\text{hours}} = 8 \times 60 \times 60 = 28800 \sec $
Current, $ i = 8.46 {\text{A}} $
The density of silver is $ 10.5{\text{g/c}}{{\text{m}}^{\text{3}}} $
The equivalent weight of silver is given by $ \dfrac{{107.8{\text{g}}}}{1} = 107.8{\text{g}} $
$ {W_{Ag}} = \dfrac{{E \times i \times t}}{{96500}} $
Now, let us substitute the values into this equation to get the weight of Silver
$ {W_{Ag}} = \dfrac{{107.8 \times 8.46 \times 28800}}{{96500}} $
Upon solving, we get
$ {W_{Ag}} = 272.18{\text{g}} $
Now, we have to find the area of the tray. So, let us use our formula
$ a = \dfrac{v}{T} $
In the above formula, thickness is known to us. But we need to find out the volume in order to calculate the area.
We know that
$ {\text{volume = }}\dfrac{{{\text{mass}}}}{{{\text{density}}}} $
We have just now calculated the mass of silver. And the density of silver is already provided to us in the question. Now, let us easily substitute to find out the volume
$ {\text{volume}} = \dfrac{{272.18}}{{10.5}} $
Upon solving, we get
$ {\text{volume}} = 25.92{\text{ ml}} $
Now, we will substitute the value of volume to get the area of the tray
$ a = \dfrac{v}{T} $
$ a = \dfrac{{25.92}}{{0.00254}} $
Upon solving, we get
$ \therefore a = {\text{10204}}{\text{.72c}}{{\text{m}}^{\text{2}}} = 1.02 \times {10^4}{\text{c}}{{\text{m}}^{\text{2}}} $
Hence, the correct option is (D.)
Note
Electrolysis is defined by passing a direct electric current through the compound in a fluid form as a process of decomposing ionic compounds into their elements. At the cathode, the cations are decreased and anions at the anode are oxidized.
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