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Question: An alloy of lead (valency\( = 2\))-thallium (valency\( = 1\)) containing \(70\% \) Pb and \(30\% \) ...

An alloy of lead (valency=2 = 2)-thallium (valency=1 = 1) containing 70%70\% Pb and 30%30\% Tl, can be electroplated onto a cathode from a perchloric acid solution. How many hours would be required to deposit 5.0 g5.0{\text{ }}g of this alloy at a current of 1.10 A1.10{\text{ }}A ? (Pb=208 = 208, Tl=204 = 204)

Explanation

Solution

First of all we will find out the amount of particular metal being deposited. Then we will find the number of equivalents. After that by using Faraday’s Law we will eventually find out the time required for deposition of the alloy.

Complete step by step answer:
In the given question it is mentioned that the alloy of Lead (Pb)-Thallium (Tl) contain 70%70\% Pb and 30%30\% Tl. We have to find out the time required to deposit a total of 5.0 g5.0{\text{ }}g of this alloy when 1.10 A1.10{\text{ }}A current flows through it.Firstly, we have to separately find the amount of particular metal being electroplated.

Amount of Lead (Pb) electroplated =5×70100=3.5 g = 5 \times \dfrac{{70}}{{100}} = 3.5{\text{ }}g
Then, Amount of Thallium (Tl) electroplated=5×30100=1.5 g = 5 \times \dfrac{{30}}{{100}} = 1.5{\text{ }}g
Secondly, we have to find the number of equivalents of metal.
Number of equivalent = weight of the metal (in grams)equivalent weight\dfrac{{{\text{weight of the metal (in grams)}}}}{{{\text{equivalent weight}}}}
Equivalent weight=Molecular weightValency = \dfrac{{{\text{Molecular weight}}}}{{{\text{Valency}}}}
Thus, Number of equivalent of Pb
 Number of equivalent of Pb=weight of Pb (in grams)equivalent weight ⇒ Number of equivalent of Pb=3.52082 ⇒ Number of equivalent of Pb=3.5104 ⇒ Number of equivalent of Pb=0.0336\text{ Number of equivalent of Pb}= \dfrac{{{\text{weight of Pb (in grams)}}}}{{{\text{equivalent weight}}}} \\\ \Rightarrow \text{ Number of equivalent of Pb} = \dfrac{{3.5}}{{\dfrac{{208}}{2}}} \\\ \Rightarrow \text{ Number of equivalent of Pb} = \dfrac{{3.5}}{{104}} \\\ \Rightarrow \text{ Number of equivalent of Pb}= 0.0336

Now, we have to find the number of equivalent of Tl
Number of equivalent of Tl=weight of Tl (in grams)equivalent weight ⇒Number of equivalent of Tl=1.52041 ⇒Number of equivalent of Tl=1.5204 ⇒Number of equivalent of Tl=0.0073 \text{Number of equivalent of Tl} = \dfrac{{{\text{weight of Tl (in grams)}}}}{{{\text{equivalent weight}}}} \\\ \Rightarrow \text{Number of equivalent of Tl}= \dfrac{{1.5}}{{\dfrac{{204}}{1}}} \\\ \Rightarrow \text{Number of equivalent of Tl}= \dfrac{{1.5}}{{204}} \\\ \Rightarrow \text{Number of equivalent of Tl} = 0.0073 \\\
Thirdly, from Faraday’s law we can find out the time required for the electroplating,
Total number of equivalents==number of faradays
⇒0.0336+0.0073=I×t96500 ⇒0.0409=1.1×t96500 ⇒0.0409×965001.1=t ⇒3588.04 s=t ∴t=59.80 min⁡ \Rightarrow 0.0336 + 0.0073 = \dfrac{{I \times t}}{{96500}} \\\ \Rightarrow 0.0409 = \dfrac{{1.1 \times t}}{{96500}} \\\ \Rightarrow \dfrac{{0.0409 \times 96500}}{{1.1}} = t \\\ \Rightarrow 3588.04{\text{ }}s = t \\\ \therefore t = 59.80{\text{ }}\min \\\
Thus, we have found out that 59.80 min⁡59.80{\text{ }}\min is required to deposit 5 g5{\text{ }}g of Pb-Tl alloy by passing 1.10 A1.10{\text{ }}A of current.

Note: From Faraday’s Law of Electrolysis we have to use a number of Faradays in order to solve the respective solution. There is a possibility to skip equivalent weight of metal which is the prime key to solve the problem. The equivalent weight should be found in terms of molecular weight and valency as done in the question.