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Question: \(Zn\) Amalgam is prepared by the electrolysis of aqueous \(ZnC{{l}_{2}}\) using \(Hg\) cathode (9 g...

ZnZn Amalgam is prepared by the electrolysis of aqueous ZnCl2ZnC{{l}_{2}} using HgHg cathode (9 gm). How much current should be passed through ZnCl2ZnC{{l}_{2}} solution for 1000 seconds to prepare a ZnZn amalgam with 25%25\% Zinc by weight ? (ZnZn atomic weight = 65.465.4)
A. 5.6A
B. 7.2A
C. 8.85 A
D. 11.2A

Explanation

Solution

The concept of Faraday’s Law of electrolysis is to be used in this question. According to the law, more the current or more the time used in electrolysis, more is the mass of substance deposited at the cathode

Complete step by step solution:
In order to answer the question, we need to learn about the Faradays Law of Electrolysis. In order to answer our question, we need to know about the Faradays Law of electrolysis. The passage of electricity through the electrolytes in their dissolved or molten state can cause chemical changes under some conditions. For example, the passing electricity through the acidified water can result in the formation of oxygen and hydrogen gases. The process of chemical decomposition of the electrolyte by the passing electricity through its dissolved or molten state is called electrolysis.
The relationship between the quantity of electricity passed and the quantity of a substance liberated at the electrode is given in the form of Faraday's laws of electrolysis. Faraday's First Law of Electrolysis. This law states that the mass of a substance liberated at the electrode is directly proportional to the quantity of electricity passed through the electrolyte.. In order to answer our question, let x be the mass of the amalgam. So, the mass of mercury is 75100x=9x=12g\dfrac{75}{100}x=9\Rightarrow x=12g
and the mass of zinc is 25100×12=3g\dfrac{25}{100}\times 12=3g.
Now, the chemical equivalent E is zinc is E=molarmassnfactorE=\dfrac{molar\,mass}{n-factor}, and also, mass=EITFmass=\dfrac{EIT}{F}.
On substituting, we obtain:
I=665.4×965001000=885100=8.85AI=\dfrac{6}{65.4}\times \dfrac{96500}{1000}=\dfrac{885}{100}=8.85A

Hence we obtain the correct answer as option C.

Note: Electrochemical equivalent of a substance is the amount of substance liberated at the electrode when current of one ampere is passed through the electrolyte for one second.