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Question: Cadmium amalgam is prepared by electrolysis of a solution of \({ CdCl }_{ 2 }\) using a mercury cath...

Cadmium amalgam is prepared by electrolysis of a solution of CdCl2{ CdCl }_{ 2 } using a mercury cathode. Find how long current of 5amperes{ 5 amperes } should be passed in order to prepare 12{ 12 } % Cd-Hg amalgam on a cathode of 2g mercury. The atomic mass of Cd = 112.40{ 112.40 }.

Explanation

Solution

Electrolysis refers to a process in which a chemical change is brought about by passing an electric current through a solution of electrolytes so that the electrolyte's ions move toward the negative and positive electrodes and react with them.

Complete step by step answer:
Amount of Cd required by 2g{ 2g } Hg to prepare 12{ 12 } % amalgam =1288×2\frac { 12 }{ 88 } { \times 2 } = 0.273g{ 0.273 g }
The half-reaction of the cathode will be:
Cd2++2eCd{ Cd }^{ 2+ }{ +2e }^{ - }{ \rightarrow Cd }
It is given that;
The molar mass of Cd = 112.40{ 112.40 }
As 1mole of an electron gives 96500C{ 96500C } charge
So, 2 moles of an electron gives 2×96500C{ 2\times 96500C } charge

Now, to deposit 1mole{ 1 mole } or 112.4{ 112.4 } g of Cd we require 2moles{ 2 moles } of electrons = 2×96500C{ 2\times 96500C } charge area will be required.
To deposit 1g{ 1g } of Cd we require = 2×96500C×112.40{ 2\times 96500C\times 112.40 } of charge area will be required
Charge required to deposit 0.273g{ 0.273 g } of Cd = 2×96500112.40×0.273{ \frac { 2\times 96500 }{ 112.40 } { \times 0.273 } } of charge area will be required.
Therefore, Charge = 468.76C{ 468.76C }
As we know, Charge = current×time{ current\times time }
By putting the values in a formula, we can calculate the time (s).
Time = chargecurrent\frac { charge }{ current }
Time = 468.76×5A{ 468.76\times 5A } = 93.75seconds{ 93.75 seconds }.

Therefore, the time required will be = 93.75seconds{ 93.75seconds }.

Note: The possibility to make a mistake is that here in this reaction 2moles{ 2 moles } of electrons are required, so don’t forget to multiply Faraday's constant with two.