Electrical Engineering ⇒ Topic : Electrolysis
|
Gaurav
| |
Electrolysis The condution of electric current through the solution of an electrolyte together with the resulting chemical changes is called electrolysis. Fig. (a) shows the process of electrolysis in a copper voltameter.When copper sulphate (CuSO4) is dissolved in water, it splits up into its components viz, the positive copper ions (Cu++) and negative sulphate ions (S04--). This process is called **ionisation. When d.c. voltage is applied across the electrodes, the negative sulphate ions (S04--) move towards the anode (+ve electrode) and positive copper ions move towards the cathode (-ve electrode) causing the following chemical changes :
Figure (a) At anode.A sulphate ion (SO4- -) on reaching the anode gives its two extra electrons to it and becomes sulphate radical. These given up electrons continue their journey towards the cathode via the external circuit. Now the sulphate radical cannot exist and, therefore, it acts chemically on the anode material to form copper sulphate according to the following reaction Cu + SO4 → CuSO4 Thus copper from anode continuously dissolves into the solution so long as this action takes place. At cathode. At the same time, a copper ion (Cu++) on reaching the cathode takes two electrons from it (these are the same electrons given by the sulphate ion at the anode and have come to cathode. Cu++ + 2e → Cu atom Thus copper from the solution (CuSO4) gets deposited on the cathode. The following points may be carefully noted :
| |
| |
Gaurav
| |
Electrolysis The condution of electric current through the solution of an electrolyte together with the resulting chemical changes is called electrolysis. Fig. (a) shows the process of electrolysis in a copper voltameter.When copper sulphate (CuSO4) is dissolved in water, it splits up into its components viz, the positive copper ions (Cu++) and negative sulphate ions (S04--). This process is called **ionisation. When d.c. voltage is applied across the electrodes, the negative sulphate ions (S04--) move towards the anode (+ve electrode) and positive copper ions move towards the cathode (-ve electrode) causing the following chemical changes :
Figure (a) At anode.A sulphate ion (SO4- -) on reaching the anode gives its two extra electrons to it and becomes sulphate radical. These given up electrons continue their journey towards the cathode via the external circuit. Now the sulphate radical cannot exist and, therefore, it acts chemically on the anode material to form copper sulphate according to the following reaction Cu + SO4 → CuSO4 Thus copper from anode continuously dissolves into the solution so long as this action takes place. At cathode. At the same time, a copper ion (Cu++) on reaching the cathode takes two electrons from it (these are the same electrons given by the sulphate ion at the anode and have come to cathode. Cu++ + 2e → Cu atom Thus copper from the solution (CuSO4) gets deposited on the cathode. The following points may be carefully noted :
| |
| |
!! OOPS Login [Click here] is required for more results / answer