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Example 9. In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold salt and the second cell contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of cop- per deposited on the cathode in the second cell. Also, calculate the magnitude of the current in ampere. Solution: We know that,

Example 9. In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold salt and the second cell contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of cop- per deposited on the cathode in the second cell. Also, calculate the magnitude of the current in ampere. Solution: We know that,

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Chapter18: Electrochemistry
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Problem 115AE: The saturated calomel electrode. abbreviated SCE. is often used as a reference electrode in making...
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Kll (Part-I) s passed Calculate NTP. passing IL ine) on- the Electrochemistry or Example 9. In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold salt and the second cell contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of cop- per deposited on the cathode in the second cell. Also, calculate the magnitude of the current in ampere. Solution: We know that, Mass of Au deposited Eq. mass of Au Mass of Cu deposited Eq. mass of Cu = Eq. mass of Au=- ; Eq. mass of Cu = 63.5 197 3 2 Mass of copper deposited 63.5 3 = 9.85 x g= 4.7625 g 197 Let Z be the electrochemical equivalent of Cu. E=ZX 96500 or Z= E 63.5 96500 2 × 96500 Applying W = Z xixt t = 5 hours = 5 x 3600 second 4.7625= 63.5 -xix5x3600 2 × 96500 or i= 4.7625 × 2 × 96500 63.5x5x3600 = 0.804 ampere Example 10. How long has a current of 3 ampere to be ap- plied through a solution of silver nitrate to coat a metal surface of 80 cm with 0.005 mm thick layer? Density of silver is 10.5 g cm -3 0.42 × 96500 108 x 3 Example 11. What current quired to liberate 10 g of chlorine one hour? Solution: Applying E=Z= 35.5=Z- or Z== Now, applying the formula W=Z: where, W = 10 g. 2= 35 96 10: i= 35. Example 12. 0.2964 g of a current of 0.5 ampere for 30 per sulphate. Calculate the an (1 faraday=96500 coulomb) Solution: Quantity of ch =0.5x30x 900 coulomb deposit copp 96500 coulomb deposit c Thus, 31.78 is the equiva At. mass= Example 13. Calculate be required to reduce 12.3 rent efficiency for the proce
Transcribed Image Text:Kll (Part-I) s passed Calculate NTP. passing IL ine) on- the Electrochemistry or Example 9. In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold salt and the second cell contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of cop- per deposited on the cathode in the second cell. Also, calculate the magnitude of the current in ampere. Solution: We know that, Mass of Au deposited Eq. mass of Au Mass of Cu deposited Eq. mass of Cu = Eq. mass of Au=- ; Eq. mass of Cu = 63.5 197 3 2 Mass of copper deposited 63.5 3 = 9.85 x g= 4.7625 g 197 Let Z be the electrochemical equivalent of Cu. E=ZX 96500 or Z= E 63.5 96500 2 × 96500 Applying W = Z xixt t = 5 hours = 5 x 3600 second 4.7625= 63.5 -xix5x3600 2 × 96500 or i= 4.7625 × 2 × 96500 63.5x5x3600 = 0.804 ampere Example 10. How long has a current of 3 ampere to be ap- plied through a solution of silver nitrate to coat a metal surface of 80 cm with 0.005 mm thick layer? Density of silver is 10.5 g cm -3 0.42 × 96500 108 x 3 Example 11. What current quired to liberate 10 g of chlorine one hour? Solution: Applying E=Z= 35.5=Z- or Z== Now, applying the formula W=Z: where, W = 10 g. 2= 35 96 10: i= 35. Example 12. 0.2964 g of a current of 0.5 ampere for 30 per sulphate. Calculate the an (1 faraday=96500 coulomb) Solution: Quantity of ch =0.5x30x 900 coulomb deposit copp 96500 coulomb deposit c Thus, 31.78 is the equiva At. mass= Example 13. Calculate be required to reduce 12.3 rent efficiency for the proce
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