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| Numerical 05 |
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| 05. In a silver-plating system, an electrolysis current of 5.0 A is used for a certain time and 0.5 moles of silver is deposited. How many moles of copper and iron will be deposited in their respective plating system if an electrolysis current of 10.0 A is passed for twice the time for silver plating? |
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| Relative atomic mass of silver = 107.3 |
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| copper = 63.54 |
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| iron = 55.85 |
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| Suggested solution: |
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| The valency of silver is one. So, silver is monoatomic. Thus, the chemical equivalent of silver is the same as the molar mass of silver. |
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| Since charge required to liberate one chemical equivalent of silver is 96500 C, therefore, the charge q required to liberate 0.5 chemical equivalent of silver is 0.5 x 96500 C, i.e., 48250 C. |
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| Let us now calculate the moles of copper deposited. |
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| I= 10A; t=2 x 9650s. Atomic mass of copper = 63.54 |
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| Since the valency of copper is 2, |
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| Clearly, one mole of copper is liberated and hence deposited. |
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| Now to calculate the moles of iron deposited, |
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| Alternative Method |
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| Double the current for double the time will deposit 0.5 x 2 x 2 mole i.e., 2 mole of silver. |
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