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How much electricity in terms of Faraday is required to produce: (i) 20.0 g of Ca from molten CaCl_(2) ? (ii) 40.0 g of Al from molten Al_(2)O_(3) ? |
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Answer» Solution :(i) Required electricity in FARADAY to obtianed 20.0 g calcium from the MOLTEN `CaCl_(2)` Ionization reaction: `CaCl_(2) to Ca^(2+) +2Cl^(-)` Reduction reaction on cathode : `Ca^(2+)+2e^(-) to Ca` `therefore 2" mol "e^(-) to 1" mol "Ca` `therefore 2F to 40g` Ca `therefore`Faraday required to produce 20 g Ca `=(20gxx2F)/(40g)=1F` (ii) Required electricity in Faraday to obtained 40.0 g ALUMINIUM from the molten `Al_(2)O_(3)`. The following reaction occurs on cathode of molten `Al_(2)O_(3)`. `Al_(2)O_(3(l))to 2Al^(3+)+3O^(2-)` and reduction `Al^(3+)+3e^(-) to Al` So, we get 1 mol Al from 3 mole `e^(-)` `therefore` We get 27.0 g Al from 3F electricity. So, the electricity required to get 40.0 g Al `=(3Fxx40.0g)/(27.0g)=4.444F` |
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