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How much electricity in terms of faraday is required to produce (i) 20.0 g of Ca from CaCl_(2) (ii) 40.0 g of Al from molten Al_(2)O_(3)? |
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Answer» SOLUTION :(i) `Cu^(2+)+2e^(-)TOCA` Thus, 1 mol of Ca, i.e., 40 g of Ca require electricity=2F`therefore20` g of Ca will require electricity=1F (ii) `AL^(3+)+3e^(-)toAl` . Thus 1 mol of Al, i.e., 27G of Al required electricity=3F `therefore40g` of Al will require electricity=`(3)/(27)xx40=4.44E` |
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