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Calculate the minimum number of kilowatt-hours of electricity required to produce 100 kg of Al by electrolysis of Al^(3+) if the required emf is 4.50 V. |
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Answer» SOLUTION :The no. of FARADAY required for electrolysis = number of eq. of AL deposited `= (10^(6))/(27//3)F = (10^(6))/(9)F`. Charge `= (10^(6))/(9) xx 96500` coulombs. `= 1.07 xx 10^(10)C`. `therefore` electric energy `= 1.07 xx 10^(10) xx 4.50 J = 4.815 xx 10^(10)J ""(because J = C xx V)` kilowatt-hours `= (4.815 xx 10^(10))/(3.6 xx 10^(6))kWh ""(because 1 kWh = 3.6 xx 10^(6)J)` `= 1.34 xx 10^(4) kWh`. |
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