Calculate the no. Of coulpmb required to deposit 40.5g AL when the electrode rxn is AL3+ +3e--->AL
Asked by Harshfarwaha | 23rd Jul, 2020, 03:27: PM
The reaction is:
Al3+ + 3e- --------------> Al
1mole 3 mole 1 mole
1 mole 3 Faradays 1 mole
1 mole of Al = 27 g is deposited by 3 faraday of electricity
Answered by Ramandeep | 23rd Jul, 2020, 06:08: PM
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