Answer:
(i)
For reduction of 1 mole of charge
needed = 3F
For reduction of 1 mole of Al3+,
electricity needed = 3 × 96500 C = 289500 C
(ii)
For reduction o 1 mole of Cu2+,
charge needed = 2F
For reduction of 1 mole of Cu2+,
electricity needed = 2 × 96500 C = 193000 C
(iii)
For reduction of 1 mole of electricity
needed = 5 F.
For reduction of 1 mole of electricity
needed = 5 × 96500 C = 482500 C
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