Calculate the number of electrons required to deposit 40.5 gram of aluminium
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Answered by
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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
40.5 g of Al is deposited by = (3/27) *40.5 F
=>3/27 *40.5 *96500 Coulombs
=>434250 Coulombs
I HOPE IT HELPS YOU :)
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
40.5 g of Al is deposited by = (3/27) *40.5 F
=>3/27 *40.5 *96500 Coulombs
=>434250 Coulombs
I HOPE IT HELPS YOU :)
Answered by
4
hello please like my answer
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