On passing 1.5 F Charge, the
1.5 F Charge, the number of moles
of alumi'um deposited at Cathode
at cathode are (Al-27)
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Answer:
The balanced chemical equation for the deposition of aluminum is shown below.
Al
3+
+3e
−
→Al
One mole of Al
3+
ions requires 3 moles of electrons.
On passing 3F charge (3 moles of electrons), the number of moles of aluminium deposited at the cathode is 1.
On passing 1.5F charge, the number of moles of aluminium deposited at cathode are 1×
3
1.5
=0.5
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