Question
How much charge is required for the following reductions:
1 mol of Al3+ to Al?
Solution
Formula required charge n × F
n = difference of charge on ions
F is constant and equal to 96487 Coulombs
Here n = 3
Hence required charge = 3 × 96487 Coulombs
= 289461 Coulombs
= 2.89 ×10 –5 Coulombs
