How much charge in coulombs is required for the reduction of one mole of $\mathrm{Al}^{3+}$ to Al?

How much charge in coulombs is required for the reduction of one mole of $\mathrm{Al}^{3+}$ to Al?
  1. $1.930 \times 10^{4} \mathrm{C}$
  2. $2 \cdot 895 \times 10^{5} \mathrm{C}$
  3. $2 \cdot 895 \times 10^{4} \mathrm{C}$
  4. $1.930 \times 10^{5} \mathrm{C}$

Solution

$\mathrm{Al}^{3 *}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}$ 1 mole of electron $=1 \mathrm{~F}=96500$ coulombs $\therefore 3 \mathrm{e}^{-}=3 \mathrm{~F}=3 \times 96500=2,89,500=2.895 \times 10^{5}$ coulombs

Asked in: MHT CET 2020 (13 Oct Shift 2)

Practice more Electrochemistry questions on Aicharya