What is the number of faraday required to produce $\cdot 0.18 \mathrm{~g}$ aluminium at cathode during…
What is the number of faraday required to produce $\cdot 0.18 \mathrm{~g}$ aluminium at cathode during electrolysis of molten $\mathrm{AlCl}_3$ ?
(Molar mass of $\mathrm{Al}=27 \mathrm{~g} \mathrm{~mol}^{-1}$ )
0.02 F
0.03 F
0.25 F
0.30 F
Solution
$\mathrm{Al}^{3+}+3 \mathrm{e}^{+} \longrightarrow \mathrm{Al}_{(\mathrm{s})}$ 1 mol of $\mathrm{Al}(27 \mathrm{~g})$ requires 3 F (3 moles of electrons)
$\therefore \quad 0.18 \mathrm{~g} \text { of Al requires }=\frac{3 \times 0.18}{27}=0.02 \mathrm{~F}$