Question

The amount of electricity required to deposit 0.9gof aluminum the electrode reaction is
$A{l}^{3+}+3e^{-}\to Al$
(atomic mass of Al = 27)

• A. $9.65\times {10}^{3}C$
• B. $1.93\times {10}^{4}C$
• C. $9.65\times {10}^{4}C$
• D. $4.34\times {10}^{5}C$

Answer 1

The correct option is A $9.65\times {10}^{3}C$

$A{l}^{3+}+3e^{-}\to Al$
27g of Al mequire = 3 $\times$ 96500C
$\therefore$ 0.9g of Al requires $=\frac{3\times 96500}{27}\times 0.9$
= 9650C.