Question

Time required to deposit one millimole of aluminium metal by the passage of $9.65$ amperes through aqueous solution of aluminium ion is :

• A. $30s$
• B. $10s$
• C. $30,000s$
• D. $10,000s$

Answer 1

The correct option is A $30s$

$A{l}^{3+}+3e^{-}\to Al$

$1mole$ requires $=3\times 96500coulomb$

$1millimole\left({10}^{-3}mole\right)requires=3\times 96500\times {10}^{-3}C$

We know that, $Q=it$
or, $t=\frac{Q}{i}$

$=\frac{3\times 96500\times {10}^{-3}}{9.65}$

$=\frac{3\times 9.65\times 10}{9.65}=30sec$.