Question

What Is The Time Required To Deposit One Millimole Of Aluminium Metal By The Passage Of $9.65$ Amperes Through Aqueous Solution Of Aluminium Ion?

Answer 1

Step 1 - Chemical Reaction

• The reaction involved in the conversion of Aluminium metal $\left(\mathrm{Al}\right)$ into Aluminium ion $\left({\mathrm{Al}}^{3+}\right)$ can be written as,

${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Al}$

• Hence, $1$ mole of ${\mathrm{Al}}^{3+}$ requires $3$ moles of electron.

Step 2 - Charge transferred

• Now the formula for Electric charge i.e. $Q=I\times t\dots \dots \dots \dots \left(1\right)$

Here, $Q$ = Electric Charge in Coloumb.

$I$ = Electric Current in Ampere = $9.65A$.

$t$ = Time of flow of current in seconds.

• The amount of Aluminium formed is given by, $x=\frac{Q}{n\times F}\dots \dots \dots \dots \left(2\right)$

Here, $F$ = Faraday;s Constant = $96500C/mol$.

$n$= Number of moles of electron = $3$

$x$ = $3$ millimoles = $3\times {10}^{-3}$ moles.

• Putting $\left(1\right)$ in $\left(2\right)$ we get,

$x=\frac{I\times t}{n\times F}$

Step 3 - Time Required

• $t=\frac{n\times F}{I}$

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

$t=30seconds$

Hence, the time required will be $30s$.