Q21.105P - Chemistry: Molecular Nature Of Matter And Change

Question

How many seconds does it take to deposit $\mathbf{65}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{g}$  of $\mathbf{Zn}$  on a steel gate when $\mathbf{21}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{A}$  is passed through a  ${\mathbf{ZnSO}}_{\mathbf{4}}$ solution?

Step-by-Step Solution

Answer

Step 1: Definition

Step 2: Calculating the time

A current of $21.0 \mathrm{A}( \mathrm{A}=\mathrm{C}/\mathrm{s})$ is passed through a  ${\mathrm{ZnSO}}_4$ solution.

We need to find the time (in seconds) it takes to deposit  $65.5 \mathrm{g}$ of  $\mathrm{Zn}$.

${\mathrm{ZnSO}}_4\to {\mathrm{Zn}}^{2+}{+\mathrm{SO}}_4^{2-}$

The half-reaction for  ${\mathrm{Zn}}^{2+}$ reduction is

${\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Zn}$

- Here we can see that for every mole of  $\mathrm{Zn}$ deposited, we need  $2$ moles of electrons.

The number of moles of $\mathrm{Zn}$ deposited is (the molar mass of $\mathrm{Zn}$ is $65.38 \mathrm{g}/\mathrm{mol}$ )

$\mathrm{Moles} \mathrm{Zn} \mathrm{deposited}=\frac{65.5 \mathrm{g}}{65.38}=1.002 \mathrm{mol} \mathrm{Zn}$

Hence, the number of moles of electrons is

$\mathrm{Moles} {\mathrm{e}}^{-}=1.002 \mathrm{mol} \mathrm{Zn}\times \frac{2 {\mathrm{mole}}^{-}}{1 \mathrm{mol} \mathrm{Zn}}=2.004 {\mathrm{mole}}^{-}$

Step 3: Calculating deposit time

- Faraday constant: charge of a mole of electrons  $\left(\mathrm{F}=96485\mathrm{C}/{\mathrm{mol}}^{-}\right)$

Now, we will calculate the charge, using the Faraday constant

Charge =  Moles ${\mathrm{e}}^{-}\times \mathrm{F}$

$=2.004 {\mathrm{mol}}^{-}\times 96485\frac{\mathrm{C}}{{\mathrm{mol}}^{-}}=193355.94\mathrm{C}$