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?

Short answer

Step 1: Definition

Step-by-step solution

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

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

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 ${\mathrm{e}}^{-}\times \mathrm{F}$

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