91Ó°ÊÓ

It's an electrochemical process that involves passing current between two electrodes through an ionized solution (the electrolyte) to deposit positive ions (anions) on the negative electrode (cathode) and negative ions (cations) on the positive electrode (cathode) (anode).

A current of $13.7\mathrm{A}(\mathrm{A}=\mathrm{C}/\mathrm{s})$is passed through a $\mathrm{Ni}{\left({\mathrm{NO}}_3\right)}_2$solution.

We need to find the time (in seconds) it takes to deposit $1.63\mathrm{g}$of $\mathrm{Ni}$.

$\mathrm{Ni}{\left({\mathrm{NO}}_3\right)}_2→{\mathrm{Ni}}^{2+}+2{\mathrm{NO}}_3-$

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

${\mathrm{Ni}}^{2+}+2{\mathrm{e}}^{-}→\mathrm{Ni}$

- Here we can see that for every mole of Ni deposited, we need $2$moles of electrons.

The number of moles of Ni deposited is (the molar mass of Ni is $58.6934\mathrm{g}/\mathrm{mol}$)

$\mathrm{Moles}\mathrm{Ni}\mathrm{deposited}=\frac{1.63\mathrm{g}}{58.6934\mathrm{g}/\mathrm{mol}}=2.777×{10}^{-2}\mathrm{mol}\mathrm{Ni}$

Hence, the number of moles of electrons is

${\mathrm{Molese}}^{-}=2.777×{10}^{-2}\mathrm{molNi}×\frac{2{\mathrm{mole}}^{-}}{1\mathrm{molNi}}=5.554×{10}^{-2}{\mathrm{mole}}^{-}$

- Faraday constant: charge of $1$mole of $\left(\mathrm{F}=96485\mathrm{C}/{\mathrm{mole}}^{-}\right)$electrons

Now, we will calculate the charge, using Faraday constant

$\mathrm{Charge}={\mathrm{Molese}}^{-}×\mathrm{F}=5.554×{10}^{-2}{\mathrm{mole}}^{-}×96485\frac{\mathrm{C}}{{\mathrm{mole}}^{-}}=5358.78\mathrm{C}$