Question

Electroplating efficiency. ${}^{\mathbf{56}}$Nickel was electrolytically plated onto a carbon electrode from a bath containing$290\mathrm{g}/{\mathrm{LNiSO}}_4.6{\mathrm{H}}_2\mathrm{O},30\mathrm{g}/\mathrm{LB}{\left(\mathrm{OH}\right)}_3$androle="math" localid="1654763379590" $8\mathrm{g}/\mathrm{LNaCl}\mathrm{at}-1.2\mathrm{V}$vs$\mathrm{Ag}\left|\begin{array}{c}\mathrm{AgCl}\end{array}\right|$The most important side reaction is reduction of${\mathrm{H}}^{+}{\mathrm{toH}}_2$.In one experiment, a carbon electrode weighing$0.4775\mathrm{g}$before deposition weighedrole="math" localid="1654763622546" $0.4798\mathrm{g}\mathrm{after}8.82\mathrm{C}$had passed through the circuit. What percentage of the current went into the reaction${\mathrm{Ni}}^2+2{\mathrm{e}}^{-}\to \mathrm{Ni}\left(\mathrm{s}\right)?$

Short answer

The percentage of the current that went into the reaction is$w=93.57\%$

Step-by-step solution

Electroplating efficiency:

Current efficiency may be used to calculate the amount of metal deposited at the cathode or dissolved at the anode under conventional electroplating circumstances. Cathode efficiency is current efficiency applied to the cathode reaction, whereas anode efficiency is current efficiency applied to the anode reaction.

Calculate how much Ni is deposited:

First we need to calculate how much is deposited:

$m\left(Ni\right)=(0.4798-0.4775)g=2.3.{10}^{-3}g$

Then we calculate of Ni:

$$\begin{gathered} \mathrm{n}\left(\mathrm{Ni}\right)=\frac{\mathrm{m}\left(\mathrm{Ni}\right)}{\mathrm{M}\left(\mathrm{Ni}\right)}=\frac{2.3.{10}^{-3}\mathrm{g}}{58.69\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}\left(\mathrm{Ni}\right)=3.92.{10}^{-5}\mathrm{mol} \end{gathered}$$

Find he percentage of the current that went into the reaction is:

Now we calculate charge of Ni:

$$\begin{gathered} \mathrm{q}\left(\mathrm{NI}\right)=\mathrm{n}\left(\mathrm{Ni}\right).\mathrm{F}.\mathrm{Z} \\ \mathrm{q}\left(\mathrm{Ni}\right)=3.92.{10}^{-5}\mathrm{mol}.96485\mathrm{C}/\mathrm{mol}.2 \\ \mathrm{q}\left(\mathrm{Ni}\right)=7.562\mathrm{C} \end{gathered}$$

The percentage of the current that went into the reaction is:

$$\begin{gathered} \mathrm{w}=\frac{7.562\mathrm{C}}{8.082\mathrm{C}}.100\% \\ \mathrm{W}=93.57\% \end{gathered}$$