91Ó°ÊÓ

• Electrolysis is a fundamental process in chemistry that involves the breakdown of an electrolyte (a solution) and the creation of positive and negative ions.
• This is the underlying concept of electrolytic cells.
• The procedure may appear hard, yet it is as simple as a walk in the park.

a)

$$\begin{gathered} \mathrm{Given}\mathrm{data}:\mathrm{E}=1.02\mathrm{V} \\ \mathrm{V}=1\mathrm{mL}\mathrm{ÒÏ}=13.53\mathrm{g}/\mathrm{mL}\mathrm{W}=? \end{gathered}$$

First, we calculate mass of Hg:

$\begin{array}{c}\mathrm{ÒÏ}=\frac{m}{V}\\ m=\mathrm{ÒÏ}â‹\dots V=13.53\mathrm{g}/\mathrm{mL}â‹\dots 1\mathrm{mL}\\ \mathrm{m}=13.53\mathrm{g}\end{array}$

Now we calculate moles of Hg:

$\begin{array}{c}n\left(\mathrm{Hg}\right)=\frac{\mathrm{m}}{\mathrm{M}}=\frac{13.53\mathrm{g}}{200.59\mathrm{g}/\mathrm{mol}}\\ \mathrm{n}\left(\mathrm{Hg}\right)=0.06745\mathrm{mol}\end{array}$

For every mole of Hg that is produced, one mole of electrons flows. So, we can write:

$n\left(Hg\right)=n\left(e^{-}\right)=0.06745mol$

So, we first calculate charge

$\begin{array}{c}\mathrm{q}=\mathrm{n}\left({\mathrm{e}}^{−}\right)â‹\dots \mathrm{F}=0.06745\mathrm{mol}×96485\mathrm{C}/\mathrm{mol}\\ \mathrm{q}=6507.9\mathrm{C}\end{array}$

And the work is:

data-custom-editor="chemistry" $\begin{array}{c}\mathrm{W}=\mathrm{q}â‹\dots \mathrm{E}=6507.9\mathrm{C}×1.02\mathrm{V}\\ \mathrm{W}=6638.1\mathrm{J}\end{array}$

b)

We have power that is 0.209 J/min, in seconds is:

$\begin{array}{c}P=0.209\mathrm{J}/\min ×\frac{1}{60}\min /\mathrm{s}\\ P=3.48×{10}^{−3}\mathrm{J}/\mathrm{s}\end{array}$

First we calculate current using the formula:

$\begin{array}{r}P=I^2â‹\dots \mathrm{R}\\ I=\sqrt{\frac{P}{R}}=\sqrt{\frac{3.48×{10}^{−3}\mathrm{J}/\mathrm{s}}{100\mathrm{Î\copyright }}}\\ I=5.9×{10}^{−3}\mathrm{A}\end{array}$

So, in each hour the charge is:

$Iâ‹\dots 1hour=5.9×{10}^{-3}C/s×3600s=21.24C$

In moles that is

$n\left(e^{-}\right)=\frac{21.24C}{96485C/mo}=2.2×{10}^{-4}mol$

For every mole of oxidized, two mole of electrons flows.

So, the n of Cd is:

$\begin{array}{c}\mathrm{n}\left(\mathrm{Cd}\right)=\frac{\mathrm{n}\left({\mathrm{e}}^{−}\right)}{2}=\frac{2.2×{10}^{−4}\mathrm{mol}}{2}\\ \mathrm{n}\left(\mathrm{Cd}\right)=1.1×{10}^{−4}\mathrm{mol}\end{array}$

The mass of Cd is

$\begin{array}{c}\mathrm{m}\left(\mathrm{Cd}\right)=\mathrm{n}\left(\mathrm{Cd}\right)â‹\dots \mathrm{M}\left(\mathrm{Cd}\right)\\ \mathrm{m}\left(\mathrm{Cd}\right)=1.1×{10}^{−4}\mathrm{mol}×112.42\mathrm{g}/\mathrm{mol}\\ \mathrm{m}\left(\mathrm{Cd}\right)=0.01237\mathrm{g}\end{array}$

So, 0.01237 g of Cd is oxidized each hour.