Question

The Weston cell is a stable voltage standard formerly used in potentiometers. (The potentiometer compares an unknown voltage with that of the standard. In contrast with the conditions of this problem, very little current may be drawn from the cell if it is to be a voltage standard.)

How much work (J)can be done by the Weston cell if the voltage is 1.02 V and1.00mLof Hg(density = 13.52g/mL) is deposited?

(b) If the cell passes current through a$\mathbf{100}\mathbf{-}\mathit{\Omega }$ resistor that dissipates heat at a rate of0.209 J/min, how many grams of are oxidized each hour? (This question is not meant to be consistent with part (a). The voltage is no longer 1.02volts.)

Short answer

(a) The work is W = 6638.1J

(b) The mass of Cd is m(Cd) = 0.01237g

Step-by-step solution

Fundamentals of electrolysis

• 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.

Determine the work

a)

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

First, we calculate mass of Hg:

$\begin{array}{c}\rho =\frac{m}{V}\\ m=\rho \cdot V=13.53\mathrm{g}/\mathrm{mL}\cdot 1\mathrm{mL}\\ \mathrm{m}=13.53\mathrm{g}\end{array}$

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

$\begin{array}{c}\mathrm{q}=\mathrm{n}\left({\mathrm{e}}^{-}\right)\cdot \mathrm{F}=0.06745\mathrm{mol}\times 96485\mathrm{C}/\mathrm{mol}\\ \mathrm{q}=6507.9\mathrm{C}\end{array}$

And the work is:

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

Determine the mass

b)

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

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

First we calculate current using the formula:

$\begin{array}{r}P=I^2\cdot \mathrm{R}\\ I=\sqrt{\frac{P}{R}}=\sqrt{\frac{3.48\times {10}^{-3}\mathrm{J}/\mathrm{s}}{100\mathrm{\Omega }}}\\ I=5.9\times {10}^{-3}\mathrm{A}\end{array}$

So, in each hour the charge is:

$I\cdot 1hour=5.9\times {10}^{-3}C/s\times 3600s=21.24C$

In moles that is

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

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

So, the n of Cd is:

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

The mass of Cd is

$\begin{array}{c}\mathrm{m}\left(\mathrm{Cd}\right)=\mathrm{n}\left(\mathrm{Cd}\right)\cdot \mathrm{M}\left(\mathrm{Cd}\right)\\ \mathrm{m}\left(\mathrm{Cd}\right)=1.1\times {10}^{-4}\mathrm{mol}\times 112.42\mathrm{g}/\mathrm{mol}\\ \mathrm{m}\left(\mathrm{Cd}\right)=0.01237\mathrm{g}\end{array}$

So, 0.01237 g of Cd is oxidized each hour.