Electrochemistry

Abstract

In chapter one, it was stated that energy can transform into different forms; one of such useful transformations is when electrical energy converts into chemical one or vice versa. Batteries transform the energy released by a chemical reaction into electricity, so we can run different devices like cell phones, computers, personal media players, and even some vehicles.

Examples of Calculations

Example 1

Calculation of cell potentials

Determine the cell potential for the following reactions:

• (a) $2{\text{Au}}^{ + } + {\text{Cu}} = {\text{Cu}}^{2 + } + 2{\text{Au}}$

• (b) ${\text{Sn}}^{4 + } + 2{\text{Ce}}^{3 + } = {\text{Sn}}^{2 + } + 2{\text{Ce}}^{4 + }$

• (c) ${\text{Fe}}^{2 + } + {\text{V}} = {\text{Fe}} + {\text{V}}^{2 + }$

• (d) $2{\text{Ag}}^{ + } + {\text{Sn}} = 2{\text{Ag}} + {\text{Sn}}^{2 + } .$

Solution

By using the potentials in Table 3.1, we have:

For part (a)

The reaction: 2Au⁺ + Cu = Cu²⁺ + 2Au can be decomposed into two half-cell reactions as:

$$ \begin{array}{*{20}l} {{\text{Au}}^{ + } + {\text{e}}^{-} = {\text{Au}};} \hfill & {E^{\text{o}} = 1.69\,{\text{V}}} \hfill \\ {{\text{Cu}} = {\text{Cu}}^{2 + } + 2{\text{e}}^{-} ;} \hfill & {E^{\text{o}} = - 0.34\,{\text{V}}} \hfill \\ \end{array} . $$

Thus the cell voltage is: 1.69 − 0.34 = 1.35 V

For part (b)

The reaction: ${\text{Sn}}^{4 + } + 2{\text{Ce}}^{3 + } = {\text{Sn}}^{2 + } + 2{\text{Ce}}^{4 + }$ can be decomposed into:

$$ \begin{array}{*{20}l} {{\text{Sn}}^{4 + } + 2{\text{e}}^{-} = {\text{Sn}}^{2 + } ;} \hfill & {E^{\text{o}} = 0.15\,{\text{V}}} \hfill \\ {{\text{Ce}}^{3 + } = {\text{Ce}}^{4 + } + {\text{e}}^{-} ;} \hfill & {E^{\text{o}} = - 1.61\,{\text{V}}} \hfill \\ \end{array} . $$

Thus the cell voltage is: 0.15 − 1.61 = −1.46 V

For part (c)

The reaction: Fe²⁺ + V = Fe + V²⁺ can be decomposed into:

$$ \begin{array}{*{20}l} {{\text{Fe}}^{2 + } + 2{\text{e}}^{-} = {\text{Fe}};} \hfill & {E^{\text{o}} = - 0.41\,{\text{V}}} \hfill \\ {{\text{V}}^{2 + } + 2{\text{e}}^{-} = {\text{V}};} \hfill & {E^{\text{o}} = - 1.19\,{\text{V}}} \hfill \\ \end{array} $$

Thus the cell voltage is: −0.41 − (−1.19) = 0.78 V

For part (d)

The reaction 2Ag⁺ + Sn = 2Ag + Sn²⁺ results from adding the reactions:

$$ \begin{array}{*{20}l} {{\text{Sn}}^{2 + } + 2{\text{e}}^{-} = {\text{Sn}};} \hfill & {E^{\text{o}} = - 0.14\,{\text{V}}} \hfill \\ {2{\text{Ag}}^{ + } + {\text{e}}^{-} = 2{\text{Ag}};} \hfill & {E^{\text{o}} = 0.80\,{\text{V}}} \hfill \\ \end{array} $$

Thus the cell voltage is: 0.80 − (−0.14) = 0.94 V

Example 2

Calculation of the equilibrium constant from standard cell voltage

Use the table of standard potentials to calculate the equilibrium constant at 25 °C for the reaction:

$$ 2{\text{Cu}} + = {\text{Cu}}2 + + {\text{Cu}} . $$

Solution

From Table 3.1, we have the following half-reactions:

$$ \begin{array}{*{20}l} {{\text{Cu}}^{ + } = {\text{Cu}}^{2 + } + {\text{e}}^{ - } ;} \hfill & {E^{\text{o}} = - 0.16\,{\text{V}}} \hfill \\ {{\text{Cu}}^{ + } + {\text{e}}^{ - } = {\text{Cu}};} \hfill & {E^{\text{o}} = 0.52\,{\text{V}}} \hfill \\ \end{array} . $$

Adding up these two reactions results in:

$$ 2{\text{Cu}}^{ + } = {\text{Cu}}^{2 + } + {\text{Cu}};\quad E^{\text{o}} = - 0.16 + 0.52 = 0.36\,{\text{V}} . $$

The free energy of the cell becomes:

$$ \Delta G^{\text{o}} = - nFE^{\text{o}} . $$

(E2.1)

On the other hand, we also know that:

$$ \Delta G^{\text{o}} = - RT\,\ln \,(K). $$

(E2.2)

Equating these expressions and noting that n = 1; thus the equilibrium constant becomes:

$$ \begin{aligned} - nFE^{\text{o}} = - RT\,\ln \,(K) \hfill \\ \ln \,(K) = \frac{{FE^{\text{o}} }}{RT} = \frac{96,500 \times 0.36}{8.314 \times 298} \hfill \\ \ln \,(K) = 14.02 \hfill \\ K = 1.23 \times 10^{6} . \hfill \\ \end{aligned} $$

Example 3

Calculation of the free energy and entropy of an electrochemical cell

The cell:

$$ {\text{Pb}}\left| {{\text{PbCl}}_{2} } \right|{\text{HCl}}\left| {\text{AgCl}} \right|{\text{Ag,}} $$

where all components are pure solids are in contact with an electrolyte made of HCl. The cell voltage is 0.49 V at 298 K. The temperature coefficient of this cell at the given temperature is −1.84 × 10⁻⁴ VK⁻¹. Calculate the free energy and the entropy changes of the cell reaction at 25 °C.

Solution

To calculate the free energy of the cell reaction, we use Eq. (3.6):

$$ \begin{aligned} \Delta G & = - nFE \\ \Delta G & = - 2 \times 96,500 \times 0.49 \\ \Delta G & = - 94,570\quad {\text{J}} .\\ \end{aligned} $$

To determine the entropy change, we use expression (3.7):

$$ \begin{aligned}\Delta S & = nF\left[ {\frac{\partial E}{\partial T}} \right]_{P} \hfill \\\Delta S & = 2 \times 96,500 \times - 1.84 \times 10^{ - 14} \hfill \\\Delta S & = - 35.5\,{\text{JK}}^{ - 1} . \hfill \\ \end{aligned} $$

Example 4

Estimation of thermodynamic quantities for an electrochemical cell.

The reaction of a cadmium–calomel cell is represented by:

$$ {\text{Cd}} + {\text{Hg}}_{2} {\text{Cl}}_{2} = {\text{Cd}}^{2 + } + 2{\text{Cl}}^{ - } + 2{\text{Hg}} . $$

All components are in their standard state. The reversible cell voltage (E ^o) varies with temperature, according to:

$$ E^{\text{o}} = 0.67 - 1.062 \times 10^{ - 4} \left( {T - 298} \right) - 2.4 \times 10^{ - 6} \left( {T - 298} \right)^{2} . $$

(E4.1)

Calculate the values of ∆G ^o, ∆S ^o, and ∆H ^o for the cell reaction at 40 °C.

Solution

We need to determine the temperature coefficient of the cell. To do so, we must take the first derivative of E ^o with respect of T:

$$ \begin{aligned} \left[ {\frac{{\partial E^{\text{o}} }}{\partial T}} \right]_{P} & = \frac{\partial }{\partial T}\left( {0.67 - 1.062 \times 10^{ - 4} \left( {T - 298} \right) - 2.4 \times 10^{ - 6} \left( {T - 298} \right)^{2} } \right) \\ \left[ {\frac{{\partial E^{\text{o}} }}{\partial T}} \right]_{P} & = 1.324 \times 10^{ - 3} - 4.8 \times 10^{ - 6} T. \\ \end{aligned} $$

(E4.2)

Now it is necessary to calculate the cell voltage (E ^o) at 40 °C (313 K) which is the test temperature. From equation (E4.1):

$$ \begin{aligned} E^{\text{o}} & = 0.67 - 1.062 \times 10^{ - 4} \left( {T - 298} \right) - 2.4 \times 10^{ - 6} \left( {T - 298} \right)^{2} \\ E^{\text{o}} & = 0.67 - 1.062 \times 10^{ - 4} \left( {313 - 298} \right) - 2.4 \times 10^{ - 6} \left( {313 - 298} \right)^{2} \\ E^{\text{o}} & = 0.667\,{\text{V}} .\\ \end{aligned} $$

Since both the standard potential and the temperature coefficient of the cell are known, it is possible to calculate ∆G ^o, ∆S ^o and ∆H ^o:

$$ \begin{aligned} \Delta G^{\text{o}} & = - nFE^{\text{o}} \\ \Delta G^{\text{o}} & = - 2 \times 96,500 \times 0.667 = - 128,731\,{\text{J}} \\ \Delta S^{\text{o}} & = nF\left[ {\frac{{\partial E^{\text{o}} }}{\partial T}} \right]_{P} \\ \Delta S^{\text{o}} & = 2 \times 96,500 \times \left( {1.324 \times 10^{ - 3} - 4.8 \times 10^{ - 6} \times 313} \right) = - 34.4\,{\text{JK}}^{ - 1} \\ \Delta H^{\text{o}} & = nF\left[ {T\left[ {\frac{{\partial E^{\text{o}} }}{\partial T}} \right]_{P} - E^{\text{o}} } \right] \\ \Delta H^{\text{o}} & = 2 \times 96,500 \times \left( {313\left( {1.324 \times 10^{ - 3} - 4.8 \times 10^{ - 6} \times 313} \right) - 0.667} \right) = - 139,539\,{\text{J}} .\\ \end{aligned} $$

It is worth noticing that from the computed enthalpy and entropy changes it is possible to calculate the Gibbs free energy associated to the cell using Eq. (1.21):

$$ \begin{aligned}\Delta G^{\text{o}} & =\Delta H^{\text{o}} - T\Delta S^{\text{o}} = - 139,539 - 313 \times - 34.4 \\\Delta G^{\text{o}} & = - 128,772\,{\text{J}} .\\ \end{aligned} $$

This value differs 41 J from the one computed using Eq. (3.6). This error can be attributed to rounding off; additionally such small difference is not significant given the order of magnitude of the Gibbs free energy value.

Example 5

Use of experimental data to calculate the thermodynamic quantities associated to a cell.

For the cell:

$$ {\text{Zn}}\left| {{\text{ZnCl}}_{2} \left( {x = 0.6} \right) + {\text{KCl}}\left( {x = 0.4} \right)} \right|{\text{Cl}}_{2} {\text{C}} . $$

The following data has been recorded:

T (°C)	E (V)
435	1.6447
436	1.6435
440	1.6420
453	1.6338
456	1.6325
461	1.6295
468	1.6235
469	1.6230
477	1.6178
490	1.6098
497	1.6052
525	1.5870
545	1.5750
546	1.5741
551	1.5695
561	1.5642
575	1.5540
583	1.5502

For this cell, E ^o is 1.5635 V at 500 °C. Answer:

• (a) What is the cell reaction?

• (b) Plot the cell voltage vs temperature and determine E at 500 °C by interpolation

• (c) Calculate the free energy, entropy and enthalpy of the cell reaction at 500 °C

• (d) Calculate the activity and the activity coefficient of ZnCl₂ in the electrolyte.

Solution

Part (a):

An electrochemical cell is represented by: anode | electrolyte | cathode; in this example we have: Zn | ZnCl₂(x = 0.6) + KCl(x = 0.4) | Cl₂ C, which means than zinc and chlorine gas are isolated at each electrode, but are in electrical contact trough the chlorides mixture. In view of this the overall cell reaction is:

$$ {\text{Zn}} + 2{\text{Cl}}^{ - } = {\text{Zn}}^{2 + } + {\text{Cl}}_{2} \uparrow + 2{\text{e}}^{ - } $$

Part (b)

Plotting the data provided: (Fig. 3.4)

Fig. 3.4

Voltage versus temperature plot

To determine the cell potential at 500 °C, we can use the data provided to interpolate this value. To do so, we need the data at 497 and 525 °C, with these points, we have:

$$ \begin{aligned} & \frac{1.5870 - 1.6052}{525 - 497}\frac{\text{V}}{{^{\text{o}} {\text{C}}}} = \frac{{1.5870 - E_{500} }}{525 - 500}\frac{\text{V}}{{^{\text{o}} {\text{C}}}} \\ & E_{500} = 1.587 - \frac{25}{28}\left( { - 0.0182} \right) = 1.6033\,{\text{V}} .\\ \end{aligned} $$

Thus the cell potential at 500 °C is 1.6033 V.

Another option is to run a linear regression on the data so we can estimate the slope of the corresponding line and its distance to the origin. Using this second approach, the calculated the cell voltage at 500 °C is 1.6029 V. This value is also plotted in the graph above (black triangle). There is good agreement between the computed values.

Part (c)

Form the linear regression on the data supplied, we can express the cell voltage as a function of the temperature and thus calculate the cell temperature coefficient:

$$ \begin{aligned} E & = - 6.45 \times 10^{ - 4} T + 1.9254 \\ \left[ {\frac{\partial E}{\partial T}} \right]_{P} & = - 6.45 \times 10^{ - 4} . \\ \end{aligned} $$

The cell free energy at 500 °C is:

$$ \begin{aligned}\Delta G & = - nFE \\\Delta G & = - 2 \times 96,500 \times 1.6029 = - 309,360\,{\text{J}} .\\ \end{aligned} $$

At 500 °C the cell entropy is:

$$ \begin{aligned}\Delta S & = nF\left[ {\frac{\partial E}{\partial T}} \right] \\\Delta S & = 2 \times 96,500 \times - 6.45 \times 10^{ - 4} = - 124.5\,{\text{JK}}^{ - 1} . \\ \end{aligned} $$

The enthalpy of the cell at 500 °C is:

$$ \begin{aligned}\Delta H & = nF\left[ {T\left[ {\frac{\partial E}{\partial T}} \right]_{P} - E} \right] \\\Delta H & = 2 \times 96,500\left( {773 \times - 6.45 \times 10^{ - 4} - 1.6029} \right) = - 405,587\,{\text{J}} .\\ \end{aligned} $$

Part (d)

The standard free energy of the cell is given by:

$$ \begin{array}{*{20}l} {\Delta G_{\text{cell}} = G_{\text{cell}} - G_{\text{cell}}^{\text{o}} } \hfill \\ {\Delta G_{\text{cell}} = - nFE_{\text{cell}} - \left( { - nFE_{\text{cell}}^{\text{o}} } \right)} \hfill \\ {\Delta G_{\text{cell}} = nF\left( {E_{\text{cell}}^{\text{o}} - E_{\text{cell}} } \right)} \hfill \\ {\Delta G_{\text{cell}} = 2 \times 96,500\left( {1.5635 - 1.6029} \right) = - 7604} \hfill \\ \end{array} . $$

With the cell free energy change, we can now estimate the activity coefficient of ZnCl₂:

$$ \begin{aligned}\Delta G_{\text{cell}} & = RT\,\ln \left( {\gamma_{{{\text{ZnCl}}_{2} }} } \right) \\ \gamma_{{{\text{ZnCl}}_{2} }} & = \exp \left( {\frac{{\Delta G_{\text{cell}} }}{RT}} \right) = \exp \left( { - \frac{7604}{8.314 \times 773}} \right) \\ \gamma_{{{\text{ZnCl}}_{2} }} & = 0.306. \\ \end{aligned} $$

Having this value, it is possible to compute the activity of zinc chloride:

$$ \begin{aligned} a_{{{\text{ZnCl}}_{2} }} & = x_{{{\text{ZnCl}}_{2} }} \times \gamma_{{{\text{ZnCl}}_{2} }} \\ a_{{{\text{ZnCl}}_{2} }} & = 0.6 \times 0.306 \\ a_{{{\text{ZnCl}}_{2} }} & = 0.184. \\ \end{aligned} $$

Further discussion on how to calculate the activity or the activity coefficient of a component in a mixture as a function of Gibbs free energy will be presented in the next chapter. In the meantime, we advise the reader to accept these formulae as shown.

Example 6

Use of experimental data to calculate thermodynamic quantities in an electrolytic cell

The cell:

$$ {\text{Pb}}\left| {{\text{PbCl}}_{2} + {\text{LiCl}}} \right|{\text{Pb}} - {\text{Bi}} $$

Yielded the following data at 700 K

x Pb	E [V]	dE/dT [V K−1]
0.848	0.00532	7.40 × 10−6
0.720	0.01148	1.44 × 10−5
0.600	0.01929	2.08 × 10−5
0.496	0.02782	2.78 × 10−5
0.414	0.03594	3.76 × 10−5
0.328	0.0454	4.64 × 10−5
0.230	0.05976	6.44 × 10−5
0.111	0.08615	1.02 × 10−4

• (a) Calculate the values of a _Pb, ∆S _Pb and ∆H _Pb for each of these compositions.

• (b) At the composition x _Pb = 0.6, if a potential of 19.29001 mV were applied to the cell so as to oppose the measured e.m.f. would the cell generate or absorb heat?

Solution

In this cell lead is dissolved into a Pb–Bi alloy; as more lead is mixed with the bismuth, the activity of the former should increase from zero (unmixed bismuth) toward unity (unmixed lead). As the solution gets richer in lead, the associated entropy and enthalpy should change accordingly.

To calculate the activity of lead in the Pb–Bi alloy, we need to use Eq. (3.6) along with Eq. (1.38):

$$ \Delta G^{\text{o}} = - nFE^{\text{o}} $$

(3.6)

$$ \Delta G^{\text{o}} = - RT\,\ln \,K. $$

(1.38)

The dissolution of lead in bismuth can be expressed as:

$$ {\text{Pb}} = {\text{Pb}}_{\text{Bi}} $$

(E6.1)

From this expression, we can write down the equilibrium constant as:

$$ K = \frac{{a_{{{\text{Pb}}_{\text{Bi}} }} }}{{a_{\text{Pb}} }}. $$

(E6.2)

The activity of pure lead is 1, therefore the equilibrium constant becomes the activity of lead in solution with bismuth, and therefore we have:

$$ - RT\,\ln \left( {a_{{{\text{Pb}}_{\text{Bi}} }} } \right) = - nFE^{\text{o}} . $$

(E6.3)

From the given data, we have that T = 700 K, n = 2. Substituting the corresponding voltage values in expression (E6.3) would yield the activity of the dissolved lead. This is shown in the table below.

To calculate the entropy of the cell as the lead content in the mixture changes, we need to take the same approach as before. In this case the values of the cell temperature coefficient must be inserted in Eq. (3.8) to compute the entropy change associated to lead dissolution in bismuth. Results are shown in the table below. The same procedure is taken to compute the enthalpy change.

Table (3.2) summarizes all the calculations of these thermodynamic quantities as a function of lead concentration in the Pb–Bi mix.

Table 3.2 Activity of lead and thermodynamic quantities as a function of lead content in bismuth

Plotting the data in Table 3.2, results in: (Fig. 3.5)

Fig. 3.5

Thermodynamic quantities as lead dissolves in bismuth at 700 K (427 °C)

For part (b)

∆H _Pb at x _Pb = 0.6 is negative, the cell would release heat.

Example 7

Estimation of energy consumption in copper deposition.

Copper refining can be done by depositing Cu²⁺ from an acidic CuSO₄ solution onto copper cathodes by applying an electric current.

Determine:

• (a) How many kWh are required for the electrolytic refining of 1 ton of copper if 0.4 V are applied to the electrolysis cell and all the current supplied is used to deposit the metal.

• (b) How would the energy requirements would change if the voltage is lowered to 0.2 V?

• (c) How much heat is liberated in each case?

Solution

For part (a):

We need to use Eq. (3.37) to determine the current supplied to the cell:

$$ m = \frac{i \times t}{F}\frac{M}{z} $$

Assuming an hour of operation and solving for i:

$$ i = \frac{Fzm}{tM} $$

Substituting values: F = 96,500 A s mole⁻¹, t = 3600 s, z = 2, m = 106 g, M = 63.54 g mole⁻¹; we obtain:

$$ i = \frac{{\left( {96,500} \right)\left( 2 \right)\left( {10^{6} } \right)}}{{\left( {3600} \right)\left( {63.54} \right)}} = 8.44 \times 10^{5} \,{\text{A}} $$

Since we know the amount of current circulating through the cell as well as the voltage supplied, we can calculate the power required:

$$ P = i{\text{V}} = \left( {8.44 \times 10^{5} } \right)\left( {0.4} \right) = 3.376 \times 10^{5} \,{\text{W}} . $$

Since the current was supplied for an hour, then the power needed is 337.6 kWh.

For part (b)

Since we cut the voltage by a half, the power needed when using only 0.2 V is 168.8 kWh.

For part (c)

All the energy supplied to the cell via electric current has to be released eventually, therefore from the First law of Thermodynamics, we can estimate the heat losses from the power supplied. Just remember that 1 kWh = 3.6 MJ, thus the liberated heat when applying 0.4 V is (337.6 kWh × 3.6 MJ kWh⁻¹) 1215.4 MJ: Similarly when 0.2 V are used, the heat released is 607.7 MJ.

Example 8

Estimation of current efficiency in a cell for copper deposition.

A set up for copper refining consists of 4 cells. After running a test for 7.33 h, with an average current of 34.2 A, results in 1089 g of deposited copper. What is the current efficiency of the process?

Solution

We need to use Eq. (3.37) to determine the theoretical amount of current needed to deposit the specified amount of copper:

$$ m = \frac{itM}{zF}. $$

Substituting values: m = 1089 g, t = 26,388 s, M = 63.54 g mole⁻¹, F = 96,500 A s mole⁻¹, z = 2 into Eq. (3.37) and solving for i:

$$ i{\text{th}} = \frac{mzF}{tM} = \frac{1089 \times 2 \times 96,500}{26,388 \times 63.54} = 125.3\,{\text{A}} . $$

Each cell uses 34.2 A; therefore the total amount of current supplied to the cells is 136.8 A. To deposit the copper the cells would only utilize 125.3 A, thus the current efficiency (η) is defined by the ratio of theoretical current used to actual current used:

$$ \begin{aligned} \eta & = \frac{{i{\text{th}}}}{{i_{\text{supl}} }} \times 100 \\ \eta & = \frac{{125.3\,{\text{A}}}}{{136.8\,{\text{A}}}} \times 100 = 91.6\% . \\ \end{aligned} $$

Example 9

Effect of process variables on the acidity of an electrolyte.

It has been suggested to produce metallic nickel powders by hydrogen reduction of a nickel sulfate solution. The reduction reaction would be:

$$ {\text{Ni}}^{2 + } + {\text{H}}_{2} = {\text{Ni}} + 2{\text{H}}^{ + } . $$

For this reaction, its standard free energy change is:∆G ^o = 640,15 − 58.948 T/J mole⁻¹.

Discuss the effect of the following variables and calculate the final value of the solution pH for the given variables.

• (a) Temperature: 100 and 200 °C

• (b) $P_{{{\text{H}}_{2} }}$: 10 and 100 atm

• (c) a_Ni ²⁺: 10⁻² and 10⁻³

Solution

We need to determine the equilibrium constant (K) of the nickel reduction reaction:

$$ {\text{Ni}}^{2 + } + {\text{H}}_{2} = {\text{Ni}} + 2{\text{H}}^{ + } ;\quad {\text{K}} = \frac{{a_{\text{Ni}} \times a_{{{\text{H}}^{ + } }}^{2} }}{{a_{{{\text{Ni}}^{2 + } }} \times P_{{{\text{H}}_{2} }} }}. $$

(E9.1)

As seen, the K value depends on the activities (concentration) of the different species involved in the reaction. In addition, we know we can express the equilibrium constant in terms of the free energy change:

$$ \Delta G^{\text{o}} = - RT\,\ln \,(K). $$

With this information, we can discuss the effect of the variables on the acidity of the resulting solution:

For part (a):

Since we are looking to estimate the effect of process temperature on the acidity of the final solution, we need to keep fixed a _Ni, a ²⁺_Ni and P _H2 and solve for a ⁺_H as a function of temperature. To simplify our calculations the fixed variables are set as 1, thus we can establish how temperature influences the acidity of the electrolyte:

If T = 100 °C = 373 K

$$ \begin{aligned}\Delta G^{\text{o}} & = 64,015 - 58.948\,{\text{T}} \\\Delta G^{\text{o}} & = 64,015 - 58.948\left( {373} \right) = 42027.4\,{\text{J}} \\ K & = \exp \left( { - \frac{{\Delta {\text{G}}^{\text{o}} }}{\text{RT}}} \right) = \exp \left( { - \frac{42027.4}{{8.314\left( {373} \right)}}} \right) \\ K & = 1.3 \times 10^{ - 6} \\ a_{{{\text{H}}^{ + } }} & = \sqrt {\frac{{K \times a_{{{\text{Ni}}^{2 + } }} \times P_{{{\text{H}}_{2} }} }}{{a_{\text{Ni}} }}} ;\quad a_{{{\text{Ni}}^{2 + } }} ,P_{{{\text{H}}_{2} }} ,a_{\text{Ni}} = 1 \\ a_{{{\text{H}}^{ + } }} & = \sqrt K = \sqrt {1.3 \times 10^{ - 6} } = 1.14 \times 10^{ - 3} . \\ \end{aligned} $$

From this, we can calculate the final pH since: pH ≈ −log (a _H ⁺)

$$ \begin{aligned} {\text{pH}} & \approx - \log \left( {a_{{{\text{H}}^{ + } }} } \right) \\ {\text{pH}} & \approx - \log \left( {1.14 \times 10^{ - 3} } \right) \\ {\text{pH}} & \approx 2.94. \\ \end{aligned} $$

If we now increase the temperature to 200 °C (473 K), we need to repeat the calculations to obtain the final pH:

$$ \begin{aligned}\Delta G^{\text{o}} & = 64,015 - 58.948\left( {473} \right) = 36,132.6\,{\text{J}} \\ \ln \,K & = - 9.19 \\ K & = 1.02 \times 10^{ - 4} \\ a_{{{\text{H}}^{ + } }} & = \sqrt K = \sqrt {1.02 \times 10^{ - 4} } = 1.01 \times 10^{ - 2} \\ {\text{pH}} & \approx - \log \left( {1.01 \times 10^{ - 2} } \right) = 1.99. \\ \end{aligned} $$

Therefore, as temperature increases, so does the acidity in the electrolyte.

For part (b):

We need to keep T, a _Ni, and a ²⁺_Ni fixed. The Temperature is set at 100 °C and the latter two variables are kept as one. By fixing the temperature, we have set the equilibrium constant as well. With this information we can estimate how the partial pressure of hydrogen affects the final acidity:

$$ \begin{aligned} & {\text{if}}\,P_{{{\text{H}}_{2} }} = 10\,{\text{atm}}; \\ & a_{{{\text{H}}^{ + } }} = \sqrt {\frac{{K \times a_{{{\text{Ni}}^{2 + } }} \times P_{{{\text{H}}_{2} }} }}{{a_{\text{Ni}} }}} ;\quad a_{{{\text{Ni}}^{2 + } }} = a_{\text{Ni}} = 1 \\ & a_{{{\text{H}}^{ + } }} = \sqrt {K \times P_{{{\text{H}}_{2} }} } = \sqrt {1.3 \times 10^{ - 6} \left( {10} \right)} = 3.61 \times 10^{ - 3} \\ & {\text{pH}} \approx - \log \left( {3.61 \times 10^{ - 3} } \right) = 2.44 \\ & {\text{if}}\,P_{{{\text{H}}_{2} }} = 100\,{\text{atm}}; \\ & a_{{{\text{H}}^{ + } }} = \sqrt {1.3 \times 10^{ - 6} \left( {100} \right)} = 0.0114 \\ & {\text{pH}} \approx - \log \left( {0.0114} \right) = 1.94. \\ \end{aligned} $$

From these calculations, it is clear that when partial pressure of hydrogen gas increases, the solution becomes more acidic.

Part (c):

In this case, the activity of nickel ions vary, which means that temperature, hydrogen pressure and concentration of H⁺ ions must be fixed. Again choosing T = 100 °C, results in setting the equilibrium constant (K = 1.3 × 10⁻⁶), thus:

$$ \begin{aligned} & {\text{if}}\,a_{{{\text{Ni}}^{2 + } }} = 10^{ - 2} ; \\ & a_{{{\text{H}}^{ + } }} = \sqrt {K \times a_{{{\text{Ni}}^{2 + } }} } \\ & a_{{{\text{H}}^{ + } }} = \sqrt {1.3 \times 10^{ - 6} \left( {10^{ - 2} } \right)} = 1.14 \times 10^{ - 4} \\ & {\text{pH}} \approx - \log \left( {1.14 \times 10^{ - 4} } \right) = 3.94 \\ & {\text{if}}\,a_{{{\text{Ni}}^{2 + } }} = 10^{ - 3} ; \\ & a_{{{\text{H}}^{ + } }} = \sqrt {1.3 \times 10^{ - 6} \left( {10^{ - 3} } \right)} = 3.6 \times 10^{ - 5} \\ & {\text{pH}} \approx - \log \left( {3.6 \times 10^{ - 5} } \right) = 4.44. \\ \end{aligned} $$

Therefore if the amount of nickel ions decrease, the less acidic the solution becomes.

Example 10

Calculation of a copper electrolytic refining plant capacity

• 1. An electrolytic copper refinery has 458 tanks of which 42 are used to produce starting sheets. Anodes weigh 240 kg. Each tank has 22 anodes and 23 cathodes. There are three pulls of cathodes per anode. The current in each tank is 5400 A, and its efficiency is 92%. 11% of the anodes are turned into scrap.

Calculate: (a) The mass of each cathode. (b) The capacity of the refinery.

• 2. A second refinery has 39 cathodes (0.95 m × 0.95 m) and 38 anodes to a cell. The anode weighs 240 kg and lasts 28 days. There are two pulls of cathodes per 28 days. The starting sheets weigh 4.54 kg. The current efficiency is of 93% and 2 V are applied to this cell.

Calculate: (c) What is the mass of the copper deposited on the cathode when pulled. (d) What is the percentage of anode scrap? (e) What is the cathode current density? (f) How many kg of copper are produces by kilowatt-day?

Solution

For part (a)

There are 22 cathodes in each tank whose weight is 240 kg. From this mass, 89% is concerted into copper cathodes, and 3 anodes are needed to obtain a cathode. Since the mass of the starting anodes is given, and the ratio anode/cathode is also known, we do not need to use Faraday’s laws to estimate the weight of each cathode.

Therefore the mass of a cathode is:

$$ 240\,{\text{kg}}_{\text{anodes}} \times 0.89 \times \frac{{1_{\text{cathode}} }}{{3_{\text{anodes}} }} = 71.2\,{\text{kg}}_{\text{cathode}} $$

For part (b)

To produce the copper cathodes there are 416 tanks available. The mass of copper that is produced in a daily basis is given by Fraday’s law. We need to assume a processing time of 24 h:

$$ \text{m}_{\text{Cu}} = n_{{\tan \,{\text{ks}}}} \times \frac{i \times t \times M}{z \times F}, $$

where m_Cu is the mass of copper in g/day; i is the current (5400 A × 0.92); t is the time (3600 s/h × 24 h); M = 63.54 g/mole; z = 2; F = 96,500 C/mole; n _{tan ks} = number of tanks for electrolysis. Substituting these values:

$$ \text{m}_{\text{Cu}} = 416 \times \frac{{\left( {5400 \times 0.92} \right) \times \left( {3600 \times 24} \right) \times 63.54}}{2 \times 96500} = 59 \times 10^{6} \frac{{{\text{g}}_{\text{Cu}} }}{\text{day}}. $$

Therefore the plant can produce 59 tons of copper per day.

For part (c):

In this case, two anodes are needed to produce a cathode, assuming 93% current efficiency and subtracting the mass of the initial sheet (4.5 kg), gives the weight of each cathode:

$$ \begin{aligned} & 240\,{\text{kg}}_{\text{anode}} \times 0.93 \times \frac{{1_{\text{cathode}} }}{{2_{\text{anodes}} }} = 111.6\,{\text{kg}}_{\text{cathode}} \\ & w_{\text{cathode}} = 111.6 - 4.5 = 107\,{\text{kg}} .\\ \end{aligned} $$

For part (d)

Since two anodes are needed to produce a cathode. The amount of copper consumed from the each anode is 107 kg × 2 = 214 kg; then the percentage of anode scrap is:

$$ \begin{aligned} \% {\text{scrap}} & = 100 \times \left( {1 - \frac{{\text{w}_{{{\text{Cu}},\;{\text{deposited}}}} }}{{\text{w}_{{{\text{Cu}},\;{\text{anode}}}} }}} \right) \\ \% {\text{scrap}} & = 100 \times \left( {1 - \frac{214}{240}} \right) = 10.8\% . \\ \end{aligned} $$

Therefore 10.8% or 26 kg of the anode is thrown away as scrap.

For part (e):

To estimate the current density, first we need to know the amount of current applied to the cell:

$$ \begin{aligned} i & = \frac{m \times z \times F}{t \times M} \times \frac{1}{\eta } = \frac{{\left( {107 \times 10^{3} } \right) \times 2 \times 96500}}{{\left( {3600 \times 24 \times 28} \right) \times 63.54}} \times \frac{1}{0.93} \\ i & = 144.5\,{\text{A}} .\\ \end{aligned} $$

The current density is the ratio current/cathode area, therefore:

$$ {\text{current}}\,{\text{density}} = \frac{i}{\text{A}} = \frac{144.5}{{0.95^{2} }} = 160\frac{\text{A}}{{{\text{m}}^{2} }}. $$

For part (f)

The power supplied to the cell is:

$$ P = i \times {\text{V}} = 144.5 \times 2 = 289\,{\text{W}} . $$

The amount of copper produced daily is:

$$ \begin{aligned} \text{m}_{\text{Cu}} & = \frac{i \times \eta \times t \times M}{z \times F} = \frac{{144.5 \times 0.93 \times \left( {3600 \times 24} \right) \times 63.54}}{2 \times 96,500} \\ \text{m}_{\text{Cu}} & = 3823\,{\text{g}}_{\text{Cu}} . \\ \end{aligned} $$

The amount of copper produced per unit of power is:

$$ \frac{{\text{m}_{\text{Cu}} }}{P} = \frac{{3823\,{\text{g}}_{\text{Cu}} }}{{289\,{\text{W}}}} = 13.22\frac{{{\text{kg}}_{\text{Cu}} }}{\text{kW}}. $$

Therefore 13.22 kg of copper are produced per kW per day.