Category: Advanced Chemistry

The three-electrode cell is used to measure the activation overpotential of a cell at various levels of current. Consider the following experiment setup:

The opening end of the Luggin capillary is brought very close to the working electrode (d≈0) and the potential of the working electrode E_eqm is measured with the potentiometer and counter electrode removed from the circuit. The two are then reconnected to the circuit and the externally applied potential is turned on to a magnitude greater than 0.57 V (e.g. 0.8V), resulting in Fe²⁺ being oxidised to Fe³⁺ at the anode and H⁺ reduced to H₂ at the cathode. The potential of the working electrode E_pol is again recorded. Assuming that the electrolyte is well-stirred, the difference between E_pol and E_eqm is the activation overpotential of the anode with respect to the current flowing at 0.8 V. By increasing the applied potential, we can tabulate the increasing levels of current, I, flowing in the circuit (see table below) and consequently record the corresponding overpotentials η_act,anode using eq61.

If the overpotential at the anode is greater than +100 mV, the second exponential in the Butler-Volmer equation tends to zero, giving:

This means that the net current density at the anode mainly comprises of anodic current density. Taking natural logarithm on both side of eq69,

Eq70 is known as the Tafel equation, named after the Swiss chemist, Julius Tafel. The plot of ln(I/A) against η is called a Tafel plot and it allows us to find the value of the transfer coefficient  from the gradient of the line and the exchange current density j₀ from the intercept. Such an experiment that monitors current at various potentials is called voltammetry.

The three-electrode cell can also be modified slightly to measure the total overpotential of a cell, . Consider the following experiment with metallic copper for both the anode and cathode, and AgCl electrode as the reference electrode.

With the potentiometer set to 0 V,

• • Measure potential of the left electrode E_{L, eqm} by connecting the reference electrode to junction X.
  • Measure potential of the right electrode E_{R, eqm} by connecting the reference electrode to junction Y.

We’d expect E_{L, eqm} = E_{R, eqm}. Next, turn the potentiometer to 1.0 V and

• • Measure potential of the left electrode (anode) E_{an, pol} by connecting the reference electrode to junction X.
  • Measure potential of the right electrode (cathode) E_{cat, pol} by connecting the reference electrode to junction Y.

We’d expect E_{an, pol} > E_{L, eqm} and E_{cat, pol} < E_{R, eqm}.

We can also measure IR drop (in mV/cm) for the above setup by setting the external applied voltage at 1.0 V and measuring the anode potential at different Luggin capillary distances, d.

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A Pourbaix diagram is a reduction potential versus pH graph. It was invented by Marcel Pourbaix, a Belgian chemist, in the 1930s. A typical Pourbaix diagram illustrates the equilibria of electrochemical and non-electrochemical reactions. It consists of plots of equations derived either from the Nernst equation (for electrochemical reaction equilibria) or the equilibrium constant relation (for non-electrochemical reaction equilibria), with all reactions related to a particular metal or non-metal. For example, the equilibria of acidified water are:

E₁ :

E₂ :

with the following Nernst equations at rtp, , and :

The Pourbaix diagram for acidified water is the E/E_SHE-pH graph of eq71 and eq72 (where E/E_SHE is the electrode potential of a reaction relative to the Standard Hydrogen Electrode):

A Pourbaix diagram is therefore a map of the relative stability of species in a reaction with respect to E/E_SHE and pH. Acidified water is unstable above the upper line with regard to the evolution of oxygen gas. Below the lower line, acidified water is again unstable, as it is converted to hydrogen gas. In other words, acidified water is relatively stable in the region between the two lines. Understanding the electrochemical stability of water is important in many industrial processes, e.g. the refining of copper, which has the following Pourbaix diagram:

The equilibria involved are:

1)

2)

3)

4)

5)

6)

7)

and their corresponding equations are:

a)

b)

c)

d)

e)

f)

g)

Equations (a), (c), (d), (e) and (g) are derived using the Nernst equation, while equations (b) and (f) originated from the equilibrium constant relation. We assume all activities of aqueous species are equivalent to their concentrations and that the diagram is plotted with all aqueous copper species concentrations of 10^-6 M.

With reference to the copper Pourbaix diagram, the horizontal line corresponds to eq(a), which is independent of pH (i.e. E doesn’t change by varying pH). The two vertical lines correspond to eq(b) and eq(f), both of which are independent of E (i.e. pH doesn’t change by varying E). The slanted lines correspond to eq(c), eq(d), eq(e) and eq(g). Furthermore, the Pourbaix diagram for the electrolysis of acidified water is superimposed on the diagram for copper, as the electrolyte consists of water.

The Pourbaix diagram allows us to select the appropriate E and pH for copper refining, which is represented by equilibrium(1). The preferred pH is between 2 and 4 (for [Cu²⁺] = 10^-6 M), to avoid the formation of oxides. Furthermore, the more negative the cathodic potential is (by application of an external potential:  see diagram below), the greater equilibrium(1) shifts to the right. However, if the potential is too negative, the deposition of copper has to compete with the evolution of hydrogen gas at the cathode, which reduces the yield. Hence, the thermodynamically preferred E value is roughly between 0.00 V and 0.16 V for 2 < pH < 4 and [Cu²⁺] = 10^-6 M. In practice, the preferred E value is also influenced by kinetics, as a result of overpotential, IR drop, etc.

Finally, the copper Pourbaix diagram is also useful for corrosion prevention. According to the diagram below, a layer of oxide is formed to prevent further corrosion of Cu (passivation) when the solution has a pH of between 8 to 12. In acidic solutions, the corrosion of Cu is prevented by bringing the potential of the metal to less than 0.1 V.

 

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Titration curves reveal important data of acid-base systems, like the pH of the stoichiometric point, pH at maximum buffer capacity, equilibrium constants, etc. They have different shapes that are governed by the equilibria of the acid, the base and water. In general, titrations are categorized into four groups:

• • Strong acid versus strong base
  • Weak acid versus strong base
  • Strong acid versus weak base
  • Weak acid versus weak base

The pH versus volume profile of each group can be expressed explicitly with a unique pH titration curve formula. Knowing the formula for all categories of titration therefore allows us to mathematically explain the shape of the curves, including the reasons for the sharp change of pH near stoichiometric points. In the following derivation of pH curve equations of the four categories of titration, we shall assume that the activity of every chemical species is equivalent to its concentration.

 

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What is the formula of the titration curve of a monoprotic weak acid versus a monoprotic weak base?

With reference to previous articles,  substitute eq6, eq10, K_w = [H⁺][OH^–] and [H⁺] = 10^-pH into eq1 (where C_a and C_b are now the concentration of a weak acid and the concentration of a weak base respectively), we have,

Eq12 is the complete pH titration curve for a monoprotic weak acid versus monoprotic weak base system. We can input it in a mathematical software to generate a curve of pH against V_b. For example, if we titrate 10 cm³ of 0.200 M of CH₃COOH (K_a = 1.75 x 10^-5) with 0.100 M of aqueous NH₃ (K_b = 1.8 x 10^-5), we have the following:

To understand the change in pH near the stoichiometric point versus the change in V_b, we assume that one drop of base is about 0.05 cm³ and substitute 19.95 cm³, 20.00 cm³ and 20.05 cm³ into eq12. The respective pH just before and just after the stoichiometric point (SP) are:

	One drop before SP	SP	One drop after SP
Volume, cm3	19.95	20.00	20.05
pH	6.91	7.01	7.10

The data shows that two drops of base cause a change of only 0.19 in pH before and after the stoichiometric point. Since the change in pH at the stoichiometric point occurs within a very narrow range, no indicator is suitable to accurately monitor the stoichiometric point, not even bromothymol blue (see diagram below for bromothymol blue colour at various pH).

Lastly, we can derive the gradient equation for a weak acid to weak base titration and investigate the inflexion point at pH 7.01 by differentiating eq12 implicitly (see this article), resulting in , i.e. a gradient that makes angle of 62.71^o with the horizontal.

 

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How does the titration curve of a strong acid versus a strong base compare with the titration curve of a weak acid versus a strong base?

Superimposing the weak acid versus strong base titration curve on the strong acid versus strong base titration curve, we have

From the graph, the two curves appears to coalesce when pH > 8. This can be rationalised by comparing eq4 and eq8:

where the two equations are approximately the same when 10^-pH → 0, i.e. at high pH. Note that the two curves do not actually coalesce and are still two separate curves when pH > 8 (discernible if the axes of the plot are scaled to a very high resolution).

 

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What is the formula of the titration curve of a polyprotic acid versus a monoprotic strong base?

Comparing eq8, eq19 and eq30 the general equation for a polyprotic acid versus a monoprotic strong base titration is:

where n is the basicity of the acid, i.e. the number of replaceable hydrogen atoms in one molecule of the acid.

 

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Charge balance equations are derived using the concept of electroneutrality, where the sum of positive charges equals to the sum of negative charges in a solution. Such equations are useful for analysing acid-base equilibria and formulating complex acid-base titration equations.

Consider a solution containing water, a strong acid of concentration C_a, and a strong base of concentration C_b, with the following equilibria:

To maintain electroneutrality, the sum of the number of moles of cations H⁺ and B⁺ must equal to that of anions A^– and OH^–. As the volume of the solution is common to all ions,

When formulating charge balance equations for aqueous compounds with multiple equilibria, we need to account for every charged species on the LHS of a particular equilibrium, which can be complicated. To avoid mistakes,  we select equilibrium expressions where species on the LHS are neutral. For example, the equilibrium equations for a diprotic acid can be presented in the following ways:

Select the first and third equilibria, i.e., we can imagine part of the initial number of moles of H₂A dissociating into H_a⁺ and HA^–, with the remaining part of the initial number of moles of H₂A dissociating into 2H_b⁺ and A^2-. For the first equilibrium, the charge balance equation is

For the third equilibrium, the charge balance equation is

Combining eq31 and eq32,

 

 

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What is the formula of the titration curve of a monoprotic strong acid versus a monoprotic strong base?

Consider a solution containing water, a strong acid (analyte) of concentration C_a and a strong base of concentration C_b with the following equilibria:

With reference to charge balance of the above equilibria, the sum of the number of moles of cations H⁺ and B⁺ must equal to that of anions A^– and OH^–. As the volume is of the solution is common to all ions,

Assume that the strong acid is fully ionised in water, at any point of the titration, the number of moles of A^– must equal to the initial number of moles of HA. As the change in volume of the solution is common to all ions,

where V_a and V_b are the volume of acid in the solution and the volume of base in the solution respectively.

Similarly, assuming that the strong base is fully ionised in water, at any point of the titration, the sum of the number of moles of B⁺ must equal to the number of moles of BOH added to the solution. As the change in volume of the solution is common to all ions,

Substitute eq2, eq3, K_w = [H⁺][OH^–] and [H⁺] = 10^-pH in eq1

Eq4 is the complete pH titration curve for a monoprotic strong acid versus monoprotic strong base system. We can input it in a mathematical software to generate a curve of pH against V_b. For example, if we titrate 10 cm³ of 0.200 M of HCl with 0.100 M of NaOH, we have the following:

To understand the sharp change in pH near the stoichiometric point versus the change in V_b, we assume that one drop of base is about 0.05 cm³ and substitute 19.95 cm³, 20.00 cm³ and 20.05 cm³ into eq4. The respective pH just before and just after the stoichiometric point (SP) are:

	One drop before SP	SP	One drop after SP
Volume, cm3	19.95	20.00	20.05
pH	3.78	7.00	10.22

The data shows that just two drops of base cause a drastic change of 6.44 in pH before and after the stoichiometric point. We can therefore use many different indicators that work within the range of the change in pH at the stoichiometric point to monitor the titration. Furthermore, we can derive the gradient equation for a monoprotic strong acid versus monoprotic strong base titration and investigate the inflexion point at pH 7 by rearranging eq4 as:

and finding by implicit differentiation to give:

At the stoichiometric point, C_bV_b – C_aV_a = 0 for a monoprotic acid-base reaction and the above equation becomes

Since [H⁺] = 10^-7 M at the stoichiometric point for a strong acid versus strong base titration, the denominator of the above equation is a very small number and hence the gradient of the pH versus V_b curve, or , is a very large value (computed to be 7238.2 cm^-3), making at angle of 89.99^o with the horizontal, i.e. a vertical line.

 

 

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What is the formula of the titration curve of a monoprotic strong acid versus a monoprotic weak base?

We make use of eq1 from a previous article except that C_b is now the concentration of a weak base. Since the weak base is partially ionised in water, at any point of the titration, the sum of the number of moles of BOH and B⁺ must equal to the number of moles of BOH if it were undissociated. As the change in volume of the solution is common to all ions,

where V_a and V_b are the volume of strong acid in the solution and the volume of weak base in the solution respectively. Substitute in eq9 where K_b is the dissociation constant of BOH and rearranging, we have,

Substitute eq10, eq2, K_w = [H⁺][OH^–] and [H⁺] = 10^-pH in eq1,

Eq11 is the complete pH titration curve for a monoprotic strong acid versus monoprotic weak base system. We can input it in a mathematical software to generate a curve of pH against V_a. For example, if we titrate 10 cm³ of 0.200 M of aqueous NH₃ (K_b = 1.8 x 10^-5) with 0.100 M of HCl, we have the following:

To understand the change in pH near the stoichiometric point versus the change in V_a, we assume that one drop of acid is about 0.05 cm³ and substitute 19.95 cm³, 20.00 cm³ and 20.05 cm³ into eq11. The respective pH just before and just after the stoichiometric point (SP) are:

	One drop before SP	SP	One drop after SP
Volume, cm3	19.95	20.00	20.05
pH	6.65	5.22	3.78

The data shows that two drops of acid cause a change of 2.87 in pH before and after the stoichiometric point. Since the change in pH at the stoichiometric point is smaller than that of a strong acid versus strong base titration, we can use fewer indicators that work within the range to monitor the titration. Lastly, we can derive the gradient equation for a strong acid to weak base titration and investigate the inflexion point at pH 5.22 by differentiating eq11 implicitly (see previous article), resulting in , i.e. a gradient that makes angle of –89.52^o with the horizontal.

 

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What is the formula of the titration curve of a monoprotic weak acid versus a monoprotic strong base?

We make use of eq1 from the previous article except that C_a is now the concentration of a weak acid. Since the weak acid is partially ionised in water, at any point of the titration, the sum of the number of moles of HA and A^– must equal to the number of moles of HA if it were undissociated. As the change in volume of the solution is common to all ions,

where V_a and V_b are the volume of weak acid in the solution and the volume of strong base in the solution respectively. Substitute in eq5 where K_a is the dissociation constant of HA and rearranging, we have,

Assuming that the strong base is fully ionised in water, at any point of the titration, the sum of the number of moles of B⁺ must equal to the number of moles of BOH added to the solution. As the change in volume of the solution is common to all ions,

Substitute eq6, eq7, K_w = [H⁺][OH^–] and [H⁺] = 10^-pH in eq1,

Eq8 is the complete pH titration curve for a monoprotic weak acid versus monoprotic strong base system. We can input it in a mathematical software to generate a curve of pH against V_b. For example, if we titrate 10 cm³ of 0.200 M of CH₃COOH (K_a = 1.75 x 10^-5) with 0.100 M of NaOH, we have the following:

To understand the change in pH near the stoichiometric point versus the change in V_b, we assume that one drop of base is about 0.05 cm³ and substitute 19.95 cm³, 20.00 cm³ and 20.05 cm³ into eq8. The respective pH just before and just after the stoichiometric point (SP) are:

	One drop before SP	SP	One drop after SP
Volume, cm3	19.95	20.00	20.05
pH	7.36	8.79	10.22

The data shows that two drops of base cause a change of 2.86 in pH before and after the stoichiometric point. Since the change in pH at the stoichiometric point is smaller than that of a strong acid versus strong base titration, we can use fewer indicators that work within the range to monitor the titration. Lastly, we can derive the gradient equation for a weak acid versus strong base titration and investigate the inflexion point at pH 8.79 by differentiating eq8 implicitly (see previous article), resulting in , i.e. a gradient that makes angle of 89.51^o with the horizontal.

 

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