3. On the titration error in donor/acceptor titrations of displacement and electronic transference reactions

Acid-base, complexation, and precipitation titrations have been extensively dealt with both in scientific and educational literature. Redox titrations, although being of primary importance, have received less attention [68]. Titration error has been the subject of several excellent papers, the emphasis being placed, however, mainly on acid-base [69–73] as well as on precipitation [74–79], and complex formation titration reactions [80, 81]. A number of papers dealing with the topic of titration error in redox titration [13, 24, 82, 83] have also been appeared. Redox titrations hold an important place in simple, fast, low-cost analysis of redox-activespecies [56]. In this chapter, a single equation is presented for symmetrical reactions for which the reduced and oxidized species of each half-reaction should be the same. Therefore the treatment is not strictly valid for such couples as Cr2O7 <sup>2</sup>�/Cr3<sup>þ</sup> or I2/I� [81].

Let SD be a particle donor (weak or strong), to be titrated with a particle acceptor TA. The analyst can choose the titrant and will always use a strong one to obtain better results. The systems implied in the titration reaction and their corresponding equilibrium constants will be given by

$$TA + bX \rightleftharpoons TD \qquad K\_T = \frac{[TD]}{[TA][X]^b} \tag{1}$$

$$SA + aX \rightleftharpoons SD \qquad K\_S = \frac{[SD]}{[SA][X]^a} \tag{2}$$

X being the particle transferred in the semi-reactions involved in the global titration reaction

$$bSD + aTA \rightleftharpoons bSA + aTD \tag{3}$$

to which corresponds the following equilibrium constant

least squares with novel weighting functions [56] for redox potentiometric data were also

The analogy between acid-base and redox-behavior, as particle exchange reactions (of protons

Titration error has been the subject of excellent publications, but the number of papers dealing with redox titration error is relatively scarce. The focus has been mainly put in acid-base, precipitation, and complex formation reaction titrations. In this chapter, an attempt is given to devise a titration error theory applicable to donor-acceptor titration of displacement and electronic transference (redox) reactions. The error will be formulated as a function of the titration parameters, in a hyperbolic sine expression way. As a matter of fact a detailed treatment of the analysis error issue is also carried out. The error analysis requires differentiating with respect to given variables leading at first sight to complex expressions, which at the

or electrons, respectively) [57, 58], has been interpreted through the years [59–67].

3. On the titration error in donor/acceptor titrations of displacement

Acid-base, complexation, and precipitation titrations have been extensively dealt with both in scientific and educational literature. Redox titrations, although being of primary importance, have received less attention [68]. Titration error has been the subject of several excellent papers, the emphasis being placed, however, mainly on acid-base [69–73] as well as on precipitation [74–79], and complex formation titration reactions [80, 81]. A number of papers dealing with the topic of titration error in redox titration [13, 24, 82, 83] have also been appeared. Redox titrations hold an important place in simple, fast, low-cost analysis of redox-activespecies [56]. In this chapter, a single equation is presented for symmetrical reactions for which the reduced and oxidized species of each half-reaction should be the same. Therefore the

Let SD be a particle donor (weak or strong), to be titrated with a particle acceptor TA. The analyst can choose the titrant and will always use a strong one to obtain better results. The systems implied in the titration reaction and their corresponding equilibrium constants will be given by

TA <sup>þ</sup> bX ⇌ TD KT <sup>¼</sup> ½ � TD

SA <sup>þ</sup> aX ⇌ SD KS <sup>¼</sup> ½ � SD

X being the particle transferred in the semi-reactions involved in the global titration reaction

<sup>2</sup>�/Cr3<sup>þ</sup> or I2/I� [81].

bSD þ aTA ⇌ bSA þ aTD ð3Þ

½ � TA ½ � <sup>X</sup> <sup>b</sup> <sup>ð</sup>1<sup>Þ</sup>

½ � SA ½ � <sup>X</sup> <sup>a</sup> <sup>ð</sup>2<sup>Þ</sup>

evaluated.

end finally appears to be compact.

122 Redox - Principles and Advanced Applications

and electronic transference reactions

treatment is not strictly valid for such couples as Cr2O7

$$K\_{\alpha\eta} = \frac{[SA]^b [TD]^a}{[SD]^b [TA]^a} = \frac{K\_T^a}{K\_S^b} \tag{4}$$

Proton free equilibria are assumed first in the presentation for the sake of clarity though redox equilibria that are independent of pH are relatively few. Effects of such factors as hydrogen ion concentration and complexing ligands may be easily incorporated [84] in the corresponding conditional constants. The particle X may be a proton, electron, cation, or an uncharged molecule. This notation is chosen to accentuate the analogy with that used in the description of acids and bases by the French School and other recognized authors [19, 29–31, 85–92]. Anyway, as TA is a strong particle acceptor, it must occur that KT >> KS.

From Eq. (3) it is readily seen that in any moment of the titration it holds that (through the entire concentration range)

$$b[TD] = a[SA] \tag{5}$$

On the other hand, we may define the (relative) titration error as

$$
\Delta T = \left(\frac{b\mathbb{C}\_T - a\mathbb{C}\_S}{a\mathbb{C}\_S}\right)\_{\text{end}} = T - 1\tag{6}
$$

where CT and CS are the analytical concentrations of titrant and analyte, respectively,

$$C\_T = [TA] + [TD] \tag{7}$$

$$\mathbf{C}\_{S} = [\mathbf{S}A] + [\mathbf{S}D] \tag{8}$$

and T, the fraction titrated, is defined as the ratio between the amount of titrant added and the initial amount of analyte at any moment of the titration. Thus, depending on the nature of the titration, ΔT might be either positive or negative. When the titration is carried out in the reverse order the same result is obtained, but the equation now bears a minus sign.

Therefore, from Eqs. (6)–(8), it follows that

$$
\Delta T = \left(\frac{b[TA] - a[SD]}{a\mathbb{C}\_S}\right)\_{\text{end}} \tag{9}
$$

By combining Eqs. (5) and (1), we have

$$\begin{bmatrix} TA \end{bmatrix} = \frac{a}{b} \frac{[SA]}{K\_T [X]^b} \tag{10}$$

and from Eq. (2) we obtain

$$\mathbb{E}[\mathbb{S}D] = \mathbb{K}\_{\mathbb{S}}[\mathbb{S}A][X]^{\mathbb{A}} \tag{11}$$

By substituting the values of ½ � TA and ½ � SD given by Eqs (10) and (11)

$$
\Delta T = \frac{[SA]}{\mathcal{C}\_S} \left( \frac{1}{K\_T [X]^b} - K\_S [X]^d \right) \tag{12}
$$

As the molarity fraction of the species ½ � SA is given by

$$f\_{\rm SA} = \frac{[\rm SA]}{\rm C\_{\rm S}} = \frac{1}{1 + [\rm X]^{\rm a} K\_{\rm S}} \tag{13}$$

Eq. (12) may be transformed into

$$
\Delta T = \frac{1}{1 + [X]^a K\_S} \left( \frac{1}{K\_T [X]^b} - K\_S [X]^a \right) \tag{14}
$$

By multiplying and dividing the right hand side of Eq. (14) by ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>ð</sup>KS=KT<sup>Þ</sup> <sup>p</sup> we get

$$
\Delta T = \frac{\sqrt{\frac{K\_s}{K\_T}}}{1 + [X]^d K\_S} \left( \frac{1}{\sqrt{K\_T K\_S} [X]^b} - \sqrt{K\_T K\_S} [X]^d \right) \tag{15}
$$

In the equivalence point, when the exact stoichiometric amount of titrant has been added, in addition to Eq. (15), the following condition is satisfied

$$a[SD] = b[TA] \tag{16}$$

and then, from Eqs. (1) and (2), it follows that when the exact stoichiometric amount of titrant has been added

$$\sqrt{K\_T K\_S} = \frac{1}{[X]\_{\text{eq}}^{(a+b)/2}} \tag{17}$$

$$pK\_{\rm eq} = \frac{\log K\_T + \log K\_S}{a+b} \tag{18}$$

and then, the potential at the equivalence point is independent of the concentration of the reactants and thus unaffected by dilution. Note, however, that Eq. (18) is not perfectly general, because the simple relation of Eq. (16) for reactants and products is not always valid. The species involved in the equilibria may be polynuclear. The p<sup>X</sup> in this instance varies with dilution.

By substituting Eq. (17) into Eq. (15)

On the Titration Curves and Titration Errors in Donor Acceptor Titrations of Displacement and Electronic… http://dx.doi.org/10.5772/intechopen.68750 125

$$
\Delta T = \frac{\sqrt{\frac{K\_s}{K\_T}}}{1 + [X]^a K\_S} [X]^{(a-b)/2} \left( \left( \frac{[X]\_{\text{eq}}}{[X]} \right)^{(a+b)/2} - \left( \frac{[X]}{[X]\_{\text{eq}}} \right)^{(a+b)/2} \right) \tag{19}
$$

A chemical error will arise because of lack of agreement between the end point and equivalence point. The difference between the end point and the equivalence point of a titration is the source of systematic error of determination. Taking into account that

$$
\Delta p X = p X\_{\text{end}} - p X\_{\text{eq}} \tag{20}
$$

and

½ �¼ SD KS½ � SA ½ � <sup>X</sup> <sup>a</sup> <sup>ð</sup>11<sup>Þ</sup>

<sup>ð</sup>KS=KT<sup>Þ</sup> <sup>p</sup> we get

ð12Þ

ð13Þ

ð14Þ

ð15Þ

ð17Þ

By substituting the values of ½ � TA and ½ � SD given by Eqs (10) and (11)

<sup>Δ</sup><sup>T</sup> <sup>¼</sup> <sup>1</sup> <sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup>

By multiplying and dividing the right hand side of Eq. (14) by ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ffiffiffiffi KS KT q

KS

<sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup>

As the molarity fraction of the species ½ � SA is given by

ΔT ¼

addition to Eq. (15), the following condition is satisfied

Eq. (12) may be transformed into

124 Redox - Principles and Advanced Applications

has been added

dilution.

By substituting Eq. (17) into Eq. (15)

<sup>Δ</sup><sup>T</sup> <sup>¼</sup> ½ � SA CS

> <sup>f</sup> SA <sup>¼</sup> ½ � SA CS

> > KS

1

KT½ � <sup>X</sup> <sup>b</sup> � KS½ � <sup>X</sup> <sup>a</sup> !

> <sup>¼</sup> <sup>1</sup> <sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup>

> > 1

1 ffiffiffiffiffiffiffiffiffiffiffi KTKS

In the equivalence point, when the exact stoichiometric amount of titrant has been added, in

and then, from Eqs. (1) and (2), it follows that when the exact stoichiometric amount of titrant

pKeq <sup>¼</sup> logKT <sup>þ</sup> log KS

and then, the potential at the equivalence point is independent of the concentration of the reactants and thus unaffected by dilution. Note, however, that Eq. (18) is not perfectly general, because the simple relation of Eq. (16) for reactants and products is not always valid. The species involved in the equilibria may be polynuclear. The p<sup>X</sup> in this instance varies with

½ � <sup>X</sup> <sup>ð</sup>aþbÞ=<sup>2</sup> eq

ffiffiffiffiffiffiffiffiffiffiffi KTKS <sup>p</sup> <sup>¼</sup> <sup>1</sup>

KT½ � <sup>X</sup> <sup>b</sup> � KS½ � <sup>X</sup> <sup>a</sup> !

<sup>p</sup> ½ � <sup>X</sup> <sup>b</sup> � ffiffiffiffiffiffiffiffiffiffiffi

!

KTKS <sup>p</sup> ½ � <sup>X</sup> <sup>a</sup>

a SD½ �¼ b TA ½ � ð16Þ

<sup>a</sup> <sup>þ</sup> <sup>b</sup> <sup>ð</sup>18<sup>Þ</sup>

KS

$$\sinh \mathbf{x} = \frac{e^{\mathbf{x}} - e^{-\mathbf{x}}}{2} \tag{21}$$

after some manipulation, the following expression may be easily obtained

$$
\Delta T = \frac{2\sqrt{\frac{K\_s}{K\_T}}}{1 + [X]^d K\_S} [X]^{(a-b)/2} \sinh\left(\ln 10(a+b)\frac{\Delta pX}{2}\right) \tag{22}
$$

or

$$
\Delta T = \mathcal{W}\sinh\left(\ln 10(a+b)\frac{\Delta pX}{2}\right) \tag{23}
$$

where the shape coefficient W is depending on the particle concentration in the end point titration when asymmetrical titrations are being considered

$$\mathcal{W} = \frac{2\sqrt{\frac{K\_s}{K\_T}}}{1 + [X]^a K\_S} [X]^{(a-b)/2} \tag{24}$$

Low values of the stoichiometric coefficients a, b, as well as low difference a�b values, and large K<sup>T</sup> values lead to lower errors.

In the vicinity of the equivalence point, þ1 >> ½X� a endKS and so

$$\mathcal{W} = 2\sqrt{\frac{K\_S}{K\_T}} [\mathbf{X}]\_{\text{end}}^{(a-b)/2} \tag{25}$$

In those cases in which the titration reaction is symmetrical, a ¼ b, and then

$$\mathcal{W} = 2\sqrt{\frac{K\_S}{K\_T}}\tag{26}$$

and the following formula is obtained for the titration error

$$
\Delta T = 2K\_{\rm eq}^{-1/(2a)} \sinh\left(\ln 10a\Delta pX\right) \tag{27}
$$

Note that the titration error may be formulated as a hyperbolic sine expression. Hyperbolic functions are of great worth in parameter estimation as shown by Asuero [93].

The methodology developed in this section of the chapter is going to be applied forward to some experimental situations characteristics of redox titration reactions. The calculations made by the hyperbolic sine method are checked with the procedure devised by de Levie [21, 26, 62]. A detailed treatment of systematic and random errors associated with the titration error is carried out in the following.

Note that the equations developed in this contribution can easily take into account the lateral reactions by using the corresponding lateral reaction coefficients and the conditional constants involved. A condition is however required, namely that the pH remains constant in the course of titration, which cannot always be achieved. Michalowski [4, 94–100] has given a general and definitive solution to the problem of redox equilibria, which does not require any restriction.

The beginnings of the rigorous GATES/Generalized Electron Balance (GEB) approach of Michalowski, which can be interpreted as a new natural law, dates back from 1992 to 1995. This approach has recently been shown repeatedly in the bibliography solving complex chemical problems and requires the use of nonlinear regression and a high level language such as MATLAB. The equations developed here, although very modest, have an obvious didactic interest and can be seen in the case of the redox equilibria as an alternative route to that given by de Levie [62].

## 4. Electronic transfer reactions

In the following, some titration curves of typical oxide-reduction reactions, involving Ce4<sup>þ</sup> and MnO4 � as titrant, are the subject of study. The transferred particle, the electron, takes the place of [X] in Eq. (22). The numerical values obtained by applying the hyperbolic sine method proposed in this contribution are checked against the method devised by de Levie [62], thus verifying the identity of the results in all cases.

## 4.1. Fe2<sup>þ</sup> titration curve with Ce4<sup>þ</sup>

The equilibrium constant and pe of the semi-reaction Ce4<sup>þ</sup> <sup>þ</sup> <sup>e</sup> <sup>¼</sup> Ce3<sup>þ</sup> (E0<sup>T</sup> <sup>¼</sup> 1.44 v) are given (Ce4<sup>þ</sup> is the acceptor) by

$$K\_{\Gamma} = \frac{\left[\text{Ce}^{3+}\right]}{\left[\text{Ce}^{4+}\right]\left[\text{e}\right]} \qquad pe = \log K\_{\Gamma} + \log \frac{\left[\text{Ce}^{4+}\right]}{\left[\text{Ce}^{3+}\right]} \tag{28}$$

Also, for the half-reaction Fe3<sup>þ</sup> <sup>þ</sup> <sup>e</sup> <sup>¼</sup> Fe2<sup>þ</sup> (E0<sup>S</sup> <sup>¼</sup> 0.68 v) (Fe2<sup>þ</sup> is the donor)

On the Titration Curves and Titration Errors in Donor Acceptor Titrations of Displacement and Electronic… http://dx.doi.org/10.5772/intechopen.68750 127

$$K\_S = \frac{\left[\text{Fe}^{2+}\right]}{\left[\text{Fe}^{3+}\right]\left[\text{e}\right]} \qquad pe = \log K\_S + \log \frac{\left[\text{Fe}^{3+}\right]}{\left[\text{Fe}^{2+}\right]} \tag{29}$$

The equilibrium constant of the overall reaction is expressed as

$$\text{Ce}^{4+} + \text{Fe}^{2+} = \text{Ce}^{3+} + \text{Fe}^{3+}, \qquad K = \frac{\left[\text{Ce}^{3+}\right]\left[\text{Fe}^{3+}\right]}{\left[\text{Ce}^{4+}\right]\left[\text{Fe}^{2+}\right]} = \frac{K\_T}{Ks} \tag{30}$$

From Eq. (22) taking into account that [X] ¼ e, and that a ¼ b ¼ 1

$$\Delta T = \frac{2\sqrt{K\_S/K\_T}}{1 + [\text{e}]K\_S} [\text{e}]^{(1-1)/2} \sinh\left(\ln 10 (1 + 1) \left(\frac{p e - p e\_{\text{eq}}}{2}\right)\right) = \frac{2\sqrt{K\_S/K\_T}}{1 + 10^{-p\varepsilon}K\_S} \sinh\left(\ln 10 (p e - p e\_{\text{eq}})\right) \tag{31}$$

From the Nernst equations applied to the two half-directions involved

$$E = E\_T^0 + 0.06 \log \frac{\left[\text{Ce}^{4+}\right]}{\left[\text{Ce}^{3+}\right]} \qquad \frac{E}{0.06} = \frac{E\_T^0}{0.06} + \log \frac{\left[\text{Ce}^{4+}\right]}{\left[\text{Ce}^{3+}\right]} \qquad pe = pe\_T^0 + \log \frac{\left[\text{Ce}^{4+}\right]}{\left[\text{Ce}^{3+}\right]} \qquad (32)$$

$$E = E\_T^0 + 0.06 \log \frac{\left[\text{Fe}^{3+}\right]}{\left[\text{Fe}^{2+}\right]} \qquad \frac{E}{0.06} = \frac{E\_T^0}{0.06} + \log \frac{\left[\text{Fe}^{3+}\right]}{\left[\text{Fe}^{2+}\right]} \qquad pe = pe\_S^0 + \log \frac{\left[\text{Fe}^{3+}\right]}{\left[\text{Fe}^{2+}\right]} \qquad (33)$$

and taking into account Eqs. (28) and (29) are reached

$$pe = \frac{E}{0.06} \tag{34}$$

$$
\log\_{T}^{0} = \frac{E\_T^0}{0.06} = \log K\_T \qquad \log K\_T = \frac{1.44}{0.06} = 24 \quad K\_T = 10^{24} \tag{35}
$$

$$
\rho v\_S^0 = \frac{E\_S^0}{0.06} = \log K\_S \qquad \log K\_S = \frac{0.68}{0.06} = 11.333 \quad K\_S = 2.154 \cdot 10^{11} \tag{36}
$$

The value of peeq is calculated, Eq. (32), from the expression

$$p e\_{\rm eq} = \frac{\log K\_T + \log K\_s}{a+b} = \frac{24 + 11.333}{1+1} = 17.667\tag{37}$$

whereby the potential at the point of equivalence is given by the expressions

$$E\_{\rm eq} = p e\_{\rm eq} \cdot 0.06 = 17.667 \cdot 0.06 = 1.06; \qquad E\_{\rm eq} = \frac{E\_T^0 + E\_S^0}{2} = \frac{1.44 + 0.68}{2} = 1.06 \tag{38}$$

Note that

<sup>Δ</sup><sup>T</sup> <sup>¼</sup> <sup>2</sup>K�1=ð2a<sup>Þ</sup>

functions are of great worth in parameter estimation as shown by Asuero [93].

carried out in the following.

126 Redox - Principles and Advanced Applications

by de Levie [62].

MnO4

4. Electronic transfer reactions

4.1. Fe2<sup>þ</sup> titration curve with Ce4<sup>þ</sup>

(Ce4<sup>þ</sup> is the acceptor) by

verifying the identity of the results in all cases.

KT <sup>¼</sup> Ce<sup>3</sup><sup>þ</sup> Ce<sup>4</sup><sup>þ</sup> ½ � <sup>e</sup>

Also, for the half-reaction Fe3<sup>þ</sup> <sup>þ</sup> <sup>e</sup> <sup>¼</sup> Fe2<sup>þ</sup> (E0<sup>S</sup> <sup>¼</sup> 0.68 v) (Fe2<sup>þ</sup> is the donor)

Note that the titration error may be formulated as a hyperbolic sine expression. Hyperbolic

The methodology developed in this section of the chapter is going to be applied forward to some experimental situations characteristics of redox titration reactions. The calculations made by the hyperbolic sine method are checked with the procedure devised by de Levie [21, 26, 62]. A detailed treatment of systematic and random errors associated with the titration error is

Note that the equations developed in this contribution can easily take into account the lateral reactions by using the corresponding lateral reaction coefficients and the conditional constants involved. A condition is however required, namely that the pH remains constant in the course of titration, which cannot always be achieved. Michalowski [4, 94–100] has given a general and definitive solution to the problem of redox equilibria, which does not require any restriction. The beginnings of the rigorous GATES/Generalized Electron Balance (GEB) approach of Michalowski, which can be interpreted as a new natural law, dates back from 1992 to 1995. This approach has recently been shown repeatedly in the bibliography solving complex chemical problems and requires the use of nonlinear regression and a high level language such as MATLAB. The equations developed here, although very modest, have an obvious didactic interest and can be seen in the case of the redox equilibria as an alternative route to that given

In the following, some titration curves of typical oxide-reduction reactions, involving Ce4<sup>þ</sup> and

The equilibrium constant and pe of the semi-reaction Ce4<sup>þ</sup> <sup>þ</sup> <sup>e</sup> <sup>¼</sup> Ce3<sup>þ</sup> (E0<sup>T</sup> <sup>¼</sup> 1.44 v) are given

pe <sup>¼</sup> log KT <sup>þ</sup> log Ce<sup>4</sup><sup>þ</sup>

Ce3<sup>þ</sup> <sup>ð</sup>28<sup>Þ</sup>

� as titrant, are the subject of study. The transferred particle, the electron, takes the place of [X] in Eq. (22). The numerical values obtained by applying the hyperbolic sine method proposed in this contribution are checked against the method devised by de Levie [62], thus

eq sinh ln 10 ð Þ aΔpX ð27Þ

$$T = \Delta T + 1\tag{39}$$

which allows us to calculate the titration curve (Figure 1)

$$p\mathbf{e} = f(T) \qquad E = f(T) \tag{40}$$

or the graph of the titration error (Figure 2)

$$
\Delta T = f(\Delta \text{pe}) \qquad \Delta T = f(E) \tag{41}
$$

The required calculations are detailed (from 0.62 to 0.82 v) in Table 1. Note that when T ¼ 0.5, the potential value, <sup>E</sup> <sup>¼</sup> 0.68 v, coincides with the normal potential of the Fe3<sup>þ</sup>/Fe2þ. Continuing the calculations would prove that when T ¼ 2, E ¼ 1.44 v, normal potential value of the Ce4þ/Ce3<sup>þ</sup> pair.

#### 4.2. Tl<sup>þ</sup> titration curve with Ce4<sup>þ</sup>

For the system Tl3<sup>þ</sup> <sup>þ</sup> 2e <sup>¼</sup> Tl<sup>þ</sup> (E0S <sup>¼</sup> 1.25 v) the equilibrium constant <sup>K</sup><sup>s</sup> and pe

$$K\_S = \frac{\left[\text{T}\text{I}^+\right]}{\left[\text{T}\text{I}^{3+}\right]\left[\text{e}\right]^2} \qquad pe = \frac{\log K\_S}{2} + \frac{1}{2}\log \frac{\left[\text{T}\text{I}^{3+}\right]}{\left[\text{T}\text{I}^+\right]} \tag{42}$$

Figure 1. Titration curve of Fe2<sup>þ</sup> with Ce4<sup>þ</sup> in acid medium (H2SO4 1 M).

On the Titration Curves and Titration Errors in Donor Acceptor Titrations of Displacement and Electronic… http://dx.doi.org/10.5772/intechopen.68750 129

Figure 2. Titration error diagram ΔT ¼ f(ΔpX).

T ¼ ΔT þ 1 ð39Þ

pe ¼ fðTÞ E ¼ fðTÞ ð40Þ

ΔT ¼ fðΔpeÞ ΔT ¼ fðEÞ ð41Þ

The required calculations are detailed (from 0.62 to 0.82 v) in Table 1. Note that when T ¼ 0.5, the potential value, <sup>E</sup> <sup>¼</sup> 0.68 v, coincides with the normal potential of the Fe3<sup>þ</sup>/Fe2þ. Continuing the calculations would prove that when T ¼ 2, E ¼ 1.44 v, normal potential value of the

<sup>2</sup> pe <sup>¼</sup> log KS

2 þ

1 2 log Tl<sup>3</sup><sup>þ</sup>

Tl<sup>þ</sup> <sup>ð</sup>42<sup>Þ</sup>

For the system Tl3<sup>þ</sup> <sup>þ</sup> 2e <sup>¼</sup> Tl<sup>þ</sup> (E0S <sup>¼</sup> 1.25 v) the equilibrium constant <sup>K</sup><sup>s</sup> and pe

KS <sup>¼</sup> Tl<sup>þ</sup> Tl3<sup>þ</sup> ½ � <sup>e</sup>

Figure 1. Titration curve of Fe2<sup>þ</sup> with Ce4<sup>þ</sup> in acid medium (H2SO4 1 M).

which allows us to calculate the titration curve (Figure 1)

or the graph of the titration error (Figure 2)

128 Redox - Principles and Advanced Applications

4.2. Tl<sup>þ</sup> titration curve with Ce4<sup>þ</sup>

Ce4þ/Ce3<sup>þ</sup> pair.

The overall reaction and its equilibrium constant are expressed as

$$\text{2Ce}^{4+} + \text{TI}^{+} = \text{2Ce}^{3+} + \text{TI}^{3+} \qquad K = \frac{\left[\text{Ce}^{3+}\right]^{2}\left[\text{TI}^{3+}\right]}{\left[\text{Ce}^{4+}\right]^{2}\left[\text{TI}^{+}\right]} = \frac{K\_{\text{T}}^{2}}{K\_{\text{S}}} \tag{43}$$

Applying the Nernst equation to the half-reaction Tl3<sup>þ</sup>/Tl<sup>þ</sup>

$$E = E\_S^0 + \frac{0.06}{2} \log \frac{\left[\text{T}^{3+}\right]}{\left[\text{T}\text{I}^+\right]} \qquad \frac{E}{0.06} = \frac{E\_S^0}{0.06} + \frac{1}{2} \log \frac{\left[\text{T}\text{I}^{3+}\right]}{\left[\text{T}\text{I}^+\right]} \qquad pe = pe\_S^0 + \frac{1}{2} \log \frac{\left[\text{T}\text{I}^{3+}\right]}{\left[\text{T}\text{I}^+\right]} \qquad (44)$$

$$p e\_S^0 = \frac{E\_S^0}{0.06} = \frac{\log K\_S}{2} \qquad \log K\_S = 2 \frac{1.25}{0.06} = 41.666 \qquad K\_S = 4.642 \cdot 10^{41} \tag{45}$$

$$p\text{e}\_{\text{eq}} = \frac{\log Kr + \log K\_s}{a+b} = \frac{24 + 41.666}{1+2} = 21.889\tag{46}$$

$$E\_{\rm eq} = p e\_{\rm eq} \cdot 0.06 = 21.889 \cdot 0.06 = 1.313; \quad E\_{\rm eq} = \frac{E\_T^0 + 2E\_S^0}{1 + 2} = \frac{1.44 + 21.25}{3} = 1.313 \qquad (47)$$

The values of KS (Eq. 45), KT (Eq. 35), pe (Eq. 34) and peeq (Eq. 46) can be replaced in Eq. (22), taking into account that a ¼ 2 and b ¼ 1


$$\begin{split} \Delta T &= \frac{2\sqrt{K\_S/K\_T}}{1+\left[\mathbf{e}\right]^2 K\_S} \left| \mathbf{e} \right|^{\frac{2}{T}} \sinh\left(\ln\left10(2+1)\left(\frac{pe-pe\_{eq}}{2}\right)\right) \\\\ &= \frac{2\sqrt{K\_S/K\_T}}{1+10^{-2p\epsilon} K\_S} \sqrt{e} \sinh\left(\frac{3}{2} \ln 10(pe-21.889)\right) \end{split} \tag{48}$$

On the Titration Curves and Titration Errors in Donor Acceptor Titrations of Displacement and Electronic… http://dx.doi.org/10.5772/intechopen.68750 131

Figure 3. Titration curves of several redox systems with Ce4<sup>þ</sup> as a titrant.
