5.3. Titration in HCl (C, V) ) NaIO (C0, V0) system

V<sup>0</sup> = 10 mL of C<sup>0</sup> = 0.01 mol/L NaIO is titrated with C = 0.1 mol/L HCl. The related curves are presented in Figures 5a-d. Initially, the reaction

Generalized Electron Balance (GEB) as the Law of Nature in Electrolytic Redox Systems http://dx.doi.org/10.5772/intechopen.69341 29

Figure 4. The (a) ΔpH/ΔΦ = (pHj+1 � pHj)/(Φj+1 � Φj), (b) ΔE/ΔΦ = (Ej+1 � Ej)/(Φj+1 � Φj) vs. Φ = (Φ<sup>j</sup> + Φj+1)/2 relationships for the NaOH (C, V) ) HIO (C0, V0) system.

$$\text{'}\text{'}\text{IO}^{-1} = \text{IO}\_3^{-1} + 2\text{I}^{-1} \tag{55}$$

and then, the reaction

5HIO <sup>þ</sup> OH�<sup>1</sup> <sup>¼</sup> <sup>2</sup>ðI2ðsÞ, I2Þ þ IO3

(a) (b) (c)

Figure 3. Plots of (a) E = E(Φ), (b) pH = pH(Φ), and (c) speciation curves for NaOH (C, V) ) HIO (C0, V0) system at V<sup>0</sup> = 10,

3HIO <sup>þ</sup> 3OH�<sup>1</sup> <sup>¼</sup> IO3

] is accompanied by an increase in [I<sup>3</sup>

]; all them disproportionate

3HIO <sup>þ</sup> 3OH�<sup>1</sup> <sup>¼</sup> IO3

curves presented in Figures 3a,b, occurring at Φ = 1. For Φ > 1, we have [I�<sup>1</sup>

Þ þ <sup>6</sup>OH�<sup>1</sup> <sup>¼</sup> IO3

 I<sup>2</sup>ðsÞ, I<sup>2</sup> þ I

In the following, at Φ ca. 0.20–0.22, a pronounced increase in [I

�1

The increase in [I

C<sup>0</sup> = 0.01, C = 0.1.

lowering of [I2] and [I<sup>3</sup>

�1

28 Redox - Principles and Advanced Applications

�1

3ðI<sup>2</sup>ðsÞ, I2, I<sup>3</sup>

5.3. Titration in HCl (C, V) ) NaIO (C0, V0) system

presented in Figures 5a-d. Initially, the reaction

�<sup>1</sup> <sup>þ</sup> 3H2O (50)

�<sup>1</sup> (52)

�<sup>1</sup> <sup>þ</sup> 3H2O (54)

]/[IO3 �1

] = 2, i.e. the

�<sup>1</sup> <sup>þ</sup> 2I�<sup>1</sup> <sup>þ</sup> 3H2O (51)

] occurs, as a result of reaction

�<sup>1</sup> <sup>þ</sup> 3H2O (53)

�1

�1 ]

�<sup>1</sup> þ ð5, <sup>5</sup>, <sup>8</sup>Þ<sup>I</sup>

�<sup>1</sup> <sup>¼</sup> <sup>I</sup><sup>3</sup>

�<sup>1</sup> <sup>þ</sup> <sup>2</sup><sup>I</sup>

This leads to the gradual disappearance of I2(s) (which is ultimately ended at Φ = 0.5347) and

At [I2(s)] > 0, we have [I2] = <sup>s</sup> = const; <sup>s</sup> = 1.33�10�<sup>3</sup> mol/L is the solubility of <sup>I</sup>2(s) in water, at 20�C. Finally, the disproportionation of HIO, affected by NaOH, can be expressed by the equation

Note that the stoichiometry of the reaction (54) is 3 : 3 = 1 : 1, which corresponds to the jump on the

stoichiometry of the products of reaction (54) equals to 1 : 2. The jumps on the curves E = E(Φ) and pH = pH(Φ) (Figures 3a,b) occur at Φ ca. 0.2 (which corresponds to the stoichiometry 1:5 of the reaction (50)) and at Φ ca. 1 (which corresponds to the stoichiometry 3:3 = 1:1 of the reaction (54)); the maxima on the corresponding derivative curves in Figures 4a,b fit the stoichiometric ratios.

V<sup>0</sup> = 10 mL of C<sup>0</sup> = 0.01 mol/L NaIO is titrated with C = 0.1 mol/L HCl. The related curves are

$$\text{25IO}^{-1} + 4\text{H}^{+1} = 2\text{I}\_2 + \text{IO}\_3^{-1} + 2\text{H}\_2\text{O} \tag{56}$$

occurs. Then I �<sup>1</sup> from Eq. (55) and I<sup>2</sup> from Eq. (56) form I<sup>3</sup> �<sup>1</sup> in the reaction I<sup>2</sup> + I �<sup>1</sup> = I<sup>3</sup> �<sup>1</sup> and [I3 �1 ] increases. At Φ = 0.4654, I2(s) appears as the solid phase

$$\text{5IO}^{-1} + \text{4H}^{+1} = \text{2I}\_{\text{2(s)}} + \text{IO}\_3^{-1} + \text{2H}\_2\text{O} \tag{57}$$

At [I2(s)] > 0, we have [I2] = const. The increase in [Cl�<sup>1</sup> ], resulted from addition of HCl, causes an increase in [I2Cl�<sup>1</sup> ], and—to a lesser extent—the increase in [ICl2 �1 ] and [ICl]. The addition of HCl lowers pH of the solution, and then [HIO] becomes larger than [IO�<sup>1</sup> ]; [HIO3] also increases. In effect, the summary concentration [HIO] + [IO�<sup>1</sup> ] after addition of an excess of HCl is higher than in the starting NaIO solution.

In the algorithm, we have allowed the participation of Cl�<sup>1</sup> ions from HCl solution in the redox reaction. However, the concentration of Cl2 and HClO as the main products of Cl�<sup>1</sup> oxidation (Figure 5b) is quite negligible. This way, one can state that the Cl�<sup>1</sup> ions practically do not participate in the redox reaction as a reducing agent. From linear combination of reactions (55) and

$$\text{5I}^{-1} + \text{IO}\_3^{-1} + \text{6H}^{+1} = \text{3I}\_{\text{2}(s)} + \text{3H}\_2\text{O} \tag{58}$$

(multiplication by 5 and 2 respectively), cancellations and division by 3, we get the reaction

$$4\,\text{5IO}^{-1} + 4\text{H}^{+1} = \text{IO}\_3^{-1} + 2\text{I}\_{2(s)} + 2\text{H}\_2\text{O} \tag{59}$$

with stoichiometry 4:5 = 0.8, which corresponds to Φ = 4:5 = 0.8, where the inflection point on the curves in Figures 5c and d are observed. The I�<sup>1</sup> and I3 �<sup>1</sup> ions are consumed in reactions (58) and

Figure 5. The speciation curves for indicated (a) iodine and (b) chlorine species Xi zi and (c) E = E(Φ), (d) pH = pH(Φ) functions for HCl (C, V) ) NaIO (C0, V0) system; V<sup>0</sup> = 10, C<sup>0</sup> = 0.01, C = 0.1.

$$\rm{6I\_3}^{-1} + \rm{IO\_3}^{-1} + \rm{6H}^{+1} = \rm{8I\_{2(s)}} + \rm{3H\_2O} \tag{60}$$

See Figure 5a.

Computer program for HCl (C, V) ) NaIO (C0, V0) system

function F = System\_NaIO\_HCl(x);

%NaIO<-HCl

% Titration of V0 mL of NaIO (C0) with V mL HCl (C).

global V Vmin Vstep Vmax V0 C C0 fi H OH pH E Kw pKw A aa

global I I3 I2 I2s HIO IO HI5O3 I5O3 H5I7O6 H4I7O6 H3I7O6 Na

global logI logI3 logI2 logI2s logHIO logIO logHI5O3 logI5O3 logH5I7O6

global logH4I7O6 logH3I7O6 logNa

global Cl Cl2 HClO ClO HCl3O2 Cl3O2 Cl4O2 Cl5O3 Cl7O4 I2Cl ICl ICl2 global logCl logCl2 logHClO logClO logHCl3O2 logCl3O2 logCl4O2 logCl5O3 global logCl7O4 logI2Cl logICl logICl2 pI pCl

E=x(1);

pH=x(2);

pI=x(3);

pCl=x(4);

H=10.^-pH;

pKw=14;

Kw=10.^-14;

OH=Kw./H;

I=10.^-pI;

Cl=10.^-pCl;

A=16.92;

ZCl=17;

ZI=53;

5I3

functions for HCl (C, V) ) NaIO (C0, V0) system; V<sup>0</sup> = 10, C<sup>0</sup> = 0.01, C = 0.1.

Computer program for HCl (C, V) ) NaIO (C0, V0) system

% Titration of V0 mL of NaIO (C0) with V mL HCl (C).

global V Vmin Vstep Vmax V0 C C0 fi H OH pH E Kw pKw A aa global I I3 I2 I2s HIO IO HI5O3 I5O3 H5I7O6 H4I7O6 H3I7O6 Na

global logI logI3 logI2 logI2s logHIO logIO logHI5O3 logI5O3 logH5I7O6

See Figure 5a.

%NaIO<-HCl

function F = System\_NaIO\_HCl(x);

30 Redox - Principles and Advanced Applications

�<sup>1</sup> <sup>þ</sup> IO3

Figure 5. The speciation curves for indicated (a) iodine and (b) chlorine species Xi

�<sup>1</sup> <sup>þ</sup> 6Hþ<sup>1</sup> <sup>¼</sup> <sup>8</sup>I<sup>2</sup>ðs<sup>Þ</sup> <sup>þ</sup> 3H2O (60)

zi and (c) E = E(Φ), (d) pH = pH(Φ)

I2=I.^2.\*10.^(2.\*A.\*(E-0.621));

I3=I.^3.\*10.^(2.\*A.\*(E-0.545));

IO=I.\*10.^(2.\*A.\*(E-0.49)+2.\*pH-2.\*pKw);

HIO=IO.\*10.^(10.6-pH);

I5O3=I.\*10.^(6.\*A.\*(E-1.08)+6.\*pH);

HI5O3=I5O3.\*10.^(0.79-pH);

H5I7O6=I.\*10.^(8.\*A.\*(E-1.24)+7.\*pH);

H4I7O6=H5I7O6.\*10.^(-3.3+pH);

H3I7O6=I.\*10.^(8.\*A.\*(E-0.37)+9.\*pH-9.\*pKw);

Cl2=Cl.^2.\*10.^(2.\*A.\*(E-1.359));

ClO=Cl.\*10.^(2.\*A.\*(E-0.88)+2.\*pH-2.\*pKw);

HClO=ClO.\*10.^(7.3-pH);

Cl3O2=Cl.\*10.^(4.\*A.\*(E-0.77)+4.\*pH-4.\*pKw); HCl3O2=Cl.\*10.^(4.\*A.\*(E-1.56)+3.\*pH); Cl4O2=Cl.\*10.^(5.\*A.\*(E-1.5)+4.\*pH); Cl5O3=Cl.\*10.^(6.\*A.\*(E-1.45)+6.\*pH); Cl7O4=Cl.\*10.^(8.\*A.\*(E-1.38)+8.\*pH); I2Cl=I2.\*10.^(0.2-pCl); ICl=I2.^0.5.\*10.^(A.\*(E-1.105)-pCl); ICl2=ICl.\*10.^(2.2-pCl); Na=C0.\*V0./(V0+V);

if I2>1.33e-3

I2s=I2-1.33e-3;

I2=1.33e-3;

aa=1;

```
else
```
aa=0;

I2s=0;

end;

%Charge balance

F=[(H-OH+Na-I-I3-IO-I5O3-H4I7O6-2.\*H3I7O6-Cl-ClO-Cl3O2-Cl5O3-Cl7O4…


%Concentration balance for I

(I+3.\*I3+2.\*(I2+aa.\*I2s)+HIO+IO+HI5O3+I5O3+H5I7O6+H4I7O6+H3I7O6+…

2.\*I2Cl+ICl+ICl2-C0.\*V0./(V0+V));

%Concentration balance for Cl

```
(Cl+2.*Cl2+HClO+ClO+HCl3O2+Cl3O2+Cl4O2+Cl5O3+Cl7O4+I2Cl+ICl…
```
+2.\*ICl2-C.\*V./(V0+V));

%Electron balance

((ZI+1).\*I+(3.\*ZI+1).\*I3+2.\*ZI.\*(I2+aa.\*I2s)+(ZI-1).\*(HIO+IO)…

+(ZI-5).\*(HI5O3+I5O3)+(ZI-7).\*(H5I7O6+H4I7O6+H3I7O6)+(ZCl+1).\*Cl+…

2.\*ZCl.\*Cl2+(ZCl-1).\*(HClO+ClO)+(ZCl-3).\*(HCl3O2+Cl3O2)…

+(ZCl-4).\*Cl4O2+(ZCl-5).\*Cl5O3+(ZCl-7).\*Cl7O4+(2.\*ZI+ZCl+1).\*I2Cl+…

```
(ZI+ZCl).*ICl+(ZI+2.*ZCl+1).*ICl2…
```

logI=log10(I);

Cl3O2=Cl.\*10.^(4.\*A.\*(E-0.77)+4.\*pH-4.\*pKw);

HCl3O2=Cl.\*10.^(4.\*A.\*(E-1.56)+3.\*pH);

Cl4O2=Cl.\*10.^(5.\*A.\*(E-1.5)+4.\*pH); Cl5O3=Cl.\*10.^(6.\*A.\*(E-1.45)+6.\*pH); Cl7O4=Cl.\*10.^(8.\*A.\*(E-1.38)+8.\*pH);

32 Redox - Principles and Advanced Applications

ICl=I2.^0.5.\*10.^(A.\*(E-1.105)-pCl);

I2Cl=I2.\*10.^(0.2-pCl);

ICl2=ICl.\*10.^(2.2-pCl);

Na=C0.\*V0./(V0+V);

I2s=I2-1.33e-3;

%Charge balance

%Concentration balance for I

2.\*I2Cl+ICl+ICl2-C0.\*V0./(V0+V));

%Concentration balance for Cl

+2.\*ICl2-C.\*V./(V0+V));

%Electron balance


F=[(H-OH+Na-I-I3-IO-I5O3-H4I7O6-2.\*H3I7O6-Cl-ClO-Cl3O2-Cl5O3-Cl7O4…

(I+3.\*I3+2.\*(I2+aa.\*I2s)+HIO+IO+HI5O3+I5O3+H5I7O6+H4I7O6+H3I7O6+…

(Cl+2.\*Cl2+HClO+ClO+HCl3O2+Cl3O2+Cl4O2+Cl5O3+Cl7O4+I2Cl+ICl…

+(ZI-5).\*(HI5O3+I5O3)+(ZI-7).\*(H5I7O6+H4I7O6+H3I7O6)+(ZCl+1).\*Cl+…

((ZI+1).\*I+(3.\*ZI+1).\*I3+2.\*ZI.\*(I2+aa.\*I2s)+(ZI-1).\*(HIO+IO)…

I2=1.33e-3;

aa=1;

aa=0; I2s=0;

else

end;

if I2>1.33e-3

logI3=log10(I3);

logI2=log10(I2);

logI2s=log10(I2s);

logHIO=log10(HIO);

logIO=log10(IO);

logHI5O3=log10(HI5O3);

logI5O3=log10(I5O3);

logH5I7O6=log10(H5I7O6);

logH4I7O6=log10(H4I7O6);

logH3I7O6=log10(H3I7O6);

logCl=log10(Cl);

logCl2=log10(Cl2);

logHClO=log10(HClO);

logClO=log10(ClO);

logHCl3O2=log10(HCl3O2);

logCl3O2=log10(Cl3O2);

logCl4O2=log10(Cl4O2);

logCl5O3=log10(Cl5O3);

logCl7O4=log10(Cl7O4);

logI2Cl=log10(I2Cl);

logICl=log10(ICl);

logICl2=log10(ICl2);

logNa=log10(Na);

% The end of program

A remark: Some notations applied in this program are as follows: [HIO3] ! HI5O3, [H3IO6 �2 ] ! H3I7O6, [ClO2 �1 ] ! Cl3O2, [ClO2] ! Cl4O2, etc.

#### 5.4. Disproportionation in static redox systems

In disproportionating redox systems, it is advisable to check an effect of dilution of the corresponding solutes on values of the corresponding variables. It is advisable to plot the desired relationships in the figures with the values pC = –logC on the abscissa, where C is the concentration [mol/L] of the solute considered. Formally, the related plots correspond to presentation, in extended logarithmic scale, the results of titration of the initial solution of this solute with pure water as a titrant T.

The static systems with C mol/L solutions of (1) HIO and (2) NaIO are shown graphically in Figures 6a-c and 7a-c, where pH, E and log ½Xi zi � values related to different concentrations C of the corresponding solutes with the values pC = �logC on the abscissa.

#### 5.4.1. C mol/L HIO

As results from the speciation diagram in Figure 6c, in more concentrated HIO solutions, i.e. at lower pC values, the predominating reactions are as follows: 5HIO = 2(I2(s), I2) + IO3 �<sup>1</sup> + 2H2O + H+1 and 5HIO = 2(I2(s), I2) + HIO3 + 2H2O; solubility of I2(s) s = 1.33∙10�<sup>3</sup> mol/L (25�C). At further dilution of HIO, the reaction 3HIO = 2I�<sup>1</sup> + IO3 �<sup>1</sup> + 3H+1 occurs in an increasing degree. This change in disproportionation scheme, more significant at pC 4–5, resulted in a change of the shapes of the plots: E = E(pC) (Figure 6a) and pH = pH(pC) (Figure 6b).

#### 5.4.2. C mol/L NaIO

The disproportionation of IO�<sup>1</sup> introduced by NaIO proceeds mainly according to the scheme 3HIO = IO3 �<sup>1</sup> + 2I�<sup>1</sup> + 3H+1 (see Figure 7c), where [I�<sup>1</sup> ]/[IO3 �1 ] ffi 2. Concentration of I2 is lower than 10�<sup>6</sup> mol/L and then solid I2(s) is not formed.

Figure 6. The plots of (a) E vs. pC, (b) pH vs. pC relationships and (c) speciation curves for indicated iodine species Xi zi in C mol/L HIO.

Generalized Electron Balance (GEB) as the Law of Nature in Electrolytic Redox Systems http://dx.doi.org/10.5772/intechopen.69341 35

Figure 7. The functions: (a) E = E(pC), (b) pH = pH(pC) and (c) speciation curves for indicated iodine species Xi zi plotted for C mol/L NaIO.

Figure 8. Plots for the KBrO3 (C, V) ) NaBr (C0, V0) system: (a) speciation diagram, (b) E = E(Φ), and (c) pH = pH(Φ) curves; V<sup>0</sup> = 10, C<sup>0</sup> = 0.01, C = 0.1.

#### 6. Symproportionating systems

A remark: Some notations applied in this program are as follows: [HIO3] ! HI5O3, [H3IO6

In disproportionating redox systems, it is advisable to check an effect of dilution of the corresponding solutes on values of the corresponding variables. It is advisable to plot the desired relationships in the figures with the values pC = –logC on the abscissa, where C is the concentration [mol/L] of the solute considered. Formally, the related plots correspond to presentation, in extended logarithmic scale, the results of titration of the initial solution of this solute with pure

The static systems with C mol/L solutions of (1) HIO and (2) NaIO are shown graphically in

zi

As results from the speciation diagram in Figure 6c, in more concentrated HIO solutions, i.e. at

H+1 and 5HIO = 2(I2(s), I2) + HIO3 + 2H2O; solubility of I2(s) s = 1.33∙10�<sup>3</sup> mol/L (25�C). At

This change in disproportionation scheme, more significant at pC 4–5, resulted in a change of

The disproportionation of IO�<sup>1</sup> introduced by NaIO proceeds mainly according to the scheme

Figure 6. The plots of (a) E vs. pC, (b) pH vs. pC relationships and (c) speciation curves for indicated iodine species Xi

]/[IO3 �1

lower pC values, the predominating reactions are as follows: 5HIO = 2(I2(s), I2) + IO3

the shapes of the plots: E = E(pC) (Figure 6a) and pH = pH(pC) (Figure 6b).

�<sup>1</sup> + 2I�<sup>1</sup> + 3H+1 (see Figure 7c), where [I�<sup>1</sup>

� values related to different concentrations C of

�<sup>1</sup> + 3H+1 occurs in an increasing degree.

] ffi 2. Concentration of I2 is lower

] ! Cl3O2, [ClO2] ! Cl4O2, etc.

the corresponding solutes with the values pC = �logC on the abscissa.

! H3I7O6, [ClO2

water as a titrant T.

5.4.1. C mol/L HIO

5.4.2. C mol/L NaIO

3HIO = IO3

C mol/L HIO.

�1

34 Redox - Principles and Advanced Applications

5.4. Disproportionation in static redox systems

Figures 6a-c and 7a-c, where pH, E and log ½Xi

further dilution of HIO, the reaction 3HIO = 2I�<sup>1</sup> + IO3

than 10�<sup>6</sup> mol/L and then solid I2(s) is not formed.

�2 ]

�<sup>1</sup> + 2H2O +

zi in

### 6.1. Titration in KBrO3 (C, V) ) NaBr (C0, V0) system

In this case, symproportionation practically does not occur (Figure 8a); concentration of HBrO, as the major product formed in the symproportionation reaction

$$\rm{BrO\_3}^{-1} + 2\rm{Br}^{-1} + 3\rm{H}^{+1} = \rm{3HBrO} \tag{61}$$

is ca. 10�<sup>6</sup> mol/L. The potential E increases monotonically (Figure 8b), whereas pH first increases, passes through maximum and then decreases (Figure 8c). The relevant pH and E changes are small. Binding the H+1 ions in reaction (61) causes a weakly alkaline reaction (Figure 8b).

#### 6.2. Titration in KBrO3 (C, V) ) NaBr (C0)+H2SO4 (C03) V0 system

The stoichiometry 1:5, i.e. Φeq = 0.2, stated for C<sup>03</sup> values indicated at the curves plotted in Figure 9 (column a), results from reaction

Figure 9. Plots for the KBrO3 (C, V) ) NaBr (C0)+H2SO4 (C03) V0 system: speciation diagrams (column a); E = E(Φ) (column b) and pH = pH(Φ) (column c), at V<sup>0</sup> = 100, C<sup>0</sup> = 0.01, C = 0.1 and indicated C<sup>03</sup> [mol/L] values for H2SO4.

$$\rm{BrO\_3}^{-1} + \rm{5Br^{-1}} + \rm{6H^{+1}} = \rm{3Br\_2} + \rm{3H\_2O} \tag{62}$$

For Φ > 0.2, an increase of efficiency of the competitive reaction

$$\text{\textbullet 2Br}^{-1} + \text{BrO}\_3^{-1} + \text{3H}^{+1} = \text{\textbullet \textbullet BrO} \tag{63}$$

is noted. A growth of C<sup>03</sup> value causes a small extension of the potential range in the jump region, on the side of higher E values (Figure 9, column b). With an increase in the C<sup>03</sup> value, the graphs of pH vs. Φ resemble two almost straight-line segments intersecting at Φeq = 0.2 (Figure 9, column c). However, the pH ranges covered by the titration curves are gradually narrowed (Figure 9, column c).

#### 7. Redox systems with two electron-active elements

In the redox systems considered above, one electron-active element was involved in disproportionation or symproportionation reactions affected by NaOH (in dynamic systems) or water (in static systems). In the HCl (C, V) ) NaIO (C0, V0) system, where HCl was used as disproportionating reagent, the possibility of oxidation of Cl�<sup>1</sup> ions was allowed a priori, in which HCl was used as a disproportionating reagent, i.e. chlorine in HCl was treated as an electron-active element. It turned out, however, that the oxidation of Cl�<sup>1</sup> occurred only in an extremely small extent, ca. 10�16/10�<sup>2</sup> ≈ 10�<sup>14</sup> part of Cl�<sup>1</sup> ions was oxidized. So, we can consider that it is virtually the disproportionation of IO�<sup>1</sup> ions originating from the NaIO.

In this section, we consider the systems where two electron-active elements are factually present.
