3. Linear dependence of algebraic equations and transformation into identity

Linear combination of algebraic equations plays a fundamental/decisive role in thermodynamics of electrolytic systems, considered according to GATES [1]. An elementary information related to linear combination, perceived (mainly) from mathematical/algebraic viewpoint, can be found in Ref. [23]. It should be noted that the results of simple addition, i.e. f<sup>1</sup> + f<sup>2</sup> and simple subtraction, i.e. f<sup>1</sup> � f<sup>2</sup> or f<sup>2</sup> � f1, are also linear combinations of any equations f<sup>1</sup> and f2.

We refer here to the problem of linear dependency of balances, analogous to the problem of dependency of linear equations, considered in elementary algebra, see the picture below. In this context, the general property of linear independency, inherently involved with redox systems, will be emphasized.

(1) For the beginning, let us take the set of linear equations:

completed by the linear combination of these equations, i.e.

$$\begin{aligned} \mathbf{c}\_1(a\_{11}\mathbf{x}\_1 + a\_{12}\mathbf{x}\_2 + a\_{13}\mathbf{x}\_3) + \mathbf{c}\_2(a\_{21}\mathbf{x}\_1 + a\_{22}\mathbf{x}\_2 + a\_{23}\mathbf{x}\_3) \\ \equiv (\mathbf{c}\_1a\_{11} + \mathbf{c}\_2a\_{21})\mathbf{x}\_1 + (\mathbf{c}\_1a\_{12} + \mathbf{c}\_2a\_{22})\mathbf{x}\_2 + (\mathbf{c}\_1a\_{13} + \mathbf{c}\_2a\_{23})\mathbf{x}\_3 = \mathbf{c}\_1b\_1 + \mathbf{c}\_2b\_2. \end{aligned} \tag{5}$$

Applying the matrix algebra, we see that the determinant

$$D = \begin{vmatrix} a\_{11} & a\_{12} & a\_{13} \\ & a\_{21} & a\_{22} & a\_{23} \\ c\_1a\_{11} + c\_2a\_{21} & c\_1a\_{12} + c\_2a\_{22} & c\_1a\_{13} + c\_2a\_{23} \end{vmatrix} \tag{6}$$

has zero value

$$D = c\_1 \cdot \begin{vmatrix} a\_{11} & a\_{12} & a\_{13} \\ a\_{21} & a\_{22} & a\_{23} \\ a\_{11} & a\_{12} & a\_{13} \end{vmatrix} + c\_2 \cdot \begin{vmatrix} a\_{11} & a\_{12} & a\_{13} \\ a\_{21} & a\_{22} & a\_{23} \\ a\_{21} & a\_{22} & a\_{23} \end{vmatrix} = c\_1 \cdot 0 + c\_2 \cdot 0 = 0 \tag{7}$$

irrespectively on the c<sup>1</sup> and c<sup>2</sup> values; at D = 0, calculation of x1, x<sup>2</sup> and x<sup>3</sup> is then impossible.

(2) Let us consider the set of <sup>G</sup> + 1 equations: <sup>f</sup>g(x) = <sup>ϕ</sup>g(x) � <sup>b</sup><sup>g</sup> = 0, where <sup>g</sup> = 0,1,…, <sup>G</sup>, <sup>x</sup><sup>T</sup> <sup>=</sup> (x1,…, xI) – transposed (T ) vector x, composed of independent (scalar) variables xi (i e < 1, I>); agi, bg ɛ R are independent (explicitly) on x. After multiplying the equations by the numbers ω<sup>g</sup> <sup>e</sup> <sup>R</sup> and addition of the resulting equations, we get the linear combination <sup>X</sup><sup>G</sup> <sup>g</sup>¼<sup>0</sup> <sup>ω</sup><sup>g</sup> � <sup>f</sup> <sup>g</sup>ðxÞ ¼ <sup>0</sup> <sup>⇔</sup> <sup>X</sup><sup>G</sup> <sup>g</sup>¼<sup>0</sup> <sup>ω</sup><sup>g</sup> � <sup>ϕ</sup>gðxÞ ¼ <sup>X</sup><sup>G</sup> <sup>g</sup>¼<sup>0</sup> <sup>ω</sup><sup>g</sup> � bg of the basic equations.

Formation of linear combinations is applicable to check the linear dependency or independency of the balances. A very useful/effective manner for checking/stating the linear dependence of the balances is the transformation of an appropriate system of equations to the identity, 0 = 0 [2, 23]. For this purpose, we will try, in all instances, to obtain the simplest form of the linear combination. To facilitate these operations, carried out by cancellation of the terms on the left and right sides of equations after changing sides of these equations, we apply the equivalent forms of the starting equations fg(x) = 0:

$$f\_{\mathcal{S}}(\mathbf{x}): \quad \varphi\_{\mathcal{S}}(\mathbf{x}) - b\_{\mathcal{S}} = 0 \Leftrightarrow \varphi\_{\mathcal{S}}(\mathbf{x}) = b\_{\mathcal{S}} \Leftrightarrow \ -f\_{\mathcal{S}}(\mathbf{x}): \quad -\varphi\_{\mathcal{S}}(\mathbf{x}) = -b\_{\mathcal{S}} \Leftrightarrow b\_{\mathcal{S}} = \varphi\_{\mathcal{S}}(\mathbf{x}) \tag{8}$$

In this notation, fg(x) will be essentially treated not as the algebraic expression on the left side of the equation fg(x) = 0, but as an equation that can be expressed in alternative forms presented above.

We refer now to the set of linear, algebraic equations

$$\sum\_{i=1}^{l} a\_{\mathfrak{g}^i} \cdot \mathbf{x}\_i = b\_{\mathfrak{g}} \Leftrightarrow \sum\_{i=1}^{l} a\_{\mathfrak{g}^i} \cdot \mathbf{x}\_i - b\_{\mathfrak{g}} = \mathbf{0}, \qquad (\mathbf{g} = 0, 1, \dots, \mathbf{G}) \tag{9}$$

where agi are the coefficients and bg is the free terms. Multiplying Eq. (9) by ωg, after subsequent summation and rearrangement, we have

Generalized Electron Balance (GEB) as the Law of Nature in Electrolytic Redox Systems http://dx.doi.org/10.5772/intechopen.69341 15

$$\sum\_{g=0}^{G} \boldsymbol{\omega}\_{\mathcal{g}} \cdot \sum\_{i=1}^{l} \mathbf{a}\_{\mathcal{g}i} \cdot \mathbf{x}\_{i} = \sum\_{g=0}^{G} \boldsymbol{\omega}\_{\mathcal{g}} \cdot \mathbf{b}\_{\mathcal{g}} \Leftrightarrow \sum\_{i=1}^{l} \mathbf{x}\_{i} \cdot \sum\_{g=0}^{G} \boldsymbol{\omega}\_{\mathcal{g}} \cdot \mathbf{a}\_{\mathcal{g}i} = \sum\_{g=0}^{G} \boldsymbol{\omega}\_{\mathcal{g}} \cdot \mathbf{b}\_{\mathcal{g}} \tag{10}$$

Assuming

c1ða11x<sup>1</sup> þ a12x<sup>2</sup> þ a13x3Þ þ c2ða21x<sup>1</sup> þ a22x<sup>2</sup> þ a23x3Þ

Applying the matrix algebra, we see that the determinant

� � � � � �

a<sup>11</sup> a<sup>12</sup> a<sup>13</sup> a<sup>21</sup> a<sup>22</sup> a<sup>23</sup> a<sup>11</sup> a<sup>12</sup> a<sup>13</sup> � � � � � � �

þ c<sup>2</sup> �

<sup>e</sup> <sup>R</sup> and addition of the resulting equations, we get the linear combination <sup>X</sup><sup>G</sup>

<sup>g</sup>¼<sup>0</sup> <sup>ω</sup><sup>g</sup> � bg of the basic equations.

� � � � � � �

irrespectively on the c<sup>1</sup> and c<sup>2</sup> values; at D = 0, calculation of x1, x<sup>2</sup> and x<sup>3</sup> is then impossible. (2) Let us consider the set of <sup>G</sup> + 1 equations: <sup>f</sup>g(x) = <sup>ϕ</sup>g(x) � <sup>b</sup><sup>g</sup> = 0, where <sup>g</sup> = 0,1,…, <sup>G</sup>, <sup>x</sup><sup>T</sup> <sup>=</sup>

agi, bg ɛ R are independent (explicitly) on x. After multiplying the equations by the numbers ω<sup>g</sup>

Formation of linear combinations is applicable to check the linear dependency or independency of the balances. A very useful/effective manner for checking/stating the linear dependence of the balances is the transformation of an appropriate system of equations to the identity, 0 = 0 [2, 23]. For this purpose, we will try, in all instances, to obtain the simplest form of the linear combination. To facilitate these operations, carried out by cancellation of the terms on the left and right sides of equations after changing sides of these equations, we apply the

f <sup>g</sup>ðxÞ : ϕgðxÞ � bg ¼ 0 ⇔ ϕgðxÞ ¼ bg ⇔ � f <sup>g</sup>ðxÞ : � ϕgðx޼�bg ⇔ bg ¼ ϕgðxÞ (8)

In this notation, fg(x) will be essentially treated not as the algebraic expression on the left side of the equation fg(x) = 0, but as an equation that can be expressed in alternative forms presented

where agi are the coefficients and bg is the free terms. Multiplying Eq. (9) by ωg, after subse-

D ¼

� � � � � � �

equivalent forms of the starting equations fg(x) = 0:

We refer now to the set of linear, algebraic equations

agi � xi <sup>¼</sup> bg <sup>⇔</sup><sup>X</sup>

I

i¼1

X I

i¼1

quent summation and rearrangement, we have

D ¼ c<sup>1</sup> �

14 Redox - Principles and Advanced Applications

(x1,…, xI) – transposed (T

<sup>g</sup>¼<sup>0</sup> <sup>ω</sup><sup>g</sup> � <sup>ϕ</sup>gðxÞ ¼ <sup>X</sup><sup>G</sup>

has zero value

<sup>⇔</sup> <sup>X</sup><sup>G</sup>

above.

� ðc1a<sup>11</sup> <sup>þ</sup> <sup>c</sup>2a21Þx<sup>1</sup> þ ðc1a<sup>12</sup> <sup>þ</sup> <sup>c</sup>2a22Þx<sup>2</sup> þ ðc1a<sup>13</sup> <sup>þ</sup> <sup>c</sup>2a23Þx<sup>3</sup> <sup>¼</sup> <sup>c</sup>1b<sup>1</sup> <sup>þ</sup> <sup>c</sup>2b2: (5)

� � � � � �

¼ c<sup>1</sup> � 0 þ c<sup>2</sup> � 0 ¼ 0 (7)

<sup>g</sup>¼<sup>0</sup> <sup>ω</sup><sup>g</sup> � <sup>f</sup> <sup>g</sup>ðxÞ ¼ <sup>0</sup>

(6)

a<sup>11</sup> a<sup>12</sup> a<sup>13</sup> a<sup>21</sup> a<sup>22</sup> a<sup>23</sup> c1a<sup>11</sup> þ c2a<sup>21</sup> c1a<sup>12</sup> þ c2a<sup>22</sup> c1a<sup>13</sup> þ c2a<sup>23</sup>

> a<sup>11</sup> a<sup>12</sup> a<sup>13</sup> a<sup>21</sup> a<sup>22</sup> a<sup>23</sup> a<sup>21</sup> a<sup>22</sup> a<sup>23</sup>

� � � � � � �

) vector x, composed of independent (scalar) variables xi (i e < 1, I>);

agi � xi � bg ¼ 0, ðg ¼ 0, 1, …, GÞ (9)

$$b\_{\mathcal{S}} = \sum\_{j=1}^{J} b\_{\mathcal{S}^j} \cdot x\_{0^j} \tag{11}$$

from Eqs. (10) and (11) we have

$$\sum\_{i=1}^{l} \mathbf{x}\_{i} \cdot \sum\_{\mathcal{g}=0}^{G} \boldsymbol{\omega}\_{\mathcal{g}} \cdot \mathbf{a}\_{\mathcal{g}^{i}} = \sum\_{j=1}^{l} \mathbf{x}\_{0j} \cdot \sum\_{\mathcal{g}=0}^{G} \boldsymbol{\omega}\_{\mathcal{g}} \cdot \mathbf{b}\_{\mathcal{g}^{j}} \tag{12}$$

Referring to the problem in question, and placing xi = Ni, x0<sup>j</sup> = N0<sup>j</sup> in Eq. (11), we write

$$b\_{\mathcal{S}} = \sum\_{j=1}^{l} b\_{\mathcal{S}^j} \cdot N\_{0^j} \tag{11a}$$

Then for ChB (Eq. (2a)), where the right side equals zero, we have

$$b\_0 = \sum\_{j=1}^{l} b\_{0\dot{\jmath}} \cdot N\_{0\dot{\jmath}} = 0 \quad \text{for} \quad b\_{0\dot{\jmath}} = 0 \quad (\dot{\jmath} = 1, ..., l) \tag{13}$$

$$\sum\_{i=1}^{l} \mathbf{N}\_{i} \cdot \sum\_{\mathbf{g}=\mathbf{0}}^{\mathbf{G}} \boldsymbol{\omega}\_{\mathbf{g}} \cdot \mathbf{a}\_{\mathbf{g}^{\circ}} = \sum\_{j=1}^{l} \mathbf{N}\_{0^{j}} \cdot \sum\_{\mathbf{g}=\mathbf{0}}^{\mathbf{G}} \boldsymbol{\omega}\_{\mathbf{g}} \cdot \mathbf{b}\_{\mathbf{g}^{\circ}} \tag{14}$$

The charge balance (ChB) is expressed by Eqs. (2a) and (2b), where Xi zi is defined by Eq. (1a) (for a static system) or Eq. (1b) (for a dynamic system). The elemental/core balances: f(H), f(O) and f(Yg) (Yg 6¼ H, O, g = 3,…, G) are written as follows:

$$f\_1 = f(H) = \sum\_{i=1}^{I} (a\_{1i} + 2n\_i) \cdot N\_i - \sum\_{j=1}^{I} b\_{1j} \cdot N\_{0j} = 0 \text{ for } Y\_1 = H,$$

$$f\_2 = f(\mathbf{O}) = \sum\_{i=1}^{I} (a\_{2i} + n\_i) \cdot N\_i - \sum\_{j=1}^{I} b\_{2j} \cdot N\_{0j} = 0 \quad \text{for } Y\_2 = O, \dots,\tag{15}$$

$$f\_3 = \sum\_{i=1}^{I} a\_{3i} \cdot N\_i - \sum\_{j=1}^{J} b\_{3j} \cdot N\_{0j} = 0, \dots \\ f\_G = \sum\_{i=1}^{I} a\_{Gi} \cdot N\_i - \sum\_{j=1}^{J} b\_{Gj} \cdot N\_{0j} = 0$$

where agi and bgj are the numbers of atoms/cores of gth kind in the ith species as a constituent of the system and in the component of jth kind, respecively. Denoting, for a moment, ω<sup>1</sup> = –1, ω<sup>2</sup> = 2, we transform the balance

$$f\_1 f\_{12} = 2 \cdot f\_2 - f\_1 = 2 \cdot f(\mathbb{O}) - f(\mathbb{H}) = \sum\_{i=2}^{l} (2a\_{2i} - a\_{1i}) \cdot N\_i - \sum\_{j=1}^{l} (2b\_{2j} - b\_{1j}) \cdot N\_{0j} = 0 \Rightarrow \tag{16}$$

$$\begin{aligned} f\_{12} &= \omega\_1 \cdot \left(\sum\_{i=2}^l (a\_{1i} \cdot N\_i - \sum\_{j=1}^l b\_{2j} \cdot N\_{0j})\right) + \omega\_2 \cdot \left(\sum\_{i=2}^l (a\_{2i} \cdot N\_i - \sum\_{j=1}^l (b\_{2j} \cdot N\_{0j})\right) \\ &= \omega\_1 \cdot f\_1^\* + \omega\_2 \cdot f\_2^\* = 0 \end{aligned} \tag{17}$$

In Eq. (17), f<sup>1</sup> \* and f<sup>2</sup> \* have the shape similar to the general expression for fg (g = 3,…, G) in Eq. (15).

In the balances related to aqueous media, the terms involved with water, i.e. N0<sup>j</sup> (for j related to H2O, as the component), N1, and all n<sup>i</sup> = ni<sup>W</sup> are not involved (in f0, f3,…, fG) or are cancelled within f<sup>12</sup> (Eq. (17)). Other species, such as CH3COOH transformable (mentally, purposefully) into C2H4O2 � C2(H2O)2, and the species in which H and O are not involved in Xi zi are also cancelled within f12.

On the basis of relations (13) and (17), the linear combination of G + 1 balances f0, f12, f3,…, f<sup>G</sup> expressed by Eq. (14) can be presented in equivalent forms:

$$\sum\_{i=2}^{l} \mathbf{N}\_{i} \cdot \left(\mathbf{z}\_{i} + \sum\_{\mathbf{g}=1}^{G} \boldsymbol{\omega}\_{\boldsymbol{\S}} \cdot \mathbf{a}\_{\boldsymbol{\S}^{i}}\right) = \sum\_{j=1}^{l} \mathbf{N}\_{0j} \cdot \sum\_{\mathbf{g}=1}^{G} \boldsymbol{\omega}\_{\boldsymbol{\S}} \cdot \mathbf{b}\_{\mathbf{g}j} \tag{18}$$

$$\begin{aligned} \sum\_{i=2}^{I} \mathbf{N}\_{i} \cdot \mathbf{z}\_{i} + \sum\_{\mathbf{g}=1}^{G} \boldsymbol{\omega}\_{\boldsymbol{\mathcal{S}}} \cdot \left( \sum\_{i=2}^{I} \mathbf{N}\_{i} \cdot \mathbf{a}\_{\boldsymbol{\mathcal{S}}} - \sum\_{j=1}^{I} \mathbf{N}\_{0\dot{\boldsymbol{\mathcal{Y}}}} \cdot \mathbf{b}\_{\boldsymbol{\mathcal{S}}} \right) &= \mathbf{0} \\ \boldsymbol{f}\_{0} + \boldsymbol{f}\_{12} + \sum\_{\mathbf{g}=3}^{G} \boldsymbol{\omega}\_{\boldsymbol{\mathcal{S}}} \cdot \boldsymbol{f}\_{\boldsymbol{\mathcal{S}}} &= \mathbf{0} \end{aligned} \tag{18a}$$

$$\begin{aligned} \mathbf{ChB} + (2 \cdot f(\mathbf{O}) - f(\mathbf{H})) + \sum\_{\mathcal{g}=3}^{G} \boldsymbol{\omega}\_{\mathcal{g}} \cdot f(Y\_{\mathcal{g}}) &= \mathbf{0} \\ \mathbf{(+1)} \cdot f(\mathbf{H}) + (-2) \cdot f(\mathbf{O}) + \sum\_{\mathcal{g}=3}^{G} (-\boldsymbol{\omega}\_{\mathcal{g}}) \cdot f(Y\_{\mathcal{g}}) - \mathbf{ChB} &= \mathbf{0} \end{aligned} \tag{18b}$$

All multipliers at Ni and N0<sup>j</sup> in Eq. (18a) are cancelled simultaneously, if we have

$$z\_i + \sum\_{\mathcal{g}=1}^G \boldsymbol{\omega}\_{\mathcal{g}} \cdot \boldsymbol{a}\_{\mathcal{g}^i} = 0 \quad \text{and} \quad \sum\_{\mathcal{g}=1}^G \boldsymbol{a}\_{\mathcal{g}} \cdot \boldsymbol{b}\_{\mathcal{g}^i} = 0 \tag{19}$$

for all i = 1,…, I and j = 1,…, J; then Eq. (18) is transformed into identity

Generalized Electron Balance (GEB) as the Law of Nature in Electrolytic Redox Systems http://dx.doi.org/10.5772/intechopen.69341 17

$$\sum\_{i=1}^{I} \mathbf{N}\_i \cdot \mathbf{0} = \sum\_{j=1}^{J} \mathbf{N}\_{0j} \cdot \mathbf{0} \Leftrightarrow \mathbf{0} \ = \begin{array}{c} \mathbf{0} \end{array} \tag{20}$$

Transformation of a set of the equations into the identity, 0 = 0, proves the linear dependence between the equations considered. Then from Eq. (18a), we have

$$f\_{12} = \sum\_{\mathcal{g}=\mathcal{3}}^{G} (-\omega\_{\mathcal{g}}) \cdot f\_{\mathcal{g}} - f\_0 \tag{18c}$$

i.e. f<sup>12</sup> is the dependent balance.

<sup>f</sup> <sup>12</sup> <sup>¼</sup> <sup>2</sup> � <sup>f</sup> <sup>2</sup> � <sup>f</sup> <sup>1</sup> <sup>¼</sup> <sup>2</sup> � <sup>f</sup>ðOÞ � <sup>f</sup>ðHÞ ¼ <sup>X</sup>

I

0 @

<sup>ð</sup>a1<sup>i</sup> � Ni �<sup>X</sup>

expressed by Eq. (14) can be presented in equivalent forms:

<sup>N</sup><sup>i</sup> � zi <sup>þ</sup><sup>X</sup>

g¼1

G

g¼3

ChB þ ð<sup>2</sup> � <sup>f</sup>ðOÞ � <sup>f</sup>ðHÞÞ þ<sup>X</sup>

ðþ1Þ � <sup>f</sup>ðHÞ þ ð�2Þ � <sup>f</sup>ðOÞ þ<sup>X</sup>

zi <sup>þ</sup><sup>X</sup> G

g¼1

for all i = 1,…, I and j = 1,…, J; then Eq. (18) is transformed into identity

0 @

Ni � zi <sup>þ</sup><sup>X</sup> G

<sup>f</sup> <sup>0</sup> <sup>þ</sup> <sup>f</sup> <sup>12</sup> <sup>þ</sup><sup>X</sup>

G

g¼1

<sup>ω</sup><sup>g</sup> � <sup>X</sup> I

ω<sup>g</sup> � f <sup>g</sup> ¼ 0

All multipliers at Ni and N0<sup>j</sup> in Eq. (18a) are cancelled simultaneously, if we have

0 @

ω<sup>g</sup> � agi

i¼2

G

g¼3

G

g¼3

<sup>ω</sup><sup>g</sup> � agi <sup>¼</sup> 0 and <sup>X</sup>

X I

i¼2

X I

i¼2

J

j¼1

� ¼ 0

i¼2

� þ ω<sup>2</sup> � f <sup>2</sup>

<sup>f</sup> <sup>12</sup> <sup>¼</sup> <sup>ω</sup><sup>1</sup> � <sup>X</sup>

16 Redox - Principles and Advanced Applications

¼ ω<sup>1</sup> � f <sup>1</sup>

\* and f<sup>2</sup>

In Eq. (17), f<sup>1</sup>

cancelled within f12.

I

<sup>ð</sup>2a2<sup>i</sup> � <sup>a</sup>1iÞ � Ni �<sup>X</sup>

<sup>A</sup> <sup>þ</sup> <sup>ω</sup><sup>2</sup> � <sup>X</sup>

J

j¼1

<sup>ð</sup>a2<sup>i</sup> � Ni �<sup>X</sup>

J

j¼1

I

0 @

\* have the shape similar to the general expression for fg (g = 3,…, G) in Eq. (15).

i¼2

ð2b2<sup>j</sup> � b1jÞ � N0<sup>j</sup> ¼ 0 ) (16)

ðb2<sup>j</sup> � N0jÞ

ω<sup>g</sup> � bgj (18)

ω<sup>g</sup> � bgj ¼ 0 (19)

1 A

(17)

zi are also

(18a)

(18b)

i¼2

b2<sup>j</sup> � N0jÞ

into C2H4O2 � C2(H2O)2, and the species in which H and O are not involved in Xi

1

In the balances related to aqueous media, the terms involved with water, i.e. N0<sup>j</sup> (for j related to H2O, as the component), N1, and all n<sup>i</sup> = ni<sup>W</sup> are not involved (in f0, f3,…, fG) or are cancelled within f<sup>12</sup> (Eq. (17)). Other species, such as CH3COOH transformable (mentally, purposefully)

On the basis of relations (13) and (17), the linear combination of G + 1 balances f0, f12, f3,…, f<sup>G</sup>

1 <sup>A</sup> <sup>¼</sup> <sup>X</sup> J

j¼1 N0<sup>j</sup> � X G

J

j¼1

ω<sup>g</sup> � fðYgÞ ¼ 0

G

g¼1

Ni � agi �<sup>X</sup>

g¼1

N0<sup>j</sup> � bgj

ð�ωgÞ � fðYgÞ � ChB ¼ 0

1 A ¼ 0 Briefly, from G + 1 starting balances: f0, f1, f2, f3, …, f<sup>G</sup> we obtain G balances: f0, f12, f3,…, fG. If f<sup>12</sup> is the dependent balance, we have G � 1 independent balances: f0, f3,…, fG; it is the case related to non-redox systems. If f<sup>12</sup> is the independent balance, we have G independent balances: f0, f12, f3,…, fG, that will be rearranged, optionally, as the set (f12, f0, f3,…, fG) related to GEB, ChB and f (Yg) (g = 3,…, G), respectively. The number of elemental/core balances f(Yg) (Yg 6¼ H, O, g = 3,…, G) and then the number of concentration balances CB(Yg) in redox systems equals K = G � 2.

As stated above, the linear combination of 2∙f(O) � f(H) with ChB and elemental/core balances f(Yg) (g = 3,…, G) provides the criterion distinguishing between non-redox and redox systems of different complexity [3, 23]. To obtain the simplest form of the linear combination, some useful/general rules allowing to select multipliers ω<sup>g</sup> of the corresponding balances are suggested. Namely, the ω<sup>g</sup> values in Eq. (10) are equal to the oxidation numbers of the electron-non-active elements in the related system. The proper linear combination of the balances is reducible (a) to the identity, 0 = 0, for non-redox systems or (b) to the simplest equation different from (not reducible to) the identity, but involving some terms related to the species and components of the system. This way we state that 2∙f(O) � f(H) is linearly dependent on the other balances in non-redox systems, i.e. it is not a new/independent balance in such systems. In redox systems, the balance 2∙f(O) � f(H) and its linear combinations with ChB and f(Yg) (g = 3,…, G) are the new equation, completing, as the generalized electron balance (GEB), the set of equations needed for resolution of electrolytic redox systems.

Static and dynamic non-redox and redox systems, with water as the main component, are considered. A static system of volume V<sup>0</sup> mL is obtained by disposable mixing different components. The dynamic system is realized according to the titrimetric mode. At defined point of the titration B (C, V) ) A (C0, V0), denoted briefly as T ) D, V mL of titrant T containing the component B (C mol/L) is added into V<sup>0</sup> mL of titrand D containing the component A (C<sup>0</sup> mol/L), and V<sup>0</sup> + V mL of D + T system/mixture is thus obtained, if the assumption of volume additivity is valid/tolerable; then D and T are sub-systems of the D + T system. The progress of the titration is represented by the fraction titrated Φ [11, 19, 47, 53, 54] value

$$
\Phi = \frac{\mathbb{C} \cdot V}{\mathbb{C}\_0 \cdot V\_0} \tag{21}
$$

For comparative purposes, we first refer to non-redox systems, where the derivation of relevant formulas will be carried out in detail also for training purposes. The regularities resulting from these examples, namely linear dependency of (linear) algebraic equations stated for nonredox electrolytic systems, will provide the reference point for further parts of this chapter, where redox systems are considered. Further examples in this chapter, related only to redox systems, will be presented in a synthetic manner, indicating the similarities and differences in the respective equations. The conclusions drawn here will provide the basis for further, important generalizations.

#### 4. Linear combination of balances for non-redox systems

#### 4.1. Examples

Example 1 (static): V<sup>0</sup> mL of CuSO4 solution is prepared from N<sup>01</sup> molecules of CuSO4�5H2O and N<sup>02</sup> molecules of H2O. The resulting solution consists of the following species: H2O (N1), H+1 (N2, n2), OH�<sup>1</sup> (N3, n3), HSO4 �<sup>1</sup> (N4, n4), SO4 �<sup>2</sup> (N5, n5), Cu+2 (N6, n6), CuOH+1 (N7, n7), Cu (OH)2 (N8, n8), Cu(OH)3 �<sup>1</sup> (N9, n9), Cu(OH)4 �<sup>2</sup> (N10, n10) and CuSO4 (N11, n11). The components and species are involved in the balances for particular elements, Yg: H, O, Cu, S, i.e.

$$\begin{aligned} f\_1 &= f(\mathbf{H}):\\ 2\mathbf{N}\_1 &+ \mathbf{N}\_2(\mathbf{1} + 2\mathbf{n}\_2) + \mathbf{N}\_3(\mathbf{1} + 2\mathbf{n}\_3) + \mathbf{N}\_4(\mathbf{1} + 2\mathbf{n}\_4) + \mathbf{N}\_5 2\mathbf{n}\_5 + \mathbf{N}\_6 2\mathbf{n}\_6 + \mathbf{N}\_7(\mathbf{1} + 2\mathbf{n}\_7) \\ &+ \mathbf{N}\_8(\mathbf{2} + 2\mathbf{n}\_8) + \mathbf{N}\_9(\mathbf{3} + 2\mathbf{n}\_9) + \mathbf{N}\_{10}(\mathbf{4} + 2\mathbf{n}\_{10}) + \mathbf{N}\_{11} 2\mathbf{n}\_{11} = 10\mathbf{N}\_{01} + 2\mathbf{N}\_{02} \end{aligned} \tag{22}$$

$$\begin{aligned} f\_2 &= f(\mathbf{O}):\\ N\_1 + N\_2 n\_2 &+ N\_3(1+n\_3) + N\_4(4+n\_4) + N\_5(4+n\_5) + N\_6 n\_6 + N\_7(1+n\_7) \end{aligned} \tag{23}$$

$$f\_{12} = \mathbf{2} \cdot f(\mathbf{O}) \cdot f(\mathbf{H}) : \ -\mathbf{N}\_2 + \mathbf{N}\_3 + 7\mathbf{N}\_4 + 8\mathbf{N}\_5 + \mathbf{N}\_7 + 2\mathbf{N}\_8 + 3\mathbf{N}\_9 + 4\mathbf{N}\_{10} + 8\mathbf{N}\_{11} = \text{ 8N}\_{\text{01}} \tag{24}$$

$$f\_0 = \text{ChB}: \qquad N\_2 - N\_3 - N\_4 - 2N\_5 + 2N\_6 + N\_7 - N\_9 - 2N\_{10} = 0\tag{25}$$

$$-2f\_3 = -2f(\mathbf{Cu}):\qquad 2\mathbf{N}\_{01} = \, 2\mathbf{N}\_6 + 2\mathbf{N}\_7 + 2\mathbf{N}\_8 + 2\mathbf{N}\_9 + 2\mathbf{N}\_{10} + \, 2\mathbf{N}\_{11} \tag{26}$$

$$-6f\_4 = -6f(\mathbb{S}) = -6f(\mathbb{S}\mathbb{O}\_4):\qquad 6\mathcal{N}\_{0\mathbb{I}} = \,\_6\mathrm{6N}\_4 + 6\mathcal{N}\_5 + 6\mathcal{N}\_{11}\tag{27}$$

Simple addition of the elemental/core balances (Eqs. (24)–(27)) gives the identity

þ N9ð3 þ n9Þ þ N10ð4 þ n10Þ þ N11ð4 þ n11Þ ¼ 9N<sup>01</sup> þ N<sup>02</sup>

$$\begin{aligned} \mathbf{2} \cdot f(\mathbf{O}) \cdot f(\mathbf{H}) + \mathbf{ChB} \cdot 2f(\mathbf{Cu}) \cdot 6f(\mathbf{S}) &= \mathbf{0} \Rightarrow \\ \mathbf{2} \cdot (+1) \cdot f(\mathbf{H}) + (-2) \cdot f(\mathbf{O}) + (+2) \cdot f(\mathbf{Cu}) + (+6) \cdot f(\mathbf{S}) \cdot \mathbf{ChB} &= \mathbf{0} \Rightarrow 0 = 0 \end{aligned} \tag{28}$$

As we see, the multipliers ω<sup>g</sup> in the transformed identity are equal to oxidation numbers of the indicated elements.

Example 2 (dynamic): V<sup>0</sup> mL of CuSO4 solution, as a titrand D composed of N<sup>01</sup> molecules of CuSO4�5H2O and N<sup>02</sup> molecules of H2O, is titrated with V mL of titrant T, composed of N<sup>03</sup> molecules of NaOH and N<sup>04</sup> molecules of H2O, added up to a defined point of the titration. In the resulting D + T mixture (two-phase system of volume ca. V<sup>0</sup> + V mL), we have the species as in Example 1 and, additionally: Cu(OH)2 (N12, n12), Na+1 (N13, n13). From the balances:

$$f\_1 = f(\mathbf{H}): \quad (\ldots) + N\_{12}(2 + 2n\_{12}) + 2N\_{13}n\_{13} = (\ldots) + N\_{03} + 2N\_{04} \tag{29}$$

$$f\_2 = f(\mathbf{O}): \quad (\ldots) + N\_{12}(2 + n\_{12}) + N\_{13}n\_{13} = (\ldots) + N\_{03} + N\_{04} \tag{30}$$

$$f\_{12} = 2 \cdot f(\mathbf{O}) \cdot f(\mathbf{H}):\ \ -\mathbf{N}\_2 + \mathbf{N}\_3 + 7\mathbf{N}\_4 + 8\mathbf{N}\_5 + \mathbf{N}\_7 + 2\mathbf{N}\_8 + 3\mathbf{N}\_9 + 4\mathbf{N}\_{10} + 8\mathbf{N}\_{11} \tag{31}$$

$$+ 2\mathbf{N}\_{12} = 8\mathbf{N}\_{01} + \mathbf{N}\_{03}$$

$$f\_0 = \text{ChB}: \quad N\_2\text{-N}\_3\text{-N}\_4\text{-2N}\_5 + 2\text{N}\_6 + N\_7\text{-N}\_9\text{-2N}\_{10} + N\_{13} = 0\tag{32}$$

$$-2f\_3 = -2f(\mathbf{Cu}):\quad 2\mathbf{N}\_{01} = 2\mathbf{N}\_6 + 2\mathbf{N}\_7 + 2\mathbf{N}\_8 + 2\mathbf{N}\_9 + 2\mathbf{N}\_{10} + 2\mathbf{N}\_{11} + 2\mathbf{N}\_{12} \tag{33}$$

$$-6f\_4 = -6f(\mathbf{S}) = -6f(\mathbf{SO}\_4):\qquad 6\mathbf{N}\_{01} = 6\mathbf{N}\_4 + 6\mathbf{N}\_5 + 6\mathbf{N}\_{11}\tag{34}$$

$$-f\_5 = -f(\text{Na}) : \begin{aligned} N\_{03} = N\_{13} \end{aligned} \tag{35}$$

That means we obtain the identity, 0 = 0, for

$$\begin{aligned} \text{R} \cdot f(\text{O}) \text{-} f(\text{H}) + \text{ChB} \cdot f(\text{Na}) - 2 \cdot f(\text{Cu}) - 6 \cdot f(\text{S}) &= 0 \quad \Rightarrow \\ \text{R} \cdot (+1) \cdot f(\text{H}) + (-2) \cdot f(\text{O}) + (+2) \cdot f(\text{Cu}) + (+6) \cdot f(\text{S}) - \text{ChB} &= 0 \end{aligned} \tag{36}$$

The terms within (…)'s in Eqs. (29) and (30) are the same as in Eqs. (22) and (23).

#### 4.2. A comment

For comparative purposes, we first refer to non-redox systems, where the derivation of relevant formulas will be carried out in detail also for training purposes. The regularities resulting from these examples, namely linear dependency of (linear) algebraic equations stated for nonredox electrolytic systems, will provide the reference point for further parts of this chapter, where redox systems are considered. Further examples in this chapter, related only to redox systems, will be presented in a synthetic manner, indicating the similarities and differences in the respective equations. The conclusions drawn here will provide the basis for further, impor-

Example 1 (static): V<sup>0</sup> mL of CuSO4 solution is prepared from N<sup>01</sup> molecules of CuSO4�5H2O and N<sup>02</sup> molecules of H2O. The resulting solution consists of the following species: H2O (N1),

nents and species are involved in the balances for particular elements, Yg: H, O, Cu, S, i.e.

2N<sup>1</sup> þ N2ð1 þ 2n2Þ þ N3ð1 þ 2n3Þ þ N4ð1 þ 2n4Þ þ N52n<sup>5</sup> þ N62n<sup>6</sup> þ N7ð1 þ 2n7Þ

N<sup>1</sup> þ N2n<sup>2</sup> þ N3ð1 þ n3Þ þ N4ð4 þ n4Þ þ N5ð4 þ n5Þ þ N6n<sup>6</sup> þ N7ð1 þ n7Þ þ N8ð2 þ n8Þ

f <sup>12</sup> ¼ 2 � fðOÞ–fðHÞ : –N<sup>2</sup> þ N<sup>3</sup> þ 7N<sup>4</sup> þ 8N<sup>5</sup> þ N<sup>7</sup> þ 2N<sup>8</sup> þ 3N<sup>9</sup> þ 4N<sup>10</sup> þ 8N<sup>11</sup> ¼ 8N<sup>01</sup> (24)

f <sup>0</sup> ¼ ChB : N<sup>2</sup> � N<sup>3</sup> � N<sup>4</sup> � 2N<sup>5</sup> þ 2N<sup>6</sup> þ N<sup>7</sup> � N<sup>9</sup> � 2N<sup>10</sup> ¼ 0 (25)

�2f <sup>3</sup> ¼ –2fðCuÞ : 2N<sup>01</sup> ¼ 2N<sup>6</sup> þ 2N<sup>7</sup> þ 2N<sup>8</sup> þ 2N<sup>9</sup> þ 2N<sup>10</sup> þ 2N<sup>11</sup> (26)

�6f <sup>4</sup> ¼ –6fðSÞ ¼ –6fðSO4Þ : 6N<sup>01</sup> ¼ 6N<sup>4</sup> þ 6N<sup>5</sup> þ 6N<sup>11</sup> (27)

ðþ1Þ � fðHÞ þ ð�2Þ � fðOÞ þ ðþ2Þ � fðCuÞ þ ðþ6Þ � fðSÞ–ChB ¼ 0 ) 0 ¼ 0

As we see, the multipliers ω<sup>g</sup> in the transformed identity are equal to oxidation numbers of the

Simple addition of the elemental/core balances (Eqs. (24)–(27)) gives the identity

þ N8ð2 þ 2n8Þ þ N9ð3 þ 2n9Þ þ N10ð4 þ 2n10Þ þ N112n<sup>11</sup> ¼ 10N<sup>01</sup> þ 2N<sup>02</sup>

�<sup>2</sup> (N5, n5), Cu+2 (N6, n6), CuOH+1 (N7, n7), Cu

(22)

(23)

(28)

�<sup>2</sup> (N10, n10) and CuSO4 (N11, n11). The compo-

�<sup>1</sup> (N4, n4), SO4

4. Linear combination of balances for non-redox systems

�<sup>1</sup> (N9, n9), Cu(OH)4

þ N9ð3 þ n9Þ þ N10ð4 þ n10Þ þ N11ð4 þ n11Þ ¼ 9N<sup>01</sup> þ N<sup>02</sup>

2 � fðOÞ–fðHÞ þ ChB–2fðCuÞ–6fðSÞ ¼ 0 )

tant generalizations.

18 Redox - Principles and Advanced Applications

4.1. Examples

H+1 (N2, n2), OH�<sup>1</sup> (N3, n3), HSO4

(OH)2 (N8, n8), Cu(OH)3

f <sup>1</sup> ¼ fðHÞ :

f <sup>2</sup> ¼ fðOÞ :

indicated elements.

The equations for f(H) and f(O) in involve the numbers ni = ni<sup>W</sup> of hydrating water molecules, attached to particular species. In the linear combinations 2∙f(O) – f(H), all the ni<sup>W</sup> are cancelled, together with N1, i.e. the ni<sup>W</sup> and N<sup>1</sup> do not enter the equation for 2∙f(O) – f(H). The ni<sup>W</sup> is not involved in ChB and in f(Yj) for Yj 6¼ H, O and then they are not introduced in the linear combinations of 2∙f(O) – f(H) with these equations. The multipliers ω<sup>g</sup> in Eq. (10) are equal to oxidation numbers of particular elements involved in the related elemental/core balances. Then, the linear combination of the balances related to a non-redox system, when put in context with Eqs. (19) and (20) gives the identity 0 = 0. This regularity is obligatory for all non-redox systems considered above (Examples 1 and 2), and elsewhere, e.g. [22, 37, 55]. Note that 2∙f(O) – f(H) involves –N<sup>2</sup> + N3, whereas ChB involves N<sup>2</sup> – N<sup>3</sup> = – (–N<sup>2</sup> + N3), and the related terms are cancelled within the combination 2∙f(O) – f(H) + ChB. Moreover, all components related to the species Xzi <sup>i</sup> � niW, not involving H and/or O in Xzi <sup>i</sup> , are cancelled too within 2∙f(O) – f(H).

Figure 1. Schemes of disproportionation and symproportionation within bromine species [5].

In further sections, we formulate the balances for redox systems containing one, two or three electron-active elements in redox systems (aqueous media), where the complete set of expressions for independent equilibrium constants interrelating concentrations of different species is involved. We start our considerations from the systems with one electron-active element involved with disproportionation and symproportionation [5, 13, 15, 16, 19]. These properties are appropriate for the elements that form compounds and species at three or more oxidation degrees. In particular, bromine (Br) forms the species on five oxidation degrees (–1, –1/3, 0, 1, 5), see Figure 1, e.g. 0 (for Br2) and 1 (for HBrO or NaBrO) e (–1, 5). There are possible transitions between different bromine species, associated with changes of the oxidation states of this element, see Figure 1.

#### 5. Disproportionating systems
