3rd order reaction of the type (A + 2B → P)

The rate law for the 3rd order reaction of the type A + 2BP is:

\frac{d[A]}{dt}=-k[A][B]^2

Using the same logic described in a previous article, we can rewrite the rate law as:

\frac{dx}{dt}=k(a-x)(b-2x)^2

where a = [A0] and b = [B0].

Integrating the above equation throughout yields

\int_{0}^{x}\frac{dx}{(a-x)(b-2x)^2}=k\int_{o}^{t}dt\; \; \; \; \; \; \; \; 15

Substituting the partial fraction expression \frac{1}{(a-x)(b-2x)^2}=\frac{2}{(2a-b)(b-2x)^2}+\frac{1}{(2a-b)^2(a-x)}-\frac{2}{(2a-b)^2(b-2x)}  in eq15 and working out some algebra gives

kt=\frac{2x}{b(2a-b)(b-2x)}+\frac{1}{(2a-b)^2}ln\frac{a(b-2x)}{b(a-x)}

 

Question

How to compute \int_{0}^{x}\frac{2}{(2a-b)(b-2x)^2}dx ?

Answer

\int_{0}^{x}\frac{2}{(2a-b)(b-2x)^2}dx=\int_{0}^{x}\frac{2b}{b(2a-b)(b-2x)^2}dx

=\int_{0}^{x}\frac{2[(b-2x)+2x]}{b(2a-b)(b-2x)^2}dx=\int_{0}^{x}\frac{2[(2a-b)(b-2x)+2x(2a-b)]}{b(2a-b)^2(b-2x)^2}dx

=\int_{0}^{x}\frac{2b(2a-b)(b-2x)-4bx(b-2a)}{b^2(2a-b)^2(b-2x)^2}dx=\frac{2x}{b(2a-b)(b-2x)}

Note that the integrand in the 5th equality can be obtained by differentiating the last term using the quotient rule.

 

 

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2nd order autocatalytic reaction

An autocatalytic reaction is one in which a product catalyses the reaction. An example is the Mn2+-catalysed oxidation of oxalic acid by potassium manganate (VII):

2MnO_4^{\; -}+16H^++5C_2O_4^{\; 2-}\rightarrow 2Mn^{2+}+10CO_2+8H_2O

The equation can be reduced to A + P → 2P, or simply, A P, with the rate law:

\frac{d[A]}{dt}=-k[A][P]\; \; \; \; \; \; \; \; 13a

Using the same logic described in a previous article, we can rewrite the rate law as:

\frac{dx}{dt}=k(a-x)(p+x)

where a = [A0] and p = [P0].

Integrating throughout, we have

\int_{0}^{x}\frac{dx}{(a-x)(p+x)}=k\int_{0}^{t}dt\; \; \; \; \; \; \; \; 14

Substituting the partial fraction expression \frac{1}{(a-x)(p+x)}=\frac{1}{a+p}\left ( \frac{1}{a-x}+\frac{1}{p+x} \right ) in eq14 and working out the algebra yields

kt=\frac{1}{a+p}ln\frac{a(p+x)}{p(a-x)}

which is equivalent to

kt=\frac{1}{[A_0]+[P_0]}ln\frac{[A_0][P]}{[P_0][A]}

 

Question

The mechanism of the above reaction is found to include the following steps:

A+P\rightleftharpoons AP

AP\rightarrow P+P\; \; \; \; \; \; (rds)

With this in mind, derive the rate law and show that it is consistent with eq13a.

Answer

We can write the rate law as:

\frac{d[P]}{dt}=2k_{rds}[AP]=2k_{rds}K[A][P]=k[A][P]

where K is the equilibrium constant for the 1st step and k = 2krdsK. Furthermore, the overall reaction is AP, which means that

\frac{d[P]}{dt}=-\frac{d[A]}{dt}=k[A][P]

 

 

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2nd order reaction of the type (A + 2B → P)

The rate law for the 2nd order reaction of the type A + 2BP is:

\frac{d[A]}{dt}=-k[A][B]

Using the same logic described in the previous article, we can rewrite the rate law as:

\frac{dx}{dt}=k(a-x)(b-2x)

where a = [A0] and b = [B0].

Rearranging the above equation and integrating throughout, we have

\int_{0}^{x}\frac{dx}{(a-x)(\frac{b}{2}-x)}=2k\int_{0}^{t}dt\; \; \; \; \; \; \; \; 13

Substituting the partial fraction expression \frac{1}{(a-x)(\frac{b}{2}-x)}=\frac{1}{a-\frac{b}{2}}\left ( \frac{1}{\frac{b}{2}-x}-\frac{1}{a-x} \right ) in eq13 and after some algebra, we have,

kt=\frac{1}{b-2a}ln\frac{a(b-2x)}{b(a-x)}

which is equivalent to

kt=\frac{1}{[B_0]-2[A_0]}ln\frac{[A_o]([B])}{[B_0]([A])}

 

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Transition state theory

The transition state theory provides a theoretical way to calculate the pre-exponential factor A and the activation energy Ea, and therefore the rate constant k of an elementary chemical reaction. This is in contrast with the Arrhenius equation, where A and Ea are obtained empirically. The theory is based on statistical thermodynamics and can be illustrated for a gas-phase bimolecular reaction using the following assumptions:

1) The reacting system, expressed as A+B\rightleftharpoons C^{\ddagger}\rightarrow P, proceeds along an energy path that includes a point of highest potential energy called the saddle point, where a distinct species known as the activated complex C^{\ddagger} is formed. The conversion of the activated complex to the product P can be seen as an asymmetric vibrational stretch:

2) A rapid pre-equilibrium is established between the activated complex and the reactants: A+B\rightleftharpoons C^{\ddagger}.

3) The rate of the reaction is attributed to the rate-determining step, which is the rate of conversion of the activated complex to the product: rate=k^{\ddagger}[C^{\ddagger}], where k^{\ddagger} is proportional to the vibrational frequency v of the activated complex.

The expression for the gas-phase rapid equilibrium from assumption 2 is:

K^{\ddagger}=\frac{\frac{p_{c^{\ddagger}}}{p^o}}{\frac{p_A}{p^o}\frac{p_B}{p^o}}=\frac{p^op_{C^{\ddagger}}}{p_Ap_B}

which in terms of molar concentration is:

K^{\ddagger}=\frac{p^o}{RT}\frac{[C^{\ddagger}]}{[A][B]}\; \; \; \; \; \; \; \; 1

Substituting eq1 into the rate equation , rate=k^{\ddagger}[C^{\ddagger}], yields:

rate=k[A][B]\; \; \; \; \; \; \; where\; k=\frac{RT}{p^o}k^{\ddagger}K^{\ddagger}

In general, the equilibrium constant, when written in terms of standard molar partition functions, is K=\left [ \prod _j\left ( \frac{q^o_{j,m}}{N_A} \right )^{v_j} \right ]e^{-\frac{\Delta_rE_0}{RT}} (see this article for derivation), where v_j is the respective stoichiometric coefficient of the J-th species. For our bimolecular reaction,

K^{\ddagger}=\frac{N_A\, q^o_{C^{\ddagger},m}}{q^o_{A,m}q^o_{B,m}}e^{-\frac{\Delta_rE_0}{RT}}\; \; \; \; \; \; \; \; 2

where \Delta_rE_0=E_0(C^{\ddagger})-E_0(A)-E_0(B).

The activated complex in our example is a linear molecule with N = 3 atoms and therefore has 3N – 5 = 4 modes of vibration (2 bending, 1 symmetric stretch and 1 asymmetric stretch). The standard molar partition function for the activated complex is the product of the partition functions of different modes of motion: q^o_{C^{\ddagger},m}=q^{o\; \; \; \; \; \; T}_{C^{\ddagger},m}q^{o\; \; \; \; \; \; V}_{C^{\ddagger},m}q^{o\; \; \; \; \; \; R}_{C^{\ddagger},m}, which we can rewrite in the form: q^o_{C^{\ddagger},m}=q^{o\; \; \; \; \; V_{asym}}_{C^{\ddagger},m}\, \bar{q}^{\, o}_{C^{\ddagger},m}, where \bar{q}^{\, o}_{C^{\ddagger},m} =q^{o\; \; \; \; \; T}_{C^{\ddagger},m}\, q^{o\; \; \; \; \; V_{others}}_{C^{\ddagger},m}\, q^{o\; \; \; \; \; R}_{C^{\ddagger},m}.

The standard molar partition function for the asymmetric vibration is q^{o\; \; \; \; \; \; V_{asym}}_{C^{\ddagger},m}=\frac{1}{1-e^{-\frac{hv}{kT}}}, where v is the vibrational frequency that leads to the conversion of the activated complex to the product. Assuming \frac{hv}{kT}\ll 1, q^{o\; \; \; \; \; \; V_{asym}}_{C^{\ddagger},m}=\frac{1}{1-e^{-\frac{hv}{kT}}}\approx\frac{kT}{hv}. Therefore,

q^o_{C^{\ddagger},m}=q^{o\; \; \; \; \; V_{asym}}_{C^{\ddagger},m}\, \bar{q}^{\, o}_{C^{\ddagger},m}\approx\frac{kT}{hv}\bar{q}^{\, o}_{C^{\ddagger},m}\; \; \; \; \; \; \; \; 3

Substituting eq3 into eq2 gives:

K^{\ddagger}=\frac{kT}{hv}\bar{K}^{\ddagger}\; \; \; \; \; \; \; \; 4

where \bar{K}^{\ddagger}=\frac{N_A\, \bar{q}^{\, o}_{C^{\ddagger},m}}{q^o_{A,m}q^o_{B,m}}e^{-\frac{\Delta_rE_0}{RT}}

Substituting eq4 into the rate constant equation k=\frac{RT}{p^o}k^{\ddagger}K^{\ddagger} yields:

k=\frac{RT}{p^o}k^{\ddagger}\frac{kT}{hv}\bar{K}^{\ddagger}\; \; \; \; \; \; \; \; 5

From assumption 3, k^{\ddagger} is proportional to the vibrational frequency v of the activated complex,

k^{\ddagger}=\kappa v\; \; \; \; \; \; \; \; 6

where \kappa is the proportionality constant called the transmission coefficient, which accounts for the notion that not every oscillation of the activated complex leads to its conversion to the product.

Substituting eq6 into eq5 results in:

k=\kappa\frac{RT}{p^o}\frac{kT}{h}\bar{K}^{\ddagger}\; \; \; \; \; \; \; \; 7

The standard Gibbs energy is \Delta G^o=-RTlnK. By analogy, we define the standard Gibbs activation energy as \Delta^{\ddagger}G^o=-RTln\bar{K}^{\ddagger} , which we shall substitute into eq7 to give:

k=\kappa\frac{RT}{p^o}\frac{kT}{h}e^{-\frac{\Delta^{\ddagger}G^o}{RT}}=\kappa\frac{RT}{p^o}\frac{kT}{h}e^{\frac{\Delta^{\ddagger}S^o}{R}}e^{-\frac{\Delta^{\ddagger}H^o}{RT}}\; \; \; \; \; \; \; \; 8

 

Question

Is the expression for standard Gibbs activation energy still valid when the equilibrium constant now excludes the factor q_{C^{\ddag},m}^{o}\, ^{V_{asym}} ?

Answer

It is an approximation that we make, which delivers an expression for the rate constant that is in good agreement with experimental values.

 

Eq7 and eq8 are different forms of the Eyring equation. Substituting eq8 in the definition of activation energy E_a=RT^2\frac{\partial lnk}{\partial T} and differentiating, noting that \Delta^{\ddagger}S^o and \Delta^{\ddagger}H^o are standard states, which are temperature-independent, gives:

E_a=\Delta^{\ddagger}H^o+2RT\; \; \; \; \; \; \; \; \; 9

Substituting eq9 into eq8 yields:

k=\kappa e^2\frac{RT}{p^o}\frac{kT}{h}e^\frac{\Delta^{\ddagger}S^o}{R}e^{-\frac{E_a}{RT}}\; \; \; \; \; \; \; \; 10

Comparing eq10 and the Arrhenius equation, we have:

A=\kappa e^2\frac{RT}{p^o}\frac{kT}{h}e^\frac{\Delta^{\ddagger}S^o}{R}

Repeating all the above steps for a gas-phase unimolecular reaction, we arrive at an activation energy of E_a=\Delta^{\ddagger}H^o+RT instead of eq9. Therefore, the transition state theory predicts that the activation energy for a gas-phase elementary reaction is

E_a=\Delta^{\ddagger}H^o+xRT

where x=1 for a unimolecular reaction and x=2 for a bimolecular reaction.

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2nd order reaction of the type (A + B → P)

The rate law for the reaction A + BP is:

\frac{d[A]}{dt}=-k[A][B]\; \; \; \; \; \; \; \; 11

Unlike the 2nd order reaction of the type AP with a rate law of \frac{d[A]}{dt}=-k[A]^2, eq11 cannot be integrated unless we express [B] in terms of [A], or [A] and [B] in terms of a third variable x. Since A and B react in a ratio of 1:1, the remaining concentrations of A and B at any time of the reaction are:

[A]=[A_0]-x\; \; \; and\; \; \; [B]=[B_0]-x

where [A0] and [B0] are the initial concentrations of A and B respectively.

Differentiating both sides of [A] = [A0] – x with respect to time yields

\frac{d[A]}{dt}=-\frac{dx}{dt}

Therefore, we can rewrite eq11 as

\frac{dx}{dt}=k\left ( a-x \right )\left ( b-x \right )\; \; \; \; \; \; \; \; 12

where a = [A0] and b = [B0]

Integrating eq12 using the partial fraction expression of \frac{1}{(a-x)(b-x)}=\frac{1}{b-a}\left ( \frac{1}{a-x}-\frac{1}{b-x} \right ) gives

\int_{0}^{x}\frac{1}{b-a}\left ( \frac{1}{a-x}-\frac{1}{b-x} \right )dx=k\int_{0}^{t}dt

After some algebra, we have,

kt=\frac{1}{b-a}ln\frac{a(b-x)}{b(a-x)}

which is equivalent to

kt=\frac{1}{[B_0]-[A_0]}ln\frac{[A_0][B]}{[B_0][A]}

 

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Equilibrium constant (derivation)

The equilibrium constant of a chemical reaction is the reaction quotient of the reaction at dynamic equilibrium.

The derivation of the equilibrium constant involves the following steps:

    1. Derive the expression for the Gibbs energy of a multi-component reaction.
    2. Derive the chemical potential of a multi-component system.
    3. Combine the two expressions.

Step 1

From eq14 of the article on the Nernst equation, the reaction Gibbs energy for a reaction is

\Delta_rG=\sum_j\mu_jv_j\; \; \; \; \; \; \; \; 1

and at standard state

\Delta_rG^{\: o}=\sum_j\mu_j^{\; \; o}v_j\; \; \; \; \; \; \; \; 2

Step 2

From eq24 of the article on the Nernst equation, the chemical potential of a multi-component system is

\mu_j=\mu_j^{\; \, o}+RTlna_j\; \; \; \; \; \; \; \; 3

Step 3

Combining eq1 and eq3

\Delta_rG=\sum_j\left ( \mu_j^{\; o}+RTlna_j \right )v_j=\sum_j\mu_j^{\; o}v_j+RT\sum_jv_jlna_j\; \; \; \; \; \; \; \; 4

Substitute eq2 in eq4

\Delta_rG=\Delta_rG^{\; o}+RT\sum_jlna_j^{\: v_j}\; \; \; \; \; \; \; \; 5

Since lnxa +lnxb +… = ln(xaxb …), eq5 becomes

\Delta_rG=\Delta_rG^{\; o}+RTln\prod _ja_j^{\: v_j}\; \; \; \; \; \; \; \; 6

Let Q=\prod_ja_j^{\; v_j}, where Q is the reaction quotient.

\Delta_rG=\Delta_rG^{\; o}+RTlnQ\; \; \; \; \; \; \; \; 7

At equilibrium, a reversible reaction is spontaneous in neither direction. Hence the change in Gibbs energy with respect to the change in the extent of the reaction, i.e. the reaction Gibbs energy \Delta_rG=\left ( \frac{\partial G}{\partial\xi} \right )_{T,p}, is zero. Eq7 becomes

\Delta_rG^{\circ}=-RTlnK\; \; \; \; \; \; \; \; 8

where K is the reaction quotient at equilibrium, i.e.

K=\left ( \prod _ja_j^{\; v_j} \right )_{eqm}\; \; \; \; \; \; \; \; 9

Since the value K of does not change for a particular reaction at constant temperature, we call it the equilibrium constant.

But how does , which is dimensionless, relate to and ? and are approximations of the thermodynamic equilibrium constant . For an ideal gas,

where .

Similarly, for solutes in dilute solutions,

Eq21a and eq21b show that and are only unitless when . Equating the two expressions gives:

 

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Nernst equation (derivation)

In 1887, Walther Nernst, a German chemist, developed the Nernst equation, which describes the relationship between an electrochemical reaction’s potential and its standard electrode potential under standard and non-standard conditions.

The derivation of the Nernst equation involves the following steps:

    1. Derive the formula for the reversible open circuit potential , Eo, of an electrochemical cell at constant temperature and pressure under standard conditions.
    2. Derive the expression for Eo in terms of chemical potentials of the reaction species at constant temperature and pressure.
    3. Derive a ‘correction factor’ for Eo in step2. This final expression, the Nernst equation, describes how the resultant Eo varies with activities of chemical species.

 

Step 1

From the definition of enthalpy, H = U + pV, we have

dH=dU+d(pV)\; \; \; \; \; \; \; \; 1

Substitute the definition of the internal energy of a system undergoing reversible change, dU = dqrev + dwrev in eq1

dH=dq_{rev}+dw_{rev}+d(pV)\; \; \; \; \; \; \; \;2

Substitute the definition of the change in Gibbs energy of a system, dG = dHd(TS) in eq2

dG+TdS+SdT=dq_{rev}+dw_{rev}+pdV+Vdp

At constant temperature and pressure, dT = dp = 0

dG=dq_{rev}+dw_{rev}+pdV-TdS\; \; \; \; \; \; \; \; 3

Substitute the definition of the change in entropy of a system, dS = dqrev/T in eq3

dG=dw_{rev}+pdV\; \; \; \; \; \; \; \; 4

Substitute the definition of total reversible work, wrev = wex + wadd, where wex is reversible expansion work and wadd is reversible additional work other than reversible expansion work, in eq4

dG=dw_{ex}+dw_{add}+pdV\; \; \; \; \; \; \; \; 5

Substitute the definition of the change in reversible expansion work, dwex = –pdV in eq5

dG=dw_{add}\; \; \; \; \; \; \; \; 6

Substitute the definition of the reversible electrical work between two points in an electric circuit, dwadd = –nFEdξ in eq6

\Delta _rG=-nFE\; \; \; \; \; \; \; \; 7

where n is the number of moles of electrons, F is the Faraday constant, E is the open circuit potential of the electrochemical cell and ΔrG = dG/ is the Gibbs energy of the electrochemical cell reaction.

When the open circuit potential of an electrochemical cell is measured at standard conditions (1 bar, 298K, 1M), we write eq7 as:

\Delta _rG^{\, o}=-nFE^{\, o}\; \; \; \; \; \; \; \; 8

 

Step 2

The chemical potential of the j-th species, in a multi-component system is

\mu_j=\left (\frac{\partial G}{\partial n_j} \right )_{T,p,n'}\; \; \; \; \; \; \; 9

where n‘ is all n other than nj.

The total differential of the Gibbs energy of the system at constant temperature and pressure is

dG=\left (\frac{\partial G}{\partial n_a} \right )_{T,p,n'}dn_a+\left (\frac{\partial G}{\partial n_b} \right )_{T,p,n'}dn_b+...\; \; \; \; \; \; \; 10

Substituting eq9 in eq10

dG=\mu_adn_a+\mu_bdn_b+...=\sum_{j}\mu_jdn_j\; \; \; \; \; \; \; \; 11

Eq11 can be expressed in another way by considering the reversible reaction

v_aa+v_bb+...\rightleftharpoons v_{\alpha}\alpha+v_{\beta}\beta+...

with the change in Gibbs energy at constant temperature and pressure of the system given by eq11. Next, let’s define

dn_j=v_jd\xi\; \; \; \; \; \; \; \; 12

where vj is the stoichiometric number of the j-th species in the reversible reaction and is the amount of substance that is being changed in the reaction.

ξ is called the extent of a reaction and has units of amount in moles. Note that the stoichiometric numbers for reactants are negative by convention and those for products are positive. Substituting eq12 in eq11

\left ( \frac{\partial G}{\partial \xi} \right )_{T,p}= \sum_{j}\mu_jv_j\; \; \; \; \; \; \; \; 13

Substitute the definition of the Gibbs energy of an electrochemical cell reaction, \Delta _rG=\left ( \frac{\partial G}{\partial \xi} \right )_{T,p} , into eq13

\Delta _rG=\sum_{j}\mu_jv_j\; \; \; \; \; \; \; \; 14

Substitute eq7 from Step1 in eq14

E=-\frac{1}{nF}\sum_{j}\mu_jv_j\; \; \; \; \; \; \; \; 15

When the open circuit potential of an electrochemical cell is measured at standard conditions (1 bar, 298K, 1M), we write eq15 as:

E^{\, o}=-\frac{1}{nF}\sum_{j}\mu^o_{\, j}v_j\; \; \; \; \; \; \; \; 16

 

Step 3

Firstly, let’s consider a system with a single pure substance. Substitute the definition of the change in Gibbs energy of a system, dG = dHd(TS), in the change of enthalpy of the system dH = dU + d(pV)

dG+SdT+TdS=dU+pdV+Vdp\; \; \; \; \; \; \; \; 17

The reversible change in internal energy of a pure substance (a system with a single pure substance only does expansion work) can be expressed as

dU=dq_{rev}+dw_{ex}\; \; \; \; \; \; \; \; 18

Substitute the definitions of entropy, dS = dqrev/T, and expansion work, –pdV, in eq18

dU=TdS-pdV\; \; \; \; \; \; \; \; 19

Substitute eq19 in eq17

dG=Vdp-SdT\; \; \; \; \; \; \; \; 20

At constant temperature, dT = 0 and eq20 becomes dG = Vdp. Integrating this new expression on both sides from p_i to p_f,

G\left ( p_f \right )=G\left ( p_i \right )+nRTln\frac{p_f}{p_i}

Let pi = 1 bar = po (i.e. standard conditions) and pf = p

G=G^{\, o}+nRTln\frac{p}{p^{\, o}}

Secondly, we extend the above working to a multi-component system of ideal gases. Since G is an extensive property, we add the Gibbs energies of all components to give the total Gibbs energy of the system:

G_{total}=\left ( G_a^{\: o}+n_aRTln\frac{p_a}{p_a^{\, o}} \right )+\left ( G_b^{\: o}+n_bRTln\frac{p_b}{p_b^{\, o}} \right )+...

Let’s assume the reference pressures of the component gases are the same since they are all ideal gases, i.e. pao = pbo = p

G_{total}=\left ( G_a^{\, o}+G_b^{\, o}+...\right )+\left ( n_aRTln\frac{p_a}{p^o}+n_bRTln\frac{p_b}{p^o}+...\right )

G_{total}=G_{total}^{\, o}+\left ( n_aRTln\frac{p_a}{p^o}+n_bRTln\frac{p_b}{p^o}+...\right )

For simplicity, let GTotal and GoTotal be G and Go respectively.

G=G^{\, o}+\left ( n_aRTln\frac{p_a}{p^o}+n_bRTln\frac{p_b}{p^o}+...\right )

Taking the partial derivative of the above with respect to nj, at constant temperature, pressure and the amount of the other components, n‘, we have

\left ( \frac{\partial G}{\partial n_j} \right )_{T,p,n'}=\left ( \frac{\partial G^{\, o}}{\partial n_j} \right )_{T,p,n'}+\left [ \frac{\partial \left ( n_aRTln\frac{p_a}{p^{\, o}}+ n_bRTln\frac{p_b}{p^{\, o}}+...\right )}{\partial n_j} \right ]_{T,p,n'}

For the last partial derivative of the above equation, only the j-th term survives.

\left ( \frac{\partial G}{\partial n_j} \right )_{T,p,n'}=\left ( \frac{\partial G^{\, o}}{\partial n_j} \right )_{T,p,n'}+RTln\frac{p_j}{p^{\, o}}\; \; \; \; \; \; \; \; 21

Substitute eq9 from Step 2 in eq21

\mu_j=\mu_j^{\, o}+RTln\frac{p_j}{p^{\, o}}\; \; \; \; \; \; \; \; 22

Similarly, for a solute in a dilute solution satisfying Henry’s law, we can write

\mu_j=\mu_j^{\, o}+RTln\frac{c_j}{c^{\, o}}\; \; \; \; \; \; \; \; 23

where cj and co are the j-th solute’s concentration and molar concentration respectively.

We can fine-tune eq22 and eq23 by replacing pj/po and cj/co with γj(pj/po) and γj(cj/co) where γj is a factor called the activity coefficient that can account for both ideal and non-ideal fluids (γ = 1 and γ < 1 for ideal and non-ideal fluids respectively). Next, we combine the modified equations into a single equation by letting aj = γj(pj/po) and aj = γj(cj/co), where aj is the activity of species j.

\mu_j=\mu_j^{\, o}+RTlna_j\; \; \; \; \; \; \; \; 24

Substituting eq24 in eq15 from Step 2,

E=-\frac{1}{nF}\sum_{j}\left ( \mu_j^{\, o}+RTlna_j \right )v_j=-\frac{1}{nF}\sum_{j} \mu_j^{\, o}v_j-\frac{RT}{nF}\sum_{j}v_jlna_j \; \; \; \; \; \; \; \; 25

Substituting eq16 from Step 2 in eq25

E=E^{\, o}-\frac{RT}{nF}\sum_{j}ln\, a_j^{\, v_j}=E^{\, o}-\frac{RT}{nF}ln\prod_{j}a_j^{\, v_j}

E=E^{\, o}-\frac{RT}{nF}lnQ\; \; \; \; \; \; \; \; 26

where Q is the reaction quotient with Q\prod _{j}a_j^{\, v_j}. 

Hence, \frac{RT}{nF}lnQ is the ‘correction factor’ for Eo, as the activities of reaction species vary. Eq26 is the Nernst equation.

For an electrochemical equation of the form Ox+ne^-\rightleftharpoons Red, the Nernst equation can be written as:

E=E^{\, o}-\frac{RT}{nF}ln\frac{a_{ Red}}{a_{ Ox}}\; \; \; \; \; \; \; \; 27

Note that since Eo is by definition a reduction potential, the reference equation is always Ox+ne^-\rightleftharpoons Red and hence eq27.

 

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pH measurement using a glass electrode

A typical experiment setup for pH measurement consists of a glass electrode (indicator electrode), a reference electrode (a regular AgCl electrode), the analyte and a potentiometer. The potentiometer ensures that current flow is negligible in the electrochemical circuit so that IR drop and overpotential are minimised, resulting in more accurate measurements. Such a method in electrochemistry where current is negligible is called potentiometry.

Comparing the setup with a circuit diagram (see above diagram), the overall potential of the electrochemical cell is:

E_{cell}=E_{glass\: elec}-E_{ext\: ref\: elec}+E_{liq\: jn}\; \; \; \; \; \; \; 31

The liquid junction potential arises due to the difference in mobility of K+ and Cl in the controlled flow of the saturated KCl solution out of the porous frit of the external reference AgCl electrode (to complete the circuit). Eanalyte = IR = 0 since current flow is negligible in potentiometric methods.

 

Question

How does the potentiometer control current in the circuit?

Answer

A potentiometer works by sliding a contact along a resistor and finding the relationship between VL and Ecell (see diagram below).

For a pH measurement experiment, the contact moves to the point where no current flows. Ecell is then calculated by:

E_{cell}=\frac{R_1R_L+R_2R_L+R_1R_2}{R_2R_L}V_L

In a modern pH meter, a computer controls the potentiometer.

 

Lastly, before a pH meter is put to use, it is calibrated using solutions of known pH. The reason for this is as follows:

Substituting eq30 from the previous article in eq31,

E_{cell}=L'-0.0592\, pH

where L’ = LEext ref elec + Eliq jn.

Since Eliq jn cannot be calculated or measured directly, the value of L’ must be determined by calibrating the pH meter with solutions of known pH.

 

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IR drop (electrochemistry)

IR drop in electrochemistry is the voltage drop due to the resistance of the analyte (and the internal KCl solution if a calomel or AgCl electrode is used) in a close circuit.

In the previous article regarding pH measurement, which involves potentiometry, no net current flows in the circuit. When a current flows, e.g. in coulometry and electrogravimetry, we have to consider two effects:

We can represent IR drop as an internal resistance of a battery in the following circuit diagram:

The general equation for a galvanic cell is:

E_{galvanic}=IR=E_{cell,open}-Ir\; \; \; \; \; \; \; \; 32

where Ecell, open = EREL.

Eq32 when applied to a Zn electrochemical cell gives the following values:

Ecell, open and Egalvanic for the above example are recorded as negative values by the voltmeters because of the way the terminals of the voltmeters are connected to the respective circuits. From eq32, \left | E_{galvanic} \right | is always less than \left | E_{cell,open} \right |.

For an electrolytic cell,

E_{applied}=E_{cell,open}-Ir\; \; \; \; \; \; \; \; 33

If no current is flowing in the electrolytic cell, eq33 reduces to Eapplied = Ecell, open. For electrolysis to proceed, \left | E_{applied} \right |>\left | E_{cell,open} \right |. For example, the variation of Eapplied for the electrolytic reduction of Zn2+ is illustrated by the following diagram:

Once again, Eapplied for the above example is recorded as a negative value by the voltmeter because of the way the voltmeter is connected to the circuit.

 

Question

With reference to the above electrochemical cell that is composed of a saturated AgCl reference electrode, a Zn working electrode and a 0.5 M Zn2+ electrolyte at rtp, what external voltage must be applied to generate an electrolytic current of 3 mA if the internal resistance is 10 Ω. What is the IR drop?

Answer

Zn^{2+}(aq)+2e^-\rightleftharpoons Zn(s)\; \; \; \; \; \; \; E^{\, o}=-0.762\: V

AgCl(s)+e^-\rightleftharpoons Ag(s)+Cl^-(aq)\; \; \; \; \; \; \; E^{\, o}=0.199\: V

Using the Nernst equation,

 

E_{cell,open}=-0.762-\frac{0.0592}{2}log \frac{1}{0.5}-0.199=-0.970\: V

Using eq33,

E_{applied}=-0.970-(0.003)(10)=-1.000\: V

Therefore, the IR drop is

 

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pH glass electrode

A pH glass electrode is an ion-selective electrode that is sensitive to H+, and therefore, used to measure pH of solutions. It is composed of a glass bulb that is H+-selective and an internal AgCl electrode.

The glass bulb is specially manufactured with a composition of 70+% SiO2 and 20+% Na2O. It has a silicate network within the glass matrix, with some Si atoms singly bonded to O atoms and other Si atoms coordinated with Na+ (see diagram above). On the outer and inner surfaces of the bulb, some of the SiO arms are also coordinated with Na+, while the rest of the SiO are uncoordinated.

Before the pH glass electrode can be used, it must be hydrated (soaked in water of pH = 7) to form two layers of gel. This is to allow H+ to diffuse into the gel layers to convert the SiONa+ groups on both surfaces to SiOH+. Since the inner surface of the bulb is already in contact with a KCl buffer solution at pH 7, both the inner and outer surfaces develop the same ion-exchange equilibrium when the electrode is dipped in water:

H^++SiO^-Na^+\rightleftharpoons Na^++SiO^-H^+

This ion-exchange occurs because H+ binds more strongly to SiO than Na+. Therefore, the position of the equilibrium is almost entirely to the right. On the inner surface, the displaced sodium ions enter the buffer solution, and on the outer surface, displaced sodium ions enter water. The hydrated surfaces now consist of a combination of SiOH+ and SiOarms, and is ready for use. The electrode is immersed in an analyte of unknown pH, establishing the following equilibrium on the outer surface:

\begin{matrix} H^+_{\; \; analyte}+SiO^-\\ outer\; activity \end{matrix} \rightleftharpoons\; \begin{matrix} SiO^-H^+_{\; \; analyte}\\ glass\; activity \end{matrix}

and the following equilibrium on the inner surface equilibrium:

\begin{matrix} H^+_{\; \; buffer}+SiO^-\\ inner\; activity \end{matrix} \rightleftharpoons\; \begin{matrix} SiO^-H^+_{\; \; buffer}\\ glass\; activity \end{matrix}

If the pH of the analyte is less than 7, the outer surface will have fewer uncoordinated SiO arms and hence, less negative charges as compared to the inner wall, whose equilibrium remains unchanged. The difference in charges on the surfaces produces a potential difference across the glass bulb. Furthermore, it confers enough activation energy for sodium ions within the glass matrix to dissociate from the silicate network. These mobile sodium ions then migrate interstitially across the bulb and along the potential gradient, to complete the circuit.

Even though the above equilibria are ion-exchanges and not redox reactions, the potential across the bulb Eb is found empirically to be Nernst-like, where:

E_{outer}=E^{\, o}_{\; glass}-\frac{RT}{nF}ln\frac{a_{glass}}{a_{\, outer}}

E_{inner}=E^{\, o}_{\; glass}-\frac{RT}{nF}ln\frac{a_{glass}}{a_{\, inner}}

E_b=E_{outer}-E_{inner}=-\frac{RT}{nF}ln\frac{a_{inner}}{a_{\, outer}}\; \; \; \; \; \; \; \; 28

 

Question

Why Eb = EouterEinner and not Eb = EinnerEouter?

Answer

We write Eb = Eouter  Einner to be consistent with the way we write Ecell ER – EL, which follows the IUPAC convention (consistency is best visualised by tracing the direction of electron flow in the circuit).

 

Comparing with a circuit diagram (see above diagram), the overall potential of the glass electrode is:

E_{glass\: electrode}=E_b-E_{int \, soln}+E_{int\: elec}

where Eint elec is the potential of the internal AgCl electrode and Eint soln is the potential drop across the saturated KCl solution.

Since no current flows in potentiometric methods,

E_{glass\: electrode}=E_b+E_{int\: elec}

Furthermore, manufacturing defect or damage of electrode over time results in the electrode recording a non-zero potential when the analyte is pH = 7. We call this an asymmetry potential, which must be included in Eglass:

E_{glass\: electrode}=E_b+E_{int\: elec}+E_{asym}\; \; \; \; \; \; \; \; 29

Substituting eq28 in eq29, where ainner, Easym and Eint elec are constants,  Eglass eletrode at rtp is:

E_{glass\, electrode}=L+0.0592\, log\, a_{outer}=L-0.0592\, pH\; \; \; \; \; \; \; \; 30

where L = Eint elec + Easym – 0.0592 log ainner.

Question

What is the function of the internal AgCl electrode?

Answer

For the pH glass electrode to work in a pH measurement experiment, there must be 2 electrodes, where redox reactions occur (to generate and receive electrons). The internal AgCl electrode serves as one of the electrodes, while the reference electrode (an external regular AgCl electrode) serves as the other electrode. The bulb, which is the pH-sensing element, can be perceived as an extension of the internal electrode.

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