Advanced Chemistry - Mono Mole

October 22, 2019September 26, 2024

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:

- IR drop
- Overpotential

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 E_cell,_open = E_R – E_L.

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

E_{cell, open} and E_galvanic 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 E_applied= E_{cell, open}. For electrolysis to proceed, $\left | E_{applied} \right |>\left | E_{cell,open} \right |$ . For example, the variation of E_applied for the electrolytic reduction of Zn²⁺ is illustrated by the following diagram:

Once again, E_applied 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

$Ir=-0.970-(-1.000)=0.03 \;V$

Next article: Polarisation

Previous article: pH measurement using a glass electrode

Content page of advanced electrochemistry

Content page of Advanced chemistry

Main content page

October 22, 2019October 27, 2024

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 Si–O^– arms are also coordinated with Na⁺, while the rest of the Si–O^– 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 Si–O^–Na⁺ groups on both surfaces to Si–O^–H⁺. 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 SiO^–H⁺and SiO^–arms, 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 Si–O^– 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 E_bis 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 E_b = E_outer – E_inner and not E_b = E_inner – E_outer?

Answer

We write E_b = E_outer – E_inner to be consistent with the way we write E_cell= E_R – E_L, 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 E_{int elec} is the potential of the internal AgCl electrode and E_{int 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 E_glass:

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

Substituting eq28 in eq29, where a_inner, E_asymand E_{int elec}are constants, E_{glass eletrode} at rtp is:

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

where L = E_{int elec}+ E_asym– 0.0592 log a_inner.

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.

Next article: pH measurement using a glass electrode

Previous article: Nernst equation

Content page of advanced electrochemistry

Content page of Advanced chemistry

Main content page

October 22, 2019December 4, 2024

Polarisation (electrochemistry)

Polarisation is the deviation of the electrode potential from its equilibrium value, E_eqm, when a certain level of current flows. It is composed of two components: activation (or kinetic) polarisation and concentration polarisation.

Activation polarisation is the change in electrode potential from its equilibrium value when a certain level of current flows, resulting from the overcoming of the activation energy of the rate-determining step in an electrochemical reaction. Concentration polarisation is the change in electrode potential from its equilibrium value when a certain level of current flows, caused by inefficiencies in electrode reactions due to sluggish mass transfer between the electrode surface and the solution.

The extent of the polarisation of an electrode or an electrochemical cell is measured by a quantity called overpotential, η, where

$\eta=\eta_{act}+\eta_{con}=E-E_{eqm}\; \; \; \; \; \; \; \; 34$

It is difficult to directly measure the individual components of overpotential when an electrochemical cell is operating. However, by making certain assumptions, it is possible to derive formulas attributing to η_actand η_con, and design experiments based on these assumptions to measure the two quantities.

Question

Why is IR drop not part of the overpotential equation?

Answer

Overpotential is a consequence of the reactions at an electrode while the IR drop is characteristic of the bulk solution.

Next article: Activation Polarisation

Previous article: IR drop

Content page of advanced electrochemistry

Content page of Advanced chemistry

Main content page

October 22, 2019September 26, 2024

Activation polarisation (electrochemistry)

Activation polarisation is the change in electrode potential from its equilibrium value when a current flows, as a result of overcoming the activation energy of the rate-determining step in an electrochemical reaction.

To understand the concept of activation polarisation, let’s analyse an experiment with the following electrode reaction:

$Ox+e^-\begin{matrix} r_{red}\\\rightleftharpoons \\ r_{ox} \end{matrix}Red\; \; \; \; \; \; \; \; 35$

where r_red is defined as the flow of the number of moles of Ox per unit area of the electrode per unit time, i.e. the rate of reduction of Ox, and r_ox is the flow of the number of moles of Red per unit area of the electrode per unit time, i.e. the rate of oxidation of Red. We represent r_red and r_ox with the following rate equations:

$r_{red}=k_c[Ox]\; \; \; \; \; \; \; \; 36$

$r_{ox}=k_a[Red]\; \; \; \; \; \; \; \; 37$

where k_c and k_a are the cathodic rate constant and anodic rate constant respectively (each with units of length per time). From the combined Faraday’s laws of electrolysis, we have

$m=\frac{Q}{F}\left ( \frac{M}{z} \right )\; \; \Rightarrow \; \; zF\frac{m}{M}=It\; \; \Rightarrow \; \; zF\frac{no.\: of\: moles}{t}=I\; \; \; \; \; \; \; \; 38$

Divide eq38 throughout by the area of electrode,

$\frac{I}{A}=zF\frac{no.\; of\; moles}{At}\; \; \Rightarrow \; \; j=zFr\; \; \; \; \; \; \; \; 39$

where j = I/A, is the current density. With reference to eq35, z = 1, and so we can rewrite eq36 and eq37 as:

$j_c=Fk_c[Ox]\; \; \; \; \; \; \; \; 40$

$j_a=Fk_a[Red]\; \; \; \; \; \; \; \; 41$

where j_c and j_a are the cathodic current density and the anodic current density respectively.

To apply the above equations, consider a piece of metal M immersed in an M⁺ solution. Over time, an electric double layer is formed, with the development of an equilibrium electrode potential between the metal and its ionic solution, such that r_red = r_ox or j_c = j_a (i.e. no net current flowing). If we connect the M⁺/M half-cell to an SHE, the circuit is closed.

Assuming that M lies below hydrogen in the electrochemical series, a net current flows towards the SHE, i.e. electrons flow towards M, the cathode. This results in a less positive electrode reduction potential and the equilibrium in eq35 shifting to the right, with r_red > r_ox. Since the electrode potential deviates from its equilibrium value when a current flows, we say that the electrode is polarised.

In terms of current density, j_c is no longer equal to j_a, and because it is a reduction reaction, j_c > j_a, i.e. a net cathodic current flows. The net current density j is given by:

$j=j_a-j_c=Fk_a[Red]-FK_c[Ox]\; \; \; \; \; \; \; \; 42$

Question

Why is the net current density j_a – j_c and not j_c – j_a?

Answer

This is to be consistent with IUPAC’s definition that a net anodic current is positive, while a net cathodic current is negative (as mentioned above, the current is cathodic when j_c > j_a, which makes j < 0).

Substituting the Eyring equation $k=Ae^{-\frac{\Delta ^{\ddagger'}G}{RT}}$ where $\Delta^{\ddagger'}G$ is the activation Gibbs energy in eq42,

$j=FA_a\left [ Red \right ]e^{-\frac{\Delta ^{\ddagger'}G_a}{RT}}-FA_c\left [ Ox \right ]e^{-\frac{\Delta ^{\ddagger'}G_c}{RT}}\; \; \; \; \; \; \; \; 43$

with

$j_a=FA_a\left [ Red \right ]e^{-\frac{\Delta ^{\ddagger'}G_a}{RT}}\; \; \; \; \; \; \; \; 44$

$j_c=FA_c\left [ Ox \right ]e^{-\frac{\Delta ^{\ddagger'}G_c}{RT}}\; \; \; \; \; \; \; \; 45$

When the electrode is at equilibrium, j_a = j_cand we can rewrite eq44 and eq45 as:

$j_0=FA_a[Red]e^{-\frac{\Delta ^\ddagger G_a}{RT}}\; \; \; \; \; \; \; \; 46$

$j_0=FA_c[Ox]e^{-\frac{\Delta ^\ddagger G_c}{RT}}\; \; \; \; \; \; \; \; 47$

where j₀ denotes the equal current densities at equilibrium and is called the exchange current density; and $\Delta^{\ddagger}G_a$ and $\Delta^{\ddagger}G_c$ are the anodic activation Gibbs energy at equilibrium and the cathodic activation Gibbs energy at equilibrium respectively.

Next, we shall investigate the energy profile of the reaction of eq35 at the cathode at equilibrium (see Fig I below). For simplicity, we assume that the transition state has equal resemblance to M⁺ and M and therefore the activation Gibbs energy required for the reduction reaction is the same as that of the oxidation reaction.

As mentioned, when the electrode is connected to an SHE, its potential becomes less positive, which in turn causes a decrease of the activation Gibbs energy for the reduction reaction (electrode has stronger attraction for M⁺) and an increase in the activation Gibbs energy for the oxidation reaction (more difficult for M⁺ to form). The resultant energy profile is shown in Fig II. Note that the peaks in Fig I and Fig II are aligned at the same Gibbs energy level for convenience.

Question

Use eq44 and eq45 to justify the change in activation Gibbs energies for the reduction and oxidation reaction at the cathode when a current flows

Answer

When a cathodic current flows, j_c > j_a, i.e. cathodic current density level increases versus that at equilibrium. Assuming that the electrolyte is well stirred such that [Ox] remains constant, $\Delta ^\ddagger G_c$ in eq45 must decrease. Simultaneously, the reduction in anodic current density at the same electrode results in an increase in $\Delta ^\ddagger G_a$ (see eq44 where [Red] is constant, as it is a solid).

By letting $\left | \Delta G \right |$ be the magnitude of the total change in Gibbs energy between the state when a current flows and the state when the electrode is at equilibrium, we have:

$\Delta ^{\ddagger}G_c=\Delta ^{\ddagger'}G_c+\alpha \left | \Delta G \right |\; \; \; \; \; \; \; \; 48$

$\Delta ^{\ddagger}G_a=\Delta ^{\ddagger'}G_a-\left ( 1- \alpha\right )\left | \Delta G \right |\; \; \; \; \; \; \; \; 49$

where $\alpha$ is a fraction of $\left | \Delta G \right |$ and is called the transfer coefficient. If you recall, the formula $\Delta _rG=-nFE$ is derived based on the definition of the reversible electrical work between two points in a circuit with a potential difference E. The change in Gibbs energy in our case is due to the change in electrode potential when a current flows, E – E_eqm and so, we have:

$\left | \Delta G \right |=-nF\left ( E-E_{eqm} \right )\; \; \; \; \; \; \; \; 50$

Substituting eq34 from the previous article in eq50, the magnitude of the change in Gibbs energy for the passage of one mole of electrons (n = 1) is:

$\left | \Delta G \right |=-F\eta\; \; \; \; \; \; \; \; 51$

Substituting eq51 in eq48 and eq49,

$\Delta^{\ddagger}G_c=\Delta^{\ddagger'}G_c-\alpha F\eta\; \; \; \; \; \; \; \; 52$

$\Delta^{\ddagger}G_a=\Delta^{\ddagger'}G_a+\left ( 1- \alpha\right )F\eta\; \; \; \; \; \; \; \; 53$

Substituting eq53 and eq52 in eq44 and eq45 respectively,

$j_a=FA_a[Red]e^{-\frac{\Delta^\ddagger G_a}{RT}}e^{\frac{(1-\alpha)F\eta}{RT}}\; \; \; \; \; \; \; \; 54$

$j_c=FA_c[Ox]e^{-\frac{\Delta^\ddagger G_c}{RT}}e^{\frac{-\alpha F\eta}{RT}}\; \; \; \; \; \; \; \; 55$

Substituting eq46 and eq47 in eq54 and eq55 respectively,

$j_a=j_0e^{(1-\alpha)f\eta}\; \; \; \; \; \; \; \; 56$

$j_c=j_0e^{-\alpha f\eta}\; \; \; \; \; \; \; \; 57$

where f = F/RT

The net current density is obtained by substituting eq56 and eq57 in eq42:

$j=j_0\left [ e^{\left ( 1-\alpha \right )f\eta}- e^{-\alpha f\eta}\right ]\; \; \; \; \; \; \; \; 58$

Eq58 is known as the Butler-Volmer equation. If the experiment that we have described so far consists of an electrolyte that is well stirred (so that mass transfer effects are minimised, i.e. no concentration polarisation), the polarisation of the electrode arising from the flow of current is attributed entirely to the change in activation energies of the reduction and oxidation reactions at the electrode, i.e. activation polarisation. Since η is a measure of activation polarisation, η in eq58 is η_act,cat, the activation overpotential at the cathode. Note that η_act,cat< 0 , because η = E – E_eqmwhere E decreases at the cathode upon polarisation.

If M is more electropositive than hydrogen, it becomes the anode when connected to the SHE. The flow of electrons away from M results in a less negative electrode reduction potential. Using the same logic, we arrive at eq58 with η being η_act,an, the activation overpotential at the anode, where η_act,an > 0.

In general, for the electrochemical cell $M_a\left | M_a^{\; +}\left | \right |M_c^{\; +} \right |M_c$ (see this article for notation),

$E_{galvanic}=\Delta\phi_R-\Delta\phi_L=\left ( E_R+\eta_R \right )-\left ( E_L+\eta_L \right )=E_{cell,open}+\eta_R-\eta_L\; \; \; \;\; 59$

where E_{cell, open} is the open-circuit potential of the cell at equilibrium, which is usually the standard electrode potential of the cell, E⁰. Since η_R < 0 and η_L> 0, we can rewrite eq59 as:

$E_{galvanic}=E_{cell,open}-\Pi \; \; \; \; \; \; \; \; 60$

where $\Pi=\left | \eta_R \right |+\eta_L$ . Eq60 shows that the relationship of E_galvanic < E_cell,open is always true when a current flows in an electrochemical cell.

For an electrolytic cell, E_cell,openis usually negative. If we also consider IR drop, the magnitude of the externally applied potential must be greater than the sum of the magnitudes of the cell open-circuit potential, potential due to IR drop and the overpotential, for electrolysis to proceed:

$E_{applied}=E_{cell,open}-Ir-\Pi\; \; \; \; \; \; \; \; 61$

Next article: Concentration Polarisation

Previous article: Polarisation

Content page of advanced electrochemistry

Content page of Advanced chemistry

Main content page

October 22, 2019September 26, 2024

Concentration polarisation (electrochemistry)

Concentration polarisation is the change in electrode potential from its equilibrium value when a certain level of current flows, due to inefficiencies in electrode reactions caused by sluggish mass transfer between the electrode surface and the solution.

When a piece of metal is immersed in its ionic solution, an equilibrium electrode potential is established with no net current flowing. If the metal is connected to another half-cell, e.g. an SHE, and the metal is below hydrogen in the electrochemical series, a net cathodic current flows.

$M^{z+}+ze^- \begin{matrix} r_{red}\\\rightleftharpoons \\ r_{ox} \end{matrix}M$

Furthermore, if the metal electrode is consuming species faster than reactants reaching the electrode surface, a concentration gradient arises between the bulk solution and the electrode surface. Since diffusion is a slow process, the depletion of reactants at the electrode surface leads to a lower electrode potential (similarly, when the electrode is producing species faster than the products leaving the electrode surface, the electrode again becomes polarised). The difference between the modified potential and the equilibrium potential is called the concentration overpotential η_con.

To derive a mathematical relationship between current (or current density) and concentration overpotential, we have to assume that activation overpotential of the cell is negligible. This is achieved by using of non-polarisable electrodes.

At equilibrium, the potential of the cathode is given by the Nernst equation:

$E=E^{\, o}+\frac{RT}{zF}ln[M^{z+}]\; \; \; \; \; \; \; \; 62$

When a current flows and polarisation of the electrode occurs, [M^z+] becomes [M^z+’] where [M^z+’] < [M^z+]. The corresponding open-circuit electrode potential is

$E'=E^{\, o}+\frac{RT}{zF}ln[M^{z+'}]\; \; \; \; \; \; \; \; 63$

The concentration overpotential of the polarised electrode is

$\eta_{con}=E'-E=\frac{RT}{zF}ln\frac{\left [ M^{z+'} \right ]}{\left [ M^{z+} \right ]}\; \; \; \; 64$

According to Fick’s first law of diffusion, the flux of ions, r, towards the cathode is

$r=-D\frac{\left [ M^{z+'} \right ]-\left [ M^{z+} \right ]}{x}\; \; \; \; \; \; \; \; 65$

where D is the coefficient of diffusion with SI units m²s^-1 and x is the distance between the bulk solution and the electrode surface.

Question

How is Fick’s first law derived?

Answer

In diffusion, particles move from a region of higher concentration to one of lower concentration. The flux of particles (i.e. the movement of the number of particles N per unit area per unit time) across a distance x is therefore proportional to the difference of [N]_x=x and [N]_x=0 divided by the change in distance, i.e.

$r=D\frac{d[N]}{dx}$

Knowing that d[N] < 0, if we defined r as positive in the direction of x, the above equation becomes

$r=-D\frac{d[N]}{dx}$

Substituting eq65 in eq39 of the previous article, the cathodic current density is:

$j=zFD\frac{\left [ M^{z+} \right ]-\left [ M^{z+'} \right ]}{x}\; \; \; \; \; \; \; \; 66$

Note that current in a cell not only flows from one electrode to the another electrode along a wire, but also continues to flow back to the first electrode through the electrolyte, and eq66 represents this flow. Rearranging eq66,

$\left [ M^{z+'} \right ]=\left [ M^{z+} \right ]-\frac{jx}{zFD}\; \; \; \; \; \; \; \; 67$

Substituting eq67 in eq64 and rearranging,

$j=\frac{zFD\left [ M^{z+} \right ]}{x}\left ( 1-e^{zf\eta_{\, con}} \right )\; \; \; \; \; \; \; \; 68$

where f = F/RT.

Concentration overpotential, like activation overpotential, leads to a lower potential than the equilibrium potential of a galvanic cell when a current flows.

$E_{galvanic}=E_{cell,open}-\Pi$

where $\Pi=\left | \eta_{con,R} \right |+\eta_{con,L}$ . In the case of an electrolytic cell, a higher applied potential is needed to maintain the desired current. If we include IR drop, the formula is:

$E_{applied}=E_{cell,open}-Ir-\Pi$

Next article: Three-electrode cell

Previous article: Activation Polarisation

Content page of advanced electrochemistry

Content page of Advanced chemistry

Main content page

October 22, 2019June 28, 2025

Three-electrode cell

A three-electrode cell is used for electrochemical measurements involving the passage of current. It is a good set-up to evaluate electrochemical quantities like overpotential and IR drop. A typical three-electrode system consists of a working electrode, a reference electrode and a counter (or auxillary) electrode (see diagram below).

Generally, the potential of a reference electrode must remain constant in order to measure the potential of a working electrode. However, electrodes become polarised when a current flows. The solution is to introduce a third electrode called a counter electrode to the cell for dynamic measurements. A high impedance voltmeter displays the potential of the working electrode and is connected between the working electrode and the reference electrode so that very negligible current flows between them. Experiments involving the polarisation of the working electrode at various current levels can therefore be conducted with respect to the counter electrode. Furthermore, the reference electrode is fitted with a movable column of electrolyte called the Luggin capillary, which allows the measurement of IR drop by varying the distance d.

It is important to note that the counter electrode serves only to complete the circuit for current flow. Ideally, it should not to interfere with reaction at the working electrode and is designed to be non-polarisable. One way to minimise the formation of overpotential at the counter electrode is to increase its surface area, A. This in turn reduces its current density (a fixed amount of current from the potentiostat is spread over a larger area), which is described by the Butler-Volmer equation:

$j=\frac{I}{A}=j_0\left [ e^{(1-\alpha)f\eta}-e^{-\alpha f\eta} \right ]$

Since j₀is an intensive property, $\left [ e^{(1-\alpha)f\eta}-e^{-\alpha f\eta} \right ]$ is small if A is large. This implies that

$e^{(1-\alpha)f\eta}\approx -e^{-\alpha f\eta}$

$(1-\alpha)f\eta\approx -\alpha f\eta\; \; \Rightarrow \; \; f\eta\approx 0$

Since f ≠ 0, η ≈ 0.

Next article: Measuring overpotential

Previous article: Concentration polarisation

Content page of advanced electrochemistry

Content page of Advanced chemistry

Main content page

October 22, 2019October 27, 2024

Measuring overpotential

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

$Fe^{3+}(aq)+e^-\rightleftharpoons Fe^{2+}(aq)\; \; \; \; \; E_{vs\: AgCl(sat'd)}=+0.57\: V$

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

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

$j=\frac{I}{A}=j_0\, e^{(1-\alpha)f\eta}\; \; \; \; \; \; \; \; 69$

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

$ln\left ( \frac{I}{A} \right )=ln j_0+(1-\alpha)f\eta\; \; \; \; \; \; \; \; 70$

Eq70 is known as the Tafel equation, named after the Swiss chemist, Julius Tafel. The plot of ln(I/A) against η is called a Tafel plot and it allows us to find the value of the transfer coefficient $\alpha$ from the gradient of the line and the exchange current density j₀from the intercept. Such an experiment that monitors current at various potentials is called voltammetry.

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

With the potentiometer set to 0 V,

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

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

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

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

$\eta_{an}=E_{an,pol}-E_{L,eqm}>0\; \; \; and\; \; \; \eta_{cat}=E_{cat,pol}-E_{R,eqm}<0$

$\Pi=\eta_{an}+\left | \eta_{cat} \right |$

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

Next article: Pourbaix diagram

Previous article: Three-electrode cell

Content page of advanced electrochemistry

Content page of Advanced chemistry

Main content page

October 22, 2019November 16, 2024

Pourbaix diagram

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

E₁ : $O_2(g)+4H^+(aq)+4e^-\rightleftharpoons 2H_2O(l)\; \; \; \; \; \; \; E^{\, o}=+1.23\, V$

E₂ : $2H^+(aq)+2e^-\rightleftharpoons H_2(g)\; \; \; \; \; \; \; E^{\, o}=0.00\, V$

with the following Nernst equations at rtp, $a_{H_2O}=1$ , $f_{O_2}=1\: atm$ and $f_{H_2}=1\: atm$ :

$E_1=1.23-\frac{(8.31)(298)}{(4)(96485)}ln\frac{1}{\left ( a_{H^+} \right )^4}=1.23-0.0592pH\; \; \; \; \; \; \; 71$

$E_2=-0.0592pH\; \; \; \; \; \; \; \; 72$

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

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

The equilibria involved are:

1) $Cu^{2+}(aq)+2e^-\rightleftharpoons Cu(s)\; \; \; \; \; E^{\, o}=+0.337\, V$

2) $CuO(s)+2H^+(aq)\rightleftharpoons Cu^{2+}(aq)+H_2O(l)$

3) $2Cu^{2+}(aq)+H_2O(l)+2e^-\rightleftharpoons Cu_2O(s)+2H^+(aq)\; \; \; \; \; E^{\, o}=+0.203\, V$

4) $Cu_2O(s)+2H^+(aq)+2e^-\rightleftharpoons 2Cu(s)+H_2O(l)\; \; \; \; \; E^{\, o}=+0.471\, V$

5) $CuO(s)+2H^+(aq)+2e^-\rightleftharpoons Cu_2O(s)+H_2O(l)\; \; \; \; \; E^{\, o}=+0.669\, V$

6) $HCuO_2^{\; -}(aq)+H^+(aq)\rightleftharpoons CuO(s)+H_2O(l)$

7) $2HCuO_2^{\; -}(aq)+4H^+(aq)+2e^-\rightleftharpoons Cu_2O(s)+3H_2O(l)\; E^{\, o}=+1.783\, V$

and their corresponding equations are:

a) $E=0.337+0.0296loga_{Cu^{2+}}$

b) $K=\frac{a_{Cu^{2+}}}{a_{H^+}^{\; \; \; \; \; 2}}\; \; \Rightarrow \; \; logK=loga_{Cu^{2+}}+2pH\; \; \; \left ( K=10^{7.89} \right )$

c) $E=0.203+0.0591loga_{Cu^{2+}}+0.0591pH$

d) $E=0.471-0.0591pH$

e) $E=0.669-0.0591pH$

f) $K=\frac{1}{a_{H^+}a_{HCuO_2^{\; -}}}\; \; \Rightarrow \; \; logK=-loga_{HCuO_2^{\; -}}+pH\; \; \; \left ( K=10^{18.83} \right )$

g) $E=1.783+0.0591loga_{HCuO_2^{\; -}}-0.1182pH$

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

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

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

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

Previous article: Measuring overpotential

Content page of advanced electrochemistry

Content page of Advanced chemistry

Main content page

September 15, 2019December 12, 2025

Complete pH titration curve: overview

Titration curves reveal important data of acid-base systems, such as the pH at the stoichiometric point, the pH at maximum buffer capacity, equilibrium constants, and more. Their shapes are governed by the equilibria of the acid, the base and water. In general, titrations are categorised into four groups:

- Strong acid versus strong base
- Weak acid versus strong base
- Strong acid versus weak base
- Weak acid versus weak base

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

next article: Monoprotic strong acid versus monoprotic strong base

Content page of complete ph titration curve

Content page of advanced chemistry

Main content page

September 15, 2019September 26, 2024

Monoprotic weak acid versus monoprotic weak base

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

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

$10^{-pH}+\frac{K_bC_bV_b10^{-pH}}{(V_a+V_b)(K_w+K_b10^{-pH})}=\frac{K_w}{10^{-pH}}+\frac{K_aC_aV_a}{(V_a+V_b)(10^{-pH}+K_a)}\; \; \; \; \; \; \; \; (12)$

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

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

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

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

Lastly, we can derive the gradient equation for a weak acid to weak base titration and investigate the inflexion point at pH 7.01 by differentiating eq12 implicitly (see this article), resulting in $\frac{dpH}{dV_b}=1.94\, cm^{-3}$ , i.e. a gradient that makes angle of 62.71^o with the horizontal.

Question

Answer

Next article: Polarisation

Previous article: pH measurement using a glass electrode

Question

Answer

Question

Answer

Next article: pH measurement using a glass electrode

Previous article: Nernst equation

Question

Answer

Next article: Activation Polarisation

Previous article: IR drop

Question

Answer

Question

Answer

Next article: Concentration Polarisation

Previous article: Polarisation

Question

Answer

Next article: Three-electrode cell

Previous article: Activation Polarisation

Next article: Measuring overpotential

Previous article: Concentration polarisation

Next article: Pourbaix diagram

Previous article: Three-electrode cell

Previous article: Measuring overpotential

next article: Monoprotic strong acid versus monoprotic strong base

next article: Comparison of strong acid versus strong base curve and weak acid versus strong base curve

Previous article: Monoprotic strong acid versus monoprotic weak base