CH212 Physical Chemistry 1 Electrochemistry (2024-25) PDF

CH212 Physical Chemistry 1 -------------------------- ### Electrochemistry: Fundamentals and Applications **Dr Brian McMillan,** **Room TG103d/TG108** **Email: brian.mcmillan\@strath.ac.uk** - The metal-electrolyte interface, electrochemical equilibrium, electrochemical potential - Ele...

CH212 Physical Chemistry 1 -------------------------- ### Electrochemistry: Fundamentals and Applications **Dr Brian McMillan,** **Room TG103d/TG108** **Email: brian.mcmillan\@strath.ac.uk** - The metal-electrolyte interface, electrochemical equilibrium, electrochemical potential - Electrochemical half-cells and cells, types of electrode, electrode and cell potentials, Nernst equation, liquid junction potential, Thermodynamics - Electrode kinetics, overpotential, the Butler-Volmer equation. - Tafel equation, exchange current density and mass transfer - Linear sweep and Cyclic voltammetry. - Corrosion and Pourbaix diagrams. - Applications of Pourbaix diagrams. Evaluation of the Gibbs energy change in the corrosion reaction. Corrosion prevention and control. - Batteries: battery characteristics; voltage, current, capacity, storage density and cycle life. - Current-voltage discharge curves. Examples: lead-acid, Ni-Cd, Ni-MH, Li^+^ ion. - Introduction to Fuel Cells: reactions and cell design; source of H~2~ fuel; applications and future prospects. **[Assessment]** - Online class test (Teaching Week 10) worth 10% overall module mark. - One question in the December Exam worth 15% overall **Part 1 -- Equilibrium Electrochemistry** **As a starting point, we will think of the interaction between a metal and a solution of its ions. This is one type of electrochemical half-cell.** - ***Metals:*** consist of fixed cations in a lattice with the electrons free to move through the lattice (*i.e.* electrons move, ions don't) - ***Solutions:*** contain ions which move through the solution, positive and negative charges are carried through the solution by these ions - - What happens when two phases which have these charged species come into contact? *e.g.* place a metal M in contact with a solution containing the metal salt, M^n+^ X^m−^. Two processes can occur: 1. **The metal can initially dissolve.** - - - 2. - - - - After the initial event, an **equilibrium** will be quickly reached at which the rate of cations moving into solution will be the same as the rate of cations depositing back onto the metal. - At this point, the metal side of the interface will have a net positive or negative charge with respect to the solution - The charge on the metal side of the interface will depend on the metal, *i.e.* whether or not it 'likes' forming cations in solution. - Thus at equilibrium, a stable interfacial potential difference, *∆φ, *will exist across the metal / solution interface: - Thermodynamics: spontaneous processes occur when they lead to a reduction in the Gibbs energy of the system. - - - These process will reach equilibrium when the Gibbs energy of the metal species is the same in both phases: When we are dealing with charged species (ions and electrons), the movement of charge also contributes to the Gibbs energy and we need to account for this. **The Chemical Potential** - *Τ*he Gibbs energy per mole of an uncharged species is called its **Chemical Potential** and this is given the symbol *μ*. - If two species in different phases, such as a metal atoms in a solid and ions in solution, are in equilibrium with each other then - So, if a species moves from the solid to the solution there will be a change in Gibbs energy of ∆*G* = *μ*~solution~* − μ*~solid~ - If a **charged** species moves across an interface where there is also an *electrical potential difference*, then this will additionally contribute to the change in energy. We therefore define an **[electrochemical potential]** to cover this: where *μ* = chemical potential of the species in J mol^-1^ *Φ * = Galvani potential of species in V (V = J/C) *n* = charge on the species *e.g.* for Cu^2+^ *n* = 2 *F* = the Faraday constant (96485 C mol^−1^) **The Interfacial Potential difference** Let's apply this to an example. Consider a metal ***M*** in contact with a solution of its ions *M^n+^* At equilibrium, the electrochemical potential of the reactants and the products must be equal. Thus, This shows that if the cations in solution have a higher chemical potential there than on the metal, they will tend to escape from the solution and deposit onto the metal. This leads to a build-up of *positive* charges on the metal and thus: These metals are more noble or less reactive such as Au, Ag, Cu *etc*. They 'prefer' to be metals. Conversely, when the chemical potential of metal atoms is higher than the ions in solution, they will tend to escape from the metal into the solution to form cations, leaving *negative* charges (electrons) on the metal. Thus: These metals are less noble or more reactive such as Na, K, Zn *etc*. They 'prefer' to be ions. In electrochemistry, ∆Φ is given the symbol *E*, which is known as the Electrode Potential and these are measurable values which you may have encountered before. Now you know where the come from! There are tables of Standard Electrode Potentials, symbol *E^o^*, which list the potential of electrode reactions under standard conditions. These are defined as: 1 All metals are in their standard states 2 All species in solution have a concentration of 1 mol dm^-3^ 3 All gases are at a pressure of 1 bar. 4 A standard temperature of 298 K is quoted (Thermodynamic Standard States only mention p). Note that it is impossible to measure the potential of an individual electrode in isolation, so the standard electrode potentials are always quoted with respect to the Standard Hydrogen Electrode (SHE) which has a *defined* potential of 0.000 V. These values will be given to you when needed -- no need to remember them. **So, how does this work in practice?** Let's take a standard Zn/Zn^2+^ electrode. This is done by placing a piece of Zn metal into a solution of 1 M Zn^2+^ ions at 25°C. If you look this reaction up in the table, you will see: There are two key points to note: Now, let's take a standard Cu electrode: If we connect a voltmeter to the pieces of metal we can measure the potential difference between the electrodes -- this is called the Cell Potential, *E~cell~*. Here it will also be *E^o^~cell~*. To prevent the build-up of charge in the solution (which will stop the process), it is also necessary to allow ions to flow between the half-cells by adding a salt bridge joining the solutions. When the two electrodes are connected by a wire, electrons will move towards the more positive electrode. This disrupts the equilibrium reactions: *Oxidation* will dominate at one of the electrodes and *Reduction* at the other. Electrons will move from the more negative Zn to the more positive Cu. Electrons arriving on the Cu will force the equilibrium to the left as the equilibrium shifts to remove the extra electrons: We end up with a *Reduction Reaction* on the Cu \ [Cu^2+^ + 2*e*^−^ ⇌ *Cu* *becomes* Cu^2+^ + 2*e*^−^ → *Cu* ]{.math.display}\ Electrons leaving the Zn causes the equilibrium to shift to the right as the equilibrium moves to replace the lost electrons: We end up with an *Oxidation Reaction* on the Zn \ [Zn^2+^ + 2*e*^−^ ⇌ *Zn* *becomes* *Zn* → Zn^2+^ + 2*e*^−^]{.math.display}\ The half-cell reactions are: \ [Cu^2+^ + 2*e*^−^ → *Cu*  *Reduction*]{.math.display}\ \ [*Zn* → Zn^2+^ + 2*e*^−  ^Oxidation]{.math.display}\ Giving [Cu^2+^ + *Zn* → *Cu*  + Zn^2+^]{.math.inline} overall There are two possible cell potentials. We could have: By convention: The electrode where **Reduction** occurs is the **Cathode** The electrode where **Oxidation** occurs is the **Anode** **Each of the electrodes is termed a 'half-cell'. An electrochemical cell requires two half-cells to be brought together.** **We can use the potential difference between the two half-cells to drive electrons through a circuit -- we can use electrochemical cells to do useful work: the amount of oxidation at the anode = the amount of reduction at the cathode.** When we connect the two half-cells together, the equilibria shift, current flows and the reactants get consumed. In this case, the reaction will stop when all of the Zn metal gets oxidised to Zn^2+^ or all of the Cu^2+^ ions deposit on the Cu metal. *As a rule, the more positive half-cell is where the reduction reaction dominates -- note that you need to use the Nernst equation to determine the potential for non-standard half-cells (more on this later).* **\ ** 1. An anode half-cell where something is oxidised 2. A cathode half-cell where something is reduced 3. An electrical connection between the anode and cothode to collect the electrons and allow them to flow and do work *i.e.* current to flow. This could be a wire, LED, any other circuit or 'load'. 4. An ionic connection between the electrodes to allow ions to flow and prevent charge build-up. This could be a salt bridge, filter paper bridge, porous pot or membrane *etc.* which allows ions to pass. So far, we have only considered one type of electrode where the metal is in contact with a solution of its own ions and the potential is sensitive to the metal **cation** concentration. In this type of cell, the metal itself is *both* a reactant *and* it allows electron exchange. This is known as an ***Electrode of the 1^st^ kind***. An ***Electrode of the 2^nd^ kind*** is where the metal ions form an insoluble salt layer on the metal surface and this system is sensitive to the concentration of the **anions**. Here is a common example: An ***Electrode of the 3^rd^ kind*** is where the metal itself is not a reactant. The electrode responds to a change in the redox state of a species which is not the same as the ion that the electrode is made of. Commonly used metals are Pt, Au, Ir, graphite (not a metal!) *etc*. these are good catalysts and generally don't react with species in solution. Here are some examples: In the same way that a steady potential is established when a metal is placed in solution of its ions, a potential is established when an inert electrode is placed in a solution containing the reactants and products of a half-reaction with ions/gases. As metals and graphite/graphene have an abundance of free electrons, they can simply act as a source or sink of electrons. This allows electrochemical reactions to take place on the surface of the metal, but the metal itself does not react. To prevent writing out all the reactions and drawing cumbersome sketches, a cell notation has been devised. The vertical line actually represents the phase boundary *e.g.* between a solid/liquid, liquid/gas *etc*. We put these together to give: We always write the ***R***eduction reaction on the ***R***ight and '\|\|' means that we have used a salt bridge between the two half cells to eliminate the liquid junction potential (more on this later). [Note that the metals themselves are always the outermost components of the notation]. Here are some additional examples involving solids, solutions and gases. Remember that in some cells, the reactions are carried out on inert metals such as Pt, Pd or sometimes graphite. Pt and Fe^2+^/Fe^3+^, AgCl/Ag, In some cells, both electrodes can share the same electrolyte: The H~2~ gas needs a metal to react on, this is usually Pt: [Liquid Junction Potential] ======================================= 1. Consider two HCl solutions at two different concentrations separated by a porous barrier. 2. At the barrier, H^+^ ions will want to cross over from the more concentrated solution to the more dilute solution, as will the Cl^−^ ions. 3. However, since the mobility of the H^+^ ions is much greater than that of Cl^−^ ions, the H^+^ ions will cross at a much faster rate. 4. this will lead to a *net* positive charge in the dilute solution and a *net* negative charge in the concentrated solution and a potential difference thus develops across the barrier 5. this potential difference then slows the movement of H^+^ ions and accelerates that of Cl^−^ ions so that eventually, both ions cross over at the same rate, but the potential difference remains. 6. the STEADY STATE POTENTIAL DIFFERENCE which exists when this occurs is called the ***Liquid Junction Potential***. The only source of electrical energy in the electrochemical cell is *E~cell~*, so some of this is wasted overcoming the liquid junction potential. A good way is to use a salt bride filled with ions of similar mobilities such as a KCl solution or gel. K^+^ and Cl^-^ ions are roughly the same size and move at similar rates. KNO~3~ can be used in Cl^-^ sensitive environments. The cell notation '\|\|' indicates that the liquid junction potential has been eliminated. Note the --ve sign as the work *leaves* the system. \ [Cu^2+^ + 2*e*^−^ → *Cu*  *Reduction*]{.math.display}\ \ [*Zn* → Zn^2+^ + 2*e*^−  ^Oxidation]{.math.display}\ This approach also enables us to work under non-standard conditions. Not every cell has solution concentrations of 1 mol dm^-3^ and gases at 1 bar partial pressure at 298 K. Remember that This is a form of the [Nernst Equation] which enables you to correct the tabulated standard potential values so that you can calculate the potential under any conditions (different concentrations, temperatures, gas pressures) for an **overall reaction**. From now on, we will be using this form of the Nernst Equation below: to correct *each individual half-cell standard potential*. This is more intuitive and leads to fewer calculation errors! For a more complicated example: [Mn*O*~4~^−^ + 8*H*^+^ + 5*e*^−^ → Mn^2+^ + 4*H*~2~O *E*^*o*^ = 1.5 *V*]{.math.inline} We apply this to each half-cell reaction individually, giving: *E~cell~ = E~reduction~* -- *E~oxidation~* and *∆G = -nFE~cell~* So: *E~cell~* +ve, *∆G* -ve, reaction spontaneous as written *E~cell~* -ve, *∆G* +ve, reaction not spontaneous as written When using the Nernst Equation: - Concentrations must be in mol dm^-3^ - Gas pressures must be in bar (if a pressure is in Pa, divide it by 1.0 × 10^5^ Pa) - Solids, including metals, and water have an effective concentration of 1. - Assume T = 298 K unless otherwise stated The Nernst equation is very important. Not only does it allow the potential of a cell or half-cell to be calculated under non-standard conditions, but you can use it to determine what potential to apply to an electrode to give a desired reaction concentration. *Let's look at some examples:* **[Simple example of 2 metal/solution interfaces]**. What is the cell potential for the following cell and is the reaction spontaneous as written? Cd\|Cd^2+^ (0.1 M) \|\| Ag^+^ (0.2 M)\|Ag **[Inert metal electrodes ]** What is the cell potential for the following cell and is the reaction spontaneous as written? **[A gas phase reaction]** What is the cell potential for the following cell and is the reaction spontaneous as written? Pt\| NO~2~ (1000 Pa)\| HNO~3~ (0.01 M) \|\| Cu(NO~3~)~2~ (0.1 M)\|Cu **[\ ]** Recall that: *∆G = ∆H - T∆S* We can therefore calculate the thermodynamic parameters of an electrochemical cell. A neat way of doing this is to measure how *E~cell~* changes with temperature over a short range as we can assume that *∆H* and *∆S* don't change much with *T* (a common assumption from Dr Moore's lectures). By looking at an overall cell reaction, it is sometimes possible to predict the sign of ∆S: Cu\|Cu^2+^\|\|Ag^+^\|Ag or Cu~(s)~ + 2Ag^+^~(aq)~ → Cu^2+^~(aq)~ + 2Ag~(s)~ ∆S Zn\|Zn^2+^ \|\|H^+^,H­­~2~\|Pt or Zn~(s)~ + 2H^+^~(aq)~ → Zn^2+^~(aq)~ + H~2(g)~ ∆S **[An Energy Level View of Electron Transfer]** In CH202, you have (or will) cover band theory. Individual ions in solution maintain their discrete atomic or molecular orbitals, in metals where many atoms are bonded together, the atomic orbitals overlap to form many energy levels of similar energy leading to the formation of an 'energy band.' When a metal and ion in solution come together, electrons are transferred between them. Electrons in the metal 'appear' to come from the top of the conduction band (the Fermi Level). This continues until the Fermi Level matches the energy levels of the ions in solution. At this point, the system is in equilibrium, with the potential described by the Nernst Equation (no net change occurs). [Fundamentals of electron transfer reactions] ========================================================= Where *A* is the electrode area. - At equilibrium, the net current density = 0, but the oxidation and reduction reactions are still happening. The magnitude of these current densities are equal, but their signs are opposite...the are called exchange current densities and are given the symbol *i~o~*. - The signs are opposite because for the oxidation process, electrons *enter* the metal surface but for the reduction process electrons *leave* the metal surface. At equilibrium, the rates at which electrons are transferred to/from the electrode are equal. Let's apply a different potential to the electrode surface... At equilibrium, the potential is +0.77 V. If we make the electrode more positive *e.g.* apply a potential of +1 V then electrons will favour moving onto the more positive electrode surface from one of ions in solution. (Fermi Level decreases). The equilibrium will shift to try to replace the lost electrons...this favours the *oxidation* reaction and a net current will flow. We define this as a **positive** current as we have made the potential more **positive**. This means that we are changing the solution composition, as we are increasing the concentration of Fe^3+^ and decreasing the concentration of Fe^2+^. By changing the electrode potential, we can do some chemistry. If we now make the electrode more negative, let's say we reduce it from =0.77 V to 0.5 V, the equilibrium will shift to try to remove the extra electrons we have added to the electrode surface (this isn't strictly correct, but it's a useful analogy. The Fermi Level increases as the electrons have more energy). This favours the ***reduction*** reaction and a net ***negative*** current will flow. We will increase \[Fe^2+^\] and reduce \[Fe^3+^\]. Note that the final concentration ratio of Fe^2+^:Fe^3+^ can be determined from the Nernst Equation for a given applied potential. - The overpotential is defined as Where: - *E*~e~ is the equilibrium potential for the electron transfer process. Not usually the standard potential -- usually need to calculate this using the Nernst Equation. - *E~app~* is the potential applied to the electrode. In the examples above, there are two values of *η*. ***η* is a measure of how far from the equilibrium potential we are.** - *α~C~* and *α~A~* are the cathodic and anodic transfer coefficients respectively, which reflect the amount of the applied potential which actually goes into reducing the activation barrier for electron transfer in the forward and reverse directions. These are typically 0.5. - *i~o~* is the exchange current density as before. - Different metals (or carbon) have different *i~o~* values. - Good electrochemical catalysts like Pt and Pd have high values of *i~o~*. - Poor electrochemical catalysts like Pb have low values of *i~o~*. - Sometimes the difference can be 10^12^ for the same reaction on different metals! - Using the best metal electrode can give you a much higher current density for the same value of *η* *i.e.* more reaction for the same energy input. A plot of how *i* varies with *η* is shown below -- note that this is an exponential relationship. Two different *i~o~* values are included. Example Calculation Determine the current density when a potential of 0.25 V is applied to an electrode immersed in a solution containing 0.1 M Sn^4+^ and 1.5 M Sn^2+^. [*E*~Sn^2+^/Sn^4+^~^*o*^ = 0.15 *V*]{.math.inline}, *i~o~* = 2.5 ×10^-3^ A dm^-2^ and *α~c~* = *α~a~* = 0.5. **[Tafel Equations]** --------------------------------- The full Butler-Volmer equation is long and cumbersome, but we can make use of the exponential relationship to our advantage... To understand this, we need to remember that when we apply a positive overpotential we favour the oxidation reaction, but some of the reverse (reduction reaction) still occurs but at a *reduced rate; vice versa* for a negative overpotential. The Butler-Volmer equation accounts for this. When the magnitude overpotential reaches 0.1V, the rate of the reverse reaction is so small that it can be neglected and we can simplify the Butler-Volmer equation. 1. At *η* values more positive than 0.1 V, the **oxidation** reaction dominates and the current density is given by: which is known as the **anodic** Tafel equation 2. At *η* values more negative than −0.1 V, the **reduction** reaction dominates and the current density is given by: which is known as the **cathodic** Tafel equation. - For an ***oxidation*** reaction with *η* more positive than 0.1 V we can use the ***anodic*** Tafel Equation - For a ***reduction*** reaction with *η* more negative than -0.1 V we can use the ***cathodic*** Tafel Equation - If *η* lies between -0.1 V and +0.1 V we need to use the full Butler-Volmer Equation. This is as illustrated: *Use the Tafel Equation to determine the current density from the previous example:* - Both the anodic and cathodic Tafel equations predict an exponential increase in the current with overpotential. - Remember, though, that electrochemical reactions happen at the electrode surface. As *η* gets larger, the current gets larger and more reactants are converted to products...this means that the concentration of reactants at the electrode surface drops. - Eventually, we run out of reactant at the electrode surface so the reaction rate becomes limited by the rate of diffusion of the reactant to the electrode surface from the bulk solution. - The maximum or 'mass transport limiting' current is given by: where *k~L~* is the mass transfer coefficient and *C* is the concentration of the reactant in the bulk solution. Be careful with units, you need the 'distance' units of *i* to match *k~L~* and *C* e.g. *i* in A **dm^-2^**, *k~L~* in **dm** s^-1^ and *C^)^* in mol **dm^-3^**. When the reaction is diffusion or mass transport limited, increasing *η* has no effect on *i*. This can be a source of wasted energy. *Can we achieve the calculated current density in our example if k~L~ = 1.4 × 10^-6^ dm s^-1^?*