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Lecture

Electrochemistry_LectureNotes.pdf

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Department
Chemistry
Course
CHEM102
Professor
Sai Yiu
Semester
Winter

Description
UNIVERSITY OF ALBERTA INTRODUCTORY UNIVERSITY CHEMISTRY II (CHEM 102) Electrochemistry Note:  Students are expected to be familiar with the concept of oxidation and reduction in section 1 of this set of lecture no.es  They will only be briefly mentioned in class time. 1 Oxidation and Reduction  Redox reaction is an electron transfer process which occurs simultaneously.  Oxidation involves electron loss and in the process, a reducing agent is oxidized. 2+ For example: Zn(s) Zn (aq) + 2e  Reduction involves electron gain and in the process, an oxidizing agent is reduced. 2+ For example: Cu (aq) +2e Cu(s) Overall reaction: Zn(s) + Cu (aq) Zn (aq) + Cu(s) Fig. 21.4 p929 Process Electron movement Main player Oxidation Reducing agent is oxidized. Its oxidation number increases Reduction Oxidizing agent is reduced. Its oxidation number decreases 1 Characterization of redox reactions (Oxidation states) The concept of electron transfer as redox reactions can only be applied to ionic compounds. However, many reactions are regarded as redox processes in spite of the fact that they involve only molecules and do not involve tranfer of electrons. 2H (2) + O (2) 2H 2(l) C(s) + O 2g) CO (2) The idea of assigning oxidation states (oxidation numbers) is another alternative definition for redox processes. The oxidation number is arbitrary assigned for convenience. It should not be related to the concept of bonding or the structure of the molecule. Rules for assigning oxidation numbers 1. The oxidation number of atoms in uncombined elements is 0. 2. In a neutral molecule, the sum of the oxidation number is 0. 3. In ions, the sum of the oxidation numbers equals the charge on the ion. 4. The oxidation number of hydrogen in all compounds, except hydrides (H ), is - +1. 5. The oxidation number of oxygen in all compounds, except peroxides, is -2. For example: 2+ 2+ Zn(s) + Cu (aq) Zn (aq) + Cu(s) Oxidation state: 0 +2 +2 0 2+ Zn is oxidized because O.S Cu is reduced because O.S increases from 0 to +2 decreases from +2 to 0 It is therefore a reducing agent It is therefore an oxidizing agent Balancing oxidation-reduction equations (Half reaction method) Steps for balancing oxidation-reduction equations: 1. You are given an unbalanced equation. Sometimes the equation is incomplete, lacking in some reactants or / and products. 2. Identify, by using oxidation numbers, the oxidizing and reducing agents from the equation. 3. Write the half-equation for the oxidizing agent (the one having a reduction in oxidation number). Write down the reactant and product on both sides of the half-equation. 4. Write the half-equation for the reducing agent (the one having an increase in oxidation number). Write down the reactant and product on both sides of the half-equation. 5. Balance each half-equation as to the number of atoms of each element. 2 (i) For acidic solution: For each excess oxygen atom on one side of an equation, balance is secured by adding one H O to the other side. Then H is used to 2 balance hydrogens. (ii) For basic solution: For each excess oxygen atom on one side of an equation, balance is secured by adding one H O to 2he same side and 2OH to the - other side. 6. Balance each half-equation as to the number of charges by adding electrons to either side of the equation. 7. Multiply each half-equation by a number so that the total number of electrons lost by the reducing agent equals the number of electrons gained by the oxidizing agent. 8. Add the two half-equations. Cancel any terms common to both sides. All electrons must be cancelled off. For example: Fe (aq) + MnO (aq)  Fe (aq) + Mn (aq) (acid solution) 4 Half equations: 2+ Reducing agent is Fe (aq) 2+ 3+ Fe (aq) Fe (aq) + e (equation 1) Oxidizing agent is MnO (aq). Since it is in acidic medium, the equation must be: 4 MnO (aq) + H (aq)  Mn (aq) 2+ 4 + Since there are 4 oxygen atoms, we will need 8H Balancing the charge should give a half equation as: MnO (a4) + 8H (aq) + 5e  Mn (aq) + 4H O (l) 2 (equation 2) Multiplying equation 1 by 5 gives: 2+ 3+ 5Fe (aq)  Fe (aq) + e (equation 3) Adding equation 2 and 3 gives: - + 2+ 2+ 3+ MnO (a4) + 8H (aq) + 5e + 5Fe (aq)  Mn (aq) + 4H O (l) + Fe (aq) +2e Answer is: MnO (aq) + 8H (aq) + 5Fe (aq)  Mn (aq) + 4H O (l) + Fe (aq) 3+ 4 2 3 Another redox equation example: - - - MnO (a4) + NO (aq)  M2O (s) + NO (aq) (Bas2c solution) 3 Half equations: - Reducing agent is NO (aq) 2 NO (2q) NO (aq) 3- Oxidizing agent is MnO (aq). 4- MnO (aq)  MnO (s) 4 2 Since there are 2 excess oxygen atoms, we will need 2H O molecules 2 Balancing the charge should give a half equation as: Since they are in basic medium, the equations should be balanced as: NO (aq) + 2OH (aq) NO (aq) + H O (l) + 2e (equation 1) 2 3 2 - - MnO (a4) + 2H O (l) 2 3e MnO (s) + 4OH (aq2 (equation 2) Multiplying equation 1 by 3 and equation 2 by 2 gives: - - - 3NO (a2) + 6OH (aq) NO (aq) + 3H O (3) + 6e 2 (equation 3) - - 2MnO (aq4 + 4H O (l) +26e 2MnO (s) + 8OH (aq)2 (equation 4) Adding equation 2 and 3 gives: 3NO (aq) + 6OH (aq) + 2MnO (aq) + 4H O (l) + 6eNO (aq) + 3H O (l) + 6e + - 2 - 4 2 3 2 2MnO (s)2+ 8OH (aq) The answer is: 3NO (a2) + 2MnO (aq) + H4O (l)NO2(aq) + 2MnO (s) 3 2OH (aq) 2 - 4 2 Voltaic cells In an oxidation and reduction process:  If the oxidizing agent and reducing agent are not in physical contact with each another, but are in separate containers called ‘half-cells’  Each half-cell contains a solution and a metallic conductor (electrode), Then the electron transfer can be used to generate electricity. Such an arrangement for converting chemical energy of a redox reaction to electrical energy is called a Galvanic Cell or a Voltaic cell. Fig. 21.5, p931 Salt bridge The two half-cells must be connected in some way that allows ions to move between them. (You need to have a complete circuit for an electric current to flow through or else imbalance of charge stops the reaction). Such an arrangement can be accomplished by: + - Insertion of a connecting electrolyte solution such as K NO 3 in a gel (a salt bridge) between the two solutions so to allow ions to flow through and completes the circuit. Reactions at electrodes An anode is the electrode at which oxidation occurs. That is, a reducing agent is oxidized and hence electrons go into the external circuit. It is called an oxidation half-cell. 2+ Zn(s) Zn (aq) + 2e (At the anode) A cathode is the electrode at which reduction occurs. That is, an oxidizing agent is reduced and hence electrons enter from the external circuit. It is called a reduction half-cell. Cu (aq) +2e Cu(s) (At the cathode) 5 Overall reaction is: 2+ 2+ Zn(s) + Cu (aq) Zn (aq) + Cu(s) 3 Cell diagram (notations of voltaic cells) A cell diagram shows the components of a galvanic cell in a symbolic way. Here are a few general conventions in the writing of a cell diagram by using line notation:  The anode, where oxidation occurs, is placed at the left side of the diagram.  The cathode, where reduction occurs, is placed at the right side of the diagram.  A boundary between different phases (for example, an electrode and a solution) is represented by a single vertical line. The state of the components must be indicated. If more than one component is present in the same phase, a comma must be used.  The boundary between half-cells which is most commonly a salt bridge, is represented by a double vertical line. Example: The first galvanic cell can be represented by 2+ 2+ Zn(s) Zn (aq) Cu (aq) Cu (s) 4 Hydrogen electrode In a hydrogen electrode, hydrogen gas is bubbled into a hydrochloric acid solution (1.0 M) on a platinum surface. The half-cell reaction is: + 2H (aq) + 2e H 2g) The cell diagram for the hydrogen electrode is: + Pt (s) │ H2(g) (1 atm) │ H (aq) (1.0 M) 6 N.B. The hydrogen electrode can both be an anode or a cathode. In case of an anode, the equation of the reaction is reversed. 5 Cell potential The fact that electrons flow from one electrode to the other indicates that there is a driving force to ‘pull’ electrons in the external circuit.  The driving force to pull electrons is called the electromotive force (emf) (E cell cell potential.  Unit of cell potential is in volts (V). A potential of one volt is the work done (1 joule) per unit charge (1 coulomb). 1V = 1 J / C  The cell potential of a cell is given by: E cell E cathodeEanode Ecell 0 (Reaction direction is spontaneous) Ecell 0 (Reaction direction is not spontaneous) 6 Standard electrode potential (Standard reduction potential)  Although it is possible to measure the cell potential, the potential of individual electrodes cannot be determined.  There is a need to set a standard. By setting the electrode potential of a standard hydrogen electrode (SHE) to zero, we can determine the relative potential of other electrodes. The standard states of an electrode:  Gas pressure = 1 atm  Solution concentration = 1 M (It is important to specify the concentration of the solution as we will see later that concentration affects the potential)  Temperature = 25 C o o The potential of the electrode is called the standard potential (E ). For SHE, E = 0.00 V An example for determining electrode potentials The cell below has a cell potential of + 0.76 V. 2+ + Zn(s) │ Zn (aq)││ H (aq) (1.0 M) │ H (g) (12atm) │ Pt (s) 7 Thus, we can use the SHE to determine the standard reduction potential of the zinc electrode. 2+ o Zn (aq) + 2e Zn (s) E Zn= - 0.76 V The electrode potential is called standard reduction potential. Standard electrode potential is expressed as the standard reduction potential If we reverse the reaction the sign of the standard electrode potential has to be changed and now the reaction is oxi
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