Electrochemistry is the branch of chemistry that deals with the use of chemical reaction to generate a potential or voltage.

Electrochemical Reactions In electrochemical reactions, electrons are transferred from one species to another. Mg (s) + HCl (aq) →

Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

Oxidation and Reduction O.S. of some element increases in the reaction. Electrons are on the right of the equation Reduction O.S. of some element decreases in the reaction. Electrons are on the left of the equation.

Oxidation and Reduction A species is oxidized when it loses electrons. Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

Oxidation and Reduction A species is reduced when it gains electrons. Here, each of the H+ gains an electron, and they combine to form H2.

Oxidation State Changes Assign oxidation states: Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g) D Fe3+ is to metallic iron. CO(g) is to carbon dioxide. Prentice-Hall © 2007

Balancing Oxidation-Reduction Equations Few can be balanced by inspection. Systematic approach required. The Half-Reaction (Ion-Electron) Method Prentice-Hall © 2007 General Chemistry: Chapter 5

Balancing in Acid Write the equations for the half-reactions. Balance all atoms except H and O. Balance oxygen using H2O. Balance hydrogen using H+. Balance charge using e-. Equalize the number of electrons. Add the half reactions. Check the balance. Prentice-Hall © 2007 General Chemistry: Chapter 5

SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq) EXAMPLE 5-6 Balancing the Equation for a Redox Reaction in Acidic Solution. The reaction described below is used to determine the sulfite ion concentration present in wastewater from a papermaking plant. Write the balanced equation for this reaction in acidic solution.. SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq) Prentice-Hall © 2007 General Chemistry: Chapter 5

EXAMPLE 5-6 Determine the oxidation states: 4+ 6+ 7+ 2+ SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq) Write the half-reactions: SO32-(aq) → SO42-(aq) + 2 e-(aq) 5 e-(aq) +MnO4-(aq) → Mn2+(aq) Balance atoms other than H and O: Already balanced for elements. Prentice-Hall © 2007 General Chemistry: Chapter 5

EXAMPLE 5-6 Balance O by adding H2O: H2O(l) + SO32-(aq) → SO42-(aq) + 2 e-(aq) 5 e-(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) Balance hydrogen by adding H+: H2O(l) + SO32-(aq) → SO42-(aq) + 2 e-(aq) + 2 H+(aq) 8 H+(aq) + 5 e-(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) Check that the charges are balanced: Add e- if necessary. Prentice-Hall © 2007

EXAMPLE 5-6 Multiply the half-reactions to balance all e-: 5 H2O(l) + 5 SO32-(aq) → 5 SO42-(aq) + 10 e-(aq) + 10 H+(aq) 16 H+(aq) + 10 e-(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l) 6 3 Add both equations and simplify: 5 SO32-(aq) + 2 MnO4-(aq) + 6H+(aq) → 5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l) Check the balance! Prentice-Hall © 2007

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Oxidation and Reduction What is reduced is the oxidizing agent. H+ oxidizes Zn by taking electrons from it. What is oxidized is the reducing agent. Zn reduces H+ by giving it electrons.

Assigning Oxidation Numbers, ON Elements in their elemental form have an oxidation number of 0. Mg Cl2 Pb The oxidation number of a monatomic ion is the same as its charge. Zn2+ F- Na+

HF NaH Nonmetals tend to have negative oxidation numbers In combination with other elements: Nonmetals tend to have negative oxidation numbers Oxygen has an oxidation number of −2, except in the peroxide ion, which has an oxidation number of −1. MgO Na2O Hydrogen is −1 when bonded to a metal and +1 when bonded to a nonmetal. HF NaH

Assigning Oxidation Numbers Fluorine always has an oxidation number of −1. HF NaF The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. ZnBr2 CuCl2

Assigning Oxidation Numbers The sum of the oxidation numbers in a neutral compound is 0. HCl H2O K2O 5. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. CO32- ClO4-

Examples Determine the oxidation number of sulfur in the following compounds: H2S SO42- S8 S2O32-

Some Redox reactions: Redox reactions involve transfer of electrons between reactants When a substance loses electrons, it undergoes oxidation; The substance oxidized is called reducing agent. When a substance gains electrons, it undergoes reduction; the substance reduced is called an oxidizing agent. Example: The reaction of a metal with either an acid or a metal salt, displacement reaction; the ion is displaced or replaced through oxidation of an element. A + BX → AX + B Mg (s) + HCl → Zn (s) + CuSO4 → ZnSO4 (s) + Cu (s)

Electrochemical Reactions

In a spontaneous Oxidation-Reduction (redox) Reaction electrons are transferred and energy is released as the reaction proceeds: Zn strip dissolves… The blue color due to Cu2+ fades. Copper metal is deposited.

20-1 Electrode Potentials and Their Measurement Cu(s) + 2Ag+(aq) Cu2+(aq) + 2 Ag(s) Cu(s) + Zn2+(aq) No reaction FIGURE 20-1 Behaviour of Ag+(aq) and Zn+(aq) in the presence of copper General Chemistry: Chapter 20

Standard Half Cell potentials, Reduction Potentials is a measure of the driving force for reduction to occur under standard conditions:1 M concentrations, P= 1atm, at 25 °C. . . Standard reduction potentials, E°red, are measured relative to a standard.

Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials.

Voltaic Cells We can use that energy to do electrical work. If we make the electrons flow through an external pathway rather than directly between reactants. We call such a setup a voltaic cell.

…the reaction is the same: Zn (s) + Cu2+ (aq) The significant difference is that the reactants, Zn (s) and Cu2+ (aq), are not in direct contact in the voltaic cell. We physically separate the reduction half of the reaction from the oxidation half of the reaction. Zn metal is in contact with a solution of Zn2+ (aq). Cu metal is in contact with a solution of Cu2+ (aq).

An electrochemical half-cell consists of a metal electrode, M, partially immersed in an aqueous solution of its ions, Mn+.

Voltaic Cells An electrochemical cell consists of two half-cells with electrodes joined by a wire and solutions joined by a salt bridge. The oxidation occurs at the anode. The reduction occurs at the cathode.

Electromotive Force (emf) Water only spontaneously flows one way in a waterfall. Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

Voltaic Cells Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

Voltaic Cells Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. Cations move toward the cathode. Anions move toward the anode.

Voltaic Cells In the cell, then, electrons leave the anode and flow through the wire to the cathode. As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

Voltaic Cells As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

Cell Diagrams and Terminology Chemistry 140 Fall 2002 Cell Diagrams and Terminology Electromotive force, Ecell The cell voltage or cell potential. Cell diagram Shows the components of the cell in a symbolic way. Anode (where oxidation occurs) on the left. Cathode (where reduction occurs) on the right. Boundary between phases shown by |. Boundary between half cells (commonly a salt bridge) shown by ||. Several memory devices have been proposed for the oxidation/anode and reduction/cathode relationships. Perhaps the simplest is that in the oxidation/anode relationship, both terms begin with a vowel: o/a; in the reduction/cathode relationship, both begin with a consonant: r/c. Another method is to remember that Cations go to the Cathode (both begin with C), that cations are formed at the anode is then obvious. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 20

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Chemistry 140 Fall 2002 This slide shows the correct symbolic description of the voltaic, or galvanic, cell in figure 20-4. Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Ecell = 1.103 V Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Standard Half Cell potentials, Reduction Potentials is a measure of the driving force for reduction to occur under standard conditions:1 M concentrations, P= 1atm, at 25 °C. . . Standard reduction potentials, E°red, are measured relative to a standard.

20-2 Standard Electrode Potentials Cell voltages, the potential differences between electrodes, are among the most precise scientific measurements. The potential of an individual electrode is difficult to establish. Arbitrary zero is chosen. The Standard Hydrogen Electrode (SHE)

The standard hydrogen electrode (SHE) Chemistry 140 Fall 2002 2 H+(a = 1) + 2 e- H2(g, 1 bar) E° = 0 V Pt|H2(g, 1 bar)|H+(a = 1) Because hydrogen is a gas at room temperature, electrodes cannot be constructed from it. The standard hydrogen electrode consists of a piece of platinum dipped into a solution containing 1 M H+(aq) with a stream of hydrogen passing over its surface. The platinum does not react but provides a surface for the reduction of H3O+(aq) to H2(g) as well as the reverse oxidation half-reaction. The standard hydrogen electrode (SHE) Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 20

Measuring standard electrode potentials Figure: 20-06 Title: Measuring standard electrode potentials Caption: (a) A SHE is the anode, and copper is the cathode. Contact between the half-cells occurs through a porous plate that acts like a salt bridge. (b) This cell has the same connections as that in part (a), but with zinc substituting for copper. However, the electron flow here is opposite that in (a), as noted by the negative voltage. This means zinc is the anode in this case. A SHE is the anode, and copper is the cathode. Contact between the half-cells occurs through a porous plate that acts like a salt bridge. (b) This cell has the same connections as that in part (a), but with zinc substituting for copper. However, the electron flow here is opposite that in (a), as noted by the negative voltage. This means zinc is the anode in this case.

Standard Electrode Potential, E° is the voltage of a cell formed from two standard electrodes E°cell = E°cathode (right) – E°anode,(left) The tendency for a reduction process to occur at an electrode. All ionic species present at a=1 (approximately 1 M). All gases are at 1 bar (approximately 1 atm).

E°cell = E°cathode - E°anode Cu2+(1M) + 2 e- → Cu(s) E°Cu2+/Cu = ? Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) E°cell = 0.340 V anode cathode Standard cell potential: the potential difference of a cell formed from two standard electrodes. E°cell = E°cathode - E°anode Copyright © 2011 Pearson Canada Inc.

Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) E°cell = 0.340 V E°cell = E°cathode - E°anode E°cell = E°Cu2+/Cu - E°H+/H2 0.340 V = E°Cu2+/Cu - 0 V E°Cu2+/Cu = +0.340 V H2(g, 1 atm) + Cu2+(1 M) → H+(1 M) + Cu(s) E°cell = 0.340 V Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 20

Voltaic Cells Two different metals (plus an electrolyte) produce a voltage when circuit is completed. Different metal pairs produce different voltages. Electrons flow from the metal higher in the activity series to the metal lower in the series through an external circuit.

Electromotive Force (emf) (the voltage produced by a voltaic cell) The potential difference between the anode and cathode in a cell is called the electromotive force (emf). It is also called the cell potential and is designated Ecell. Electromotive force (emf) is the force required to push electrons through the external circuit

Standard Half Cell potentials, Reduction Potentials is a measure of the driving force for reduction to occur under standard conditions:1 M concentrations, P= 1atm, at 25 °C. . . Standard reduction potentials, E°red, are measured relative to a standard.

General Chemistry: Chapter 20 TABLE 20.1 Some Selected Standard Electrode (Reduction) Potentials at 25°C Reduction Half-Reaction E°, V Acidic Solution Slide 49 of 53 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 20

General Chemistry: Chapter 20 TABLE 20.1 Some Selected Standard Electrode (Reduction) Potentials at 25°C (continued) Reduction Half-Reaction E°, V Acidic Solution Slide 50 of 53 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 20

General Chemistry: Chapter 20 TABLE 20.1 Some Selected Standard Electrode (Reduction) Potentials at 25°C (continued) Reduction Half-Reaction E°, V Acidic Solution Basic Solution Slide 51 of 53 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 20

Corrosion: Unwanted Voltaic Cells

Figure: 20-20 Title: Protection of iron against electrolytic corrosion Caption: In the anodic reaction, the metal that is more easily oxidized loses electrons to produce metal ions: in (a), this is iron; in (b), it is zinc. In the cathodic reaction, oxygen gas, which is dissolved in a thin film of water on the metal, is reduced to OH-. Rusting of iron occurs in (a), but not in (b). When iron corrodes, Fe2+ and OH- ions from the half-reactions initiate these further reactions: Fe2+ + 2OH--> Fe(OH)2(s) 4Fe(OH)2(s) + O2 + 2H2O --> 4Fe(OH)3(s) 2Fe(OH)3(s) --> Fe2O3 • H2O(s) + 2H2O(l)

Figure: 20-20-02UN Title: Magnesium sacrificial anodes Caption: The small cylindrical bars of magnesium attached to the steel ship provide cathodic protection against corrosion. Galvanization is another type of cathodic protection, the use of magnesium here is more cost effective.

Cell Potentials  Ered = −0.76 V  Ered = +0.34 V For the oxidation in this cell, For the reduction, Ered = −0.76 V  Ered = +0.34 V 

Cell Potentials Ecell  = Ered (cathode) − (anode) = +0.34 V − (−0.76 V) = +1.10 V

Oxidizing and Reducing Agents Cathode is always the electrode with more positive reduction potential. The greater the difference between the two reduction potential, the greater the voltage of the cell.

A voltaic cell is based on the following two standard half-reactions: Cd2+ (aq) + 2e- Cd (s) Sn2+ (aq) + 2e- Sn(s) By using the data from Standard Reduction Potentials table, determine: the half reactions that occur at the cathode and the anode the standard cell potential. Cd2+(aq) + 2 e- -----> Cd(s) -0.40 V Answer: 0.267 V.

Ecell as a function of Concentration Cell potential depends on concentration of reactants, products and temperature. The Nernst Equation allows us to calculate the voltage produced by any electrochemical cell given Eo values for its electrodes and the concentrations of reactants and products.

At 25oC the Nernst equation is simplified to E = Eo - (0.0592/n) × logQ Eo = standard electrochemical cell potential n = number of electrons passed from anode to cathode. Q = the equilibrium expression of the electrochemical reaction.

Example: If we have the reaction: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) If [Cu2+] = 5.0 M and [Zn2+] = 0.050M:

Calculate the voltage produced by the cell Sn(s)|Sn2+||Ag+|Ag(s) at 25oC given: [Sn2+] = 0.15 M [Ag+] = 1.7 M Answer: 0.70 Volt.

Concentration Cells Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell, be 0, but Q would not. Ecell  Therefore, as long as the concentrations are different, E will not be 0.

Applications of Oxidation-Reduction Reactions

+4 Pb + PbO2 + 2H2SO4 → 2PbSO4 + H2O 0 +2 +2 Figure: 20-15 Title: A lead-acid (storage) cell Caption: The composition of the electrodes is described in the text. The cell reaction that occurs as the cell is discharged in given in equation (20.24). The voltage of the cell is 2.02 V. In this depiction, two anode plates are connected in parallel fashion, as are two cathode plates. Car batteries may have up to 6 cells connected in parallel, as opposed to the 2 shown here. +4 +2 Pb + PbO2 + 2H2SO4 → 2PbSO4 + H2O 0 +2

Hydrogen Fuel Cells

General Chemistry: Chapter 20 Chemistry 140 Fall 2002 Fuel Cells O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq) 2{H2(g) + 2 OH-(aq) → 2 H2O(l) + 2 e-} 2H2(g) + O2(g) → 2 H2O(l) E°cell = E°O2/OH- - E°H2O/H2 = 0.401 V – (-0.828 V) = 1.229 V  is the efficiency value used to evaluate fuel cell performance. FIGURE 20-18  = ΔG°/ ΔH° = 0.83 A hydrogen-oxygen fuel cell Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 20

The flow of electrons from anode to cathode is spontaneous The flow of electrons from anode to cathode is spontaneous. What is the “driving force”? • Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode. • Potential difference is the difference in electrical potential. • The potential difference is measured in volts.

Standard Cell Potentials The cell potential at standard conditions can be found through this equation: Ecell  = Ered (cathode) − Ered (anode) The cell potential for all spontaneous reactions at Standard conditions is a positive value. The standard conditions include 1 M concentrations, and P= 1 atm (g), for reactants and products at 25 °C.