Chapter 18. Electrochemistry - PDF Free Download

Chapter 18 Electrochemistry

Oxidation-Reduction Reactions Review of Terms Oxidation-reduction (redox) reactions always involve a transfer of electrons from one species to another. Oxidation number - the charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. An oxidizing agent is a species that oxidizes another species; it is itself reduced. A reducing agent is a species that reduces another species; it is itself oxidized. Half-reaction method method of balancing redox reactions based on separating the chemical process into the oxidation and reduction half reactions. 2

Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3. The oxidation number of oxygen is usually 2. In H 2 O 2 and O 2 2- it is 1. 3

4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is 1. 5. Group IA metal ions are always +1, IIA metal ions are always +2 and fluorine is always 1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. Oxidation numbers of all the atoms in HCO 3-? HCO 3 - O = -2 H = +1 3(-2) + 1 +? = -1 C = +4 4

The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Acidic Solution 1. Write separate equations for the oxidation and reduction half reactions. 2. For each half reaction: a) Balance all the elements except H and O. b) Balance O using H 2 O. c) Balance H using H +. d) Balance the charge using electrons. 3. If necessary, multiply one or both balanced half reactions by an integer to equalize the number of electrons transferred in the two half reactions. 4. Add the half reactions, and cancel identical species. 5. Check that the elements and charges are balanced. 5

The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Acidic Solution 6

A Problem To Consider 18.1. Balance the oxidation reduction reaction that occurs between Fe 2+ and MnO 4 in acidic solution using halfreaction method. MnO 4 + Fe 2+ Mn 2+ + Fe 3+ 7

A Problem To Consider 18.1. (Continued) Start by assigning oxidation numbers and identifying the oxidizing and the reducing agents: +7 2 +2 +2 +3 MnO 4 + Fe 2+ Mn 2+ + Fe 3+ The oxidation numbers are given above each of the elements. Fe 2+ is oxidized from +2 to +3. Fe 2+ is the reducing agent. Mn in MnO 4 is reduced from +7 to +2. MnO 4 is the oxidizing agent. 8

A Problem To Consider 18.1. (Continued) Write the half reactions: Oxidation half-reaction: Fe 2+ Oxidation half-reaction is balanced. Fe 3+ + 1 e Reduction half-reaction: MnO 4 + 5 e Mn 2+ Reduction half-reaction is not balanced. 9

A Problem To Consider 18.1. (Continued) Balance the reduction half-reaction: a) Balance O by adding H 2 O to the right side of the half-reaction. b) Balance H by adding H + to the left side of the half-reaction. Reduction half-reaction: 8 H + + MnO 4 + 5 e Mn 2+ + 4 H 2 O Now, reduction half-reaction is balanced! 10

18.1. (Continued) A Problem To Consider Oxidation half-reaction: Fe 2+ Fe 3+ + 1 e Reduction half-reaction: 8 H + + MnO 4 + 5 e Mn 2+ + 4 H 2 O To have the same number of electrons gained and lost, multiply the oxidation half reaction by 5: Oxidation half-reaction: 5 Fe 2+ 5 Fe 3+ + 5 e Reduction half-reaction: 8 H + + MnO 4 + 5 e Mn 2+ + 4 H 2 O 11

18.1. (Continued) A Problem To Consider Oxidation half-reaction: 5 Fe 2+ 5 Fe 3+ + 5 e Reduction half-reaction: 8 H + + MnO 4 + 5 e Mn 2+ + 4 H 2 O Add the half reactions together: 5 Fe 2+ + 8 H + + MnO 4 + 5 e 5 Fe 3+ + 5 e + Mn 2+ + 4 H 2 O 5 Fe 2+ + 8 H + + MnO 4 5 Fe 3+ + Mn 2+ + 4 H 2 O reducing agent oxidizing agent 12

A Problem To Consider 18.2. Balance the oxidation reduction reaction that occurs between Cr 2 O 7 2 and SO 3 2 in acidic solution using halfreaction method. Cr 2 O 7 2 + SO 3 2 Cr 3+ + SO 4 2 13

18.2. (Continued) A Problem To Consider Start by assigning oxidation numbers and identifying the oxidizing and the reducing agents: +6 2 +4 2 +3 +6 2 Cr 2 O 2 7 + SO 2 3 Cr 3+ 2 + SO 4 The oxidation numbers are given above each of the elements. S in SO 3 2 is oxidized from +4 to +6. SO 3 2 is the reducing agent. Cr in Cr 2 O 7 2 is reduced from +6 to +3. Cr 2 O 7 2 is the oxidizing agent. 14

18.2. (Continued) A Problem To Consider Write the half reactions: Oxidation half-reaction: SO 3 2 SO 4 2 + 2 e Reduction half-reaction: Cr 2 O 7 2 + 6 e 2 Cr 3+ Both half-reactions are unbalanced. 15

A Problem To Consider 18.2. (Continued) Balance the oxidation half-reaction: a) Balance O by adding H 2 O to the left side of the half-reaction. b) Balance H by adding H + to the right side of the half-reaction. Oxidation half-reaction: H 2 O + SO 3 2 SO 4 2 + 2 e + 2 H + Now, oxidation half-reaction is balanced! 16

A Problem To Consider 18.2. (Continued) Balance the reduction half-reaction: a) Balance O by adding H 2 O to the right side of the half-reaction. b) Balance H by adding H + to the left side of the half-reaction. Reduction half-reaction: 14 H + + Cr 2 O 7 2 + 6 e 2 Cr 3+ + 7 H 2 O Now, reduction half-reaction is balanced! 17

18.2. (Continued) A Problem To Consider Oxidation half-reaction: H 2 O + SO 3 2 SO 4 2 + 2 e + 2 H + Reduction half-reaction: 14 H + + Cr 2 O 7 2 + 6 e 2 Cr 3+ + 7 H 2 O To have the same number of electrons gained and lost, multiply the oxidation half reaction by 3: Oxidation half-reaction: 3 H 2 O + 3 SO 3 2 3 SO 4 2 + 6 e + 6 H + Reduction half-reaction: 14 H + + Cr 2 O 7 2 + 6 e 2 Cr 3+ + 7 H 2 O 18

18.2. (Continued) A Problem To Consider Oxidation half-reaction: 3 H 2 O + 3 SO 2 3 3 SO 2 4 + 6 e + 6 H + Reduction half-reaction: 14 H + + Cr 2 O 2 7 + 6 e 2 Cr 3+ + 7 H 2 O Add the half reactions together: 3 H 2 O + 3 SO 2 3 + 14 H + + Cr 2 O 2 7 + 6 e 3 SO 2 4 + 6 e + 6 H + + 2 Cr 3+ + 7 H 2 O 3 SO 2 3 + 8 H + + Cr 2 O 2 7 3 SO 2 4 + 2 Cr 3+ + 4 H 2 O reducing agent oxidizing agent 19

The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Basic Solution 1. Use the half reaction method as specified for acidic solutions to obtain the final balanced equation as if H + ions were present. 2. To both sides of the equation obtained above, add a number of OH ions that is equal to the number of H + ions. (We want to eliminate H + by forming H 2 O.) 3. Form H 2 O on the side containing both H + and OH ions, and eliminate the number of H 2 O molecules that appear on both sides of the equation. 4. Check that elements and charges are balanced. 20

The Half Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Basic Solution 21

A Problem To Consider 18.3. Balance the oxidation reduction reaction that occurs between I and MnO 4 in basic solution using half-reaction method. MnO 4 + I MnO 2 + I 2 22

18.3. (Continued) A Problem To Consider Start by assigning oxidation numbers and identifying the oxidizing and the reducing agents: +7 2 1 +4 2 0 MnO 4 + I MnO 2 + I 2 The oxidation numbers are given above each of the elements. I is oxidized from 1 to 0. I is the reducing agent. Mn in MnO 4 is reduced from +7 to +4. MnO 4 is the oxidizing agent. 23

A Problem To Consider 18.3. (Continued) Write the half reactions: Oxidation half-reaction: 2 I I 2 + 2 e Oxidation half-reaction is balanced. Reduction half-reaction: MnO 4 + 3 e MnO 2 Reduction half-reaction is not balanced. 24

A Problem To Consider 18.3. (Continued) Balance the reduction half-reaction: a) Balance O by adding H 2 O to the right side of the half-reaction. b) Balance H by adding H + to the left side of the half-reaction. Reduction half-reaction: 4 H + + MnO 4 + 3 e MnO 2 + 2 H 2 O Now, reduction half-reaction is balanced! 25

18.3. (Continued) A Problem To Consider Oxidation half-reaction: 2 I I 2 + 2 e Reduction half-reaction: 4 H + + MnO 4 + 3 e MnO 2 + 2 H 2 O To have the same number of electrons gained and lost, multiply the oxidation half reaction by 3, and multiply the reduction half reaction by 2: Oxidation half-reaction: 6 I 3 I 2 + 6 e Reduction half-reaction: 8 H + + 2 MnO 4 + 6 e 2 MnO 2 + 4 H 2 O 26

18.3. (Continued) A Problem To Consider Oxidation half-reaction: 6 I 3 I 2 + 6 e Reduction half-reaction: 8 H + + 2 MnO 4 + 6 e 2 MnO 2 + 4 H 2 O Add the half reactions together: 6 I + 8 H + + 2 MnO 4 + 6 e 3 I 2 + 6 e + 2 MnO 2 + 4 H 2 O 6 I + 8 H + + 2 MnO 4 3 I 2 + 2 MnO 2 + 4 H 2 O 27

18.3. (Continued) A Problem To Consider 6 I + 8 H + + 2 MnO 4 3 I 2 + 2 MnO 2 + 4 H 2 O Add eight OH ions to both sides of the equation and form H 2 O on the side containing both H + and OH ions: 6 I + 8 H 2 O + 2 MnO 4 3 I 2 + 2 MnO 2 + 4 H 2 O + 8 OH Eliminate four H 2 O molecules from both sides of the equation: 6 I + 4 H 2 O + 2 MnO 4 3 I 2 + 2 MnO 2 + 8 OH reducing agent oxidizing agent 28

A Problem To Consider 18.4. Balance the oxidation reduction reaction that occurs between Se and Cr(OH) 3 in basic solution using half-reaction method. Se + Cr(OH) 3 Se + SeO 3 2 29

18.4. (Continued) A Problem To Consider Start by assigning oxidation numbers and identifying the oxidizing and the reducing agents: 0 +3 2 +1 0 +4 2 2 Se + Cr(OH) 3 Cr + SeO 3 The oxidation numbers are given above each of the elements. Se is oxidized from 0 to +4. Se is the reducing agent. Cr in Cr(OH) 3 is reduced from +3 to 0. Cr(OH) 3 is the oxidizing agent. 30

18.4. (Continued) A Problem To Consider Write the half reactions: Oxidation half-reaction: Se SeO 3 2 + 4 e Reduction half-reaction: Cr(OH) 3 + 3 e Cr Both half-reactions are unbalanced. 31

A Problem To Consider 18.4. (Continued) Balance the oxidation half-reaction: a) Balance O by adding H 2 O to the left side of the half-reaction. b) Balance H by adding H + to the right side of the half-reaction. Oxidation half-reaction: 3 H 2 O + Se SeO 3 2 + 4 e + 6 H + Now, oxidation half-reaction is balanced! 32

A Problem To Consider 18.4. (Continued) Balance the reduction half-reaction: a) Balance O by adding H 2 O to the right side of the half-reaction. b) Balance H by adding H + to the left side of the half-reaction. Reduction half-reaction: 3 H + + Cr(OH) 3 + 3 e Cr + 3 H 2 O Now, reduction half-reaction is balanced! 33

18.4. (Continued) A Problem To Consider Oxidation half-reaction: 3 H 2 O + Se SeO 3 2 + 4 e + 6 H + Reduction half-reaction: 3 H + + Cr(OH) 3 + 3 e Cr + 3 H 2 O To have the same number of electrons gained and lost, multiply the oxidation half reaction by 3, and multiply the reduction half reaction by 4: Oxidation half-reaction: 9 H 2 O + 3 Se 3 SeO 3 2 + 12 e + 18 H + Reduction half-reaction: 12 H + + 4 Cr(OH) 3 + 12 e 4 Cr + 12 H 2 O 34

18.4. (Continued) A Problem To Consider Oxidation half-reaction: 9 H 2 O + 3 Se 3 SeO 2 3 + 12 e + 18 H + Reduction half-reaction: 12 H + + 4 Cr(OH) 3 + 12 e 4 Cr + 12 H 2 O Add the half reactions together: 9 H 2 O + 3 Se + 12 H + + 4 Cr(OH) 3 + 12 e 3 SeO 2 3 + 12 e + 18 H + + 4 Cr + 12 H 2 O 3 Se + 4 Cr(OH) 3 3 SeO 2 3 + 6 H + + 4 Cr + 3 H 2 O 35

18.4. (Continued) A Problem To Consider 3 Se + 4 Cr(OH) 3 3 SeO 3 2 + 6 H + + 4 Cr + 3 H 2 O Add six OH ions to both sides of the equation and form H 2 O on the side containing both H + and OH ions: 6 OH + 3 Se + 4 Cr(OH) 3 3 SeO 3 2 + 6 H 2 O + 4 Cr + 3 H 2 O 6 OH + 3 Se + 4 Cr(OH) 3 3 SeO 3 2 + 4 Cr + 9 H 2 O reducing agent oxidizing agent 36

Electrochemistry Branch of chemistry that deals with the interconversion of electrical energy and chemical energy. 37

Electrochemical Cells An electrochemical cell is a system consisting of electrodes that dip into an electrolyte in which a chemical reaction either uses or generates an electric current. A voltaic, or galvanic, cell is an electrochemical cell in which a spontaneous redox reaction generates an electric current. An electrolytic cell is an electrochemical cell in which an electric current drives an otherwise nonspontaneous redox reaction. 38

Voltaic (Galvanic) Cell Consists of two half cells that are electrically connected. Each half cell is the portion of the electrochemical cell in which a half reaction takes place. Cells must also have an internal cell connection, such as a porous disk or a salt bridge, to allow ions to flow without extensive mixing of the solutions. Porous disk contains tiny passages that allow hindered flow of ions. Salt bridge contains a strong electrolyte held in a Jello like matrix. 39

Zinc/Copper Voltaic Cell Half-Cells A simple half-cell can be made from a metal strip dipped into a solution of its metal ion: For example, the zinc-zinc ion half cell consists of a zinc strip dipped into a solution of a zinc salt. Another simple half-cell consists of a copper strip dipped into a solution of a copper salt. Because zinc has a greater tendency to lose electrons than copper, zinc atoms in the zinc electrode lose electrons to form zinc ions. The electrons flow through the external circuit to the copper electrode where copper ions gain the electrons to become copper metal 40

Atomic View of Zinc/Copper Voltaic Cell 41

Zinc/Copper Voltaic Cell Internal Connection The two half-cells must also be connected internally to allow ions to flow between them. Without this internal connection, too much positive charge builds up in the zinc half-cell (and too much negative charge in the copper half-cell) causing the reaction to stop. A salt bridge is a tube of an electrolyte in a gel that is connected to the two half-cells of a voltaic cell. The salt bridge allows the flow of ions but prevents the mixing of the different solutions that would allow direct reaction of the cell reactants. 42

Zinc/Copper Voltaic Cell Half-Cell Reactions There are two half-cell reactions: 1) oxidation half-reaction: Zn (s) Zn 2+ (aq) + 2e The electrode at which oxidation takes place is called the anode. The anode in a voltaic cell has a negative sign because electrons flow from it. Note that electrons are given up at the anode and thus flow from it to the cathode where reduction occurs. 43

Zinc/Copper Voltaic Cell Half-Cell Reactions There are two half-cell reactions: 2) reduction half-reaction: Cu 2+ (aq) + 2e Cu (s) The electrode at which reduction takes place is called the cathode. The cathode in a voltaic cell has a positive sign. Note that electrons are given up at the anode and thus flow from it to the cathode where reduction occurs. 44

Zinc/Copper Voltaic Cell Cell Reaction oxidation half-reaction: Zn (s) Zn 2+ (aq) + 2e reduction half-reaction: Cu 2+ (aq) + 2e Cu (s) The sum of the two half-reactions: Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) is the net reaction that occurs in the voltaic cell; it is called the cell reaction. 45

The electrode at which oxidation takes place is called the anode. The electrode at which reduction takes place is called the cathode. Electrons flow through the external circuit from the anode (zinc electrode) to the cathode (copper electrode). In the salt bridge, cations move toward the cathode and anions move toward the anode to maintain charge balance. 46

Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction 47

Notation for Voltaic Cells Zinc/Copper Voltaic Cell It is convenient to have a shorthand way of designating particular voltaic cells. The cell consisting of the zinc-zinc ion half-cell and the copper-copper ion half-cell, is written: anode 2+ Zn (s) Zn (aq) Cu 2+ (aq) Cu (s) cathode The anode (oxidation half-cell) is written on the right. The cathode (reduction half-cell) is written on the left. 49

Notation for Voltaic Cells Zinc/Copper Voltaic Cell It is convenient to have a shorthand way of designating particular voltaic cells. The cell consisting of the zinc-zinc ion half-cell and the copper-copper ion half-cell, is written: 2+ Zn (s) Zn (aq) Cu 2+ (aq) Cu (s) The cell terminals are at the extreme ends in the cell notation. 50

Notation for Voltaic Cells Zinc/Copper Voltaic Cell It is convenient to have a shorthand way of designating particular voltaic cells. The cell consisting of the zinc-zinc ion half-cell and the copper-copper ion half-cell, is written: 2+ Zn (s) Zn (aq) Cu 2+ (aq) Cu (s) A single vertical bar indicates a phase boundary, such as between a solid terminal and the electrode solution. 51

Notation for Voltaic Cells Zinc/Copper Voltaic Cell It is convenient to have a shorthand way of designating particular voltaic cells. The cell consisting of the zinc-zinc ion half-cell and the copper-copper ion half-cell, is written: 2+ Zn (s) Zn (aq) Cu 2+ (aq) Cu (s) salt bridge The two half-cells are connected by a salt bridge, denoted by two vertical bars. 52

Notation for Voltaic Cells Hydrogen Electrode When the half-reaction involves a gas, an inert material such as platinum serves as a terminal and an electrode surface on which the reaction occurs. For example, in the hydrogen electrode hydrogen bubbles over a platinum plate immersed in an acidic solution. The cathode half-reaction is: + 2 H (aq) + 2e H 2(g) 53

Notation for Voltaic Cells Hydrogen Electrode The notation for the hydrogen electrode, written as a cathode, is: + H (aq) H 2(g) Pt (s) To write such an electrode as an anode, you simply reverse the notation: + Pt (s) H 2(g) H (aq) 54

Notation for Voltaic Cells Complete Description To fully specify a voltaic cell, it is necessary to give the concentrations of solutions and the pressure of gases. In the cell notation, these are written in parentheses. For example zinc-hydrogen voltaic cell notation: Zn (s) หZn 2+ (aq) (1.0 M)ฮH + (aq) (1.0 M) H 2 g (1.0 atm) Pt (s) 55

A Problem To Consider 18.5. Give the overall cell reaction for the voltaic cell: Cd (s) หCd 2+ (aq) (1.0 M)ฮH + (aq) (1.0 M) H 2 g (1.0 atm) Pt (s) oxidation half-reaction: Cd (s) Cd 2+ (aq) + 2e + reduction half-reaction: 2 H (aq) + 2e H 2(g) The sum of the two half-reactions: + Cd (s) + 2 H (aq) Cd 2+ (aq) + H 2(g) 56

Electromotive Force The movement of electrons is analogous to the pumping of water from one point to another. Water moves from a point of high pressure to a point of lower pressure. Thus, a pressure difference is required. The work expended in moving the water through a pipe depends on the volume of water and the pressure difference 57

Electromotive Force The movement of electrons is analogous to the pumping of water from one point to another. An electric charge moves from a point of high electrical potential (high electrical pressure) to one of lower electrical potential. The work expended in moving the electrical charge through a conductor depends on the amount of charge and the potential difference. 58

Electromotive Force Potential difference (emf)is the difference in electric potential (electrical pressure) between two points in the circuit: emf = potential difference (V) = work (J) charge (C) The volt, V, is the SI unit of electrical potential. It is equivalent to 1 joule of work produced or required when 1 coulomb of charge is transferred between two points in the circuit: 1 V = 1 J C 59

Voltaic Cell Potential The maximum potential difference between the electrodes of a voltaic cell is referred to as cell potential (E cell ), or the electromotive force (emf) of the cell. E cell = w q work charge When a cell produces current, the cell potential, E cell, is positive, and the current can be used to do work, w, outside of the system. Thus the cell potential, E cell, and work, w, have the opposite signs. 60

Voltaic Cell Potential The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential All three terms are used interchangeably. Zn (s) Zn 2+ (1 M) Cu 2+ (1 M) Cu (s) anode cathode 61

Maximum Cell Potential The maximum work in a cell, w max, would be obtained at maximum potential, E max : w max = q E max Actual potential difference (voltage) across the electrodes is always less than the maximum possible voltage of the cell, because the flow of electrons reduces the electrical pressure. Thus, a cell voltage has its maximum value when no current flows. E max can be measured by an electronic digital voltmeter, which draws negligible current. 62

Maximum vs. Actual Work The maximum work in a cell, w max, would be obtained at maximum potential, E max : w max = q E max However, in any real spontaneous process the energy is always wasted. The actual work, w, is always less than the calculated maximum work, w max. Actual work w = q E Actual potential 63

Voltaic Cell Potential The Faraday constant, F, is the magnitude of charge on one mole of electrons: F = 96500 C = 9.65 10 4 C Let n be the number of moles of electrons transferred in the overall cell reaction. Then the quantity of charge transferred (q) would be: q = n F Therefore, the maximum work attainable by a voltaic cell is: w max = n F E max 64

A Problem To Consider 18.6. The emf of the electrochemical cell below is 0.650 V. The overall cell reaction is: 2+ + Hg 2(aq) + H 2(g) 2 Hg (l) + 2 H (aq) a) What is the voltaic cell notation for this cell? b) Calculate the maximum electrical work of this cell when 0.500 g H 2 is consumed. 65

A Problem To Consider 18.6. The emf of the electrochemical cell below is 0.650 V. The overall cell reaction is: 2+ + Hg 2(aq) + H 2(g) 2 Hg (l) + 2 H (aq) a) What is the voltaic cell notation for this cell? + oxidation half-reaction: H 2(g) 2 H (aq) + 2e (anode) 2+ reduction half-reaction: Hg 2(aq) + 2e 2 Hg (aq) (cathode) Voltaic cell notation: + Pt (s) ቚH 2 g ቚH (aq) 2+ ቛHg 2(aq) ቚHg (l) 66

A Problem To Consider 18.6. The emf of the electrochemical cell below is 0.650 V. The overall cell reaction is: 2+ + Hg 2(aq) + H 2(g) 2 Hg (l) + 2 H (aq) b) Calculate the maximum electrical work of this cell when 0.500 g H 2 is consumed. To answer question b) we need to know how many moles of electrons were transferred when 0.500 g H 2 is consumed. Stoichiometric calculation based on the oxidation half-reaction is required. + H 2(g) 2 H (aq) + 2e 67

18.6. (Continued) A Problem To Consider b) Calculate the maximum electrical work of this cell when 0.500 g H 2 is consumed. 0.500 g x mol electrons + H 2(g) 2 H (aq) + 2e 1 mol 2.02 g/mol 2 mol electrons When 0.500 g of H 2 is consumed x mol of electrons are transferred, when 2.02 g of H 2 is consumed 2 mol of electrons are transferred. 0.500 g H 2 2.02 g H 2 = x 2 mol e x = 0.500 g H 2 2 mol e 2.02 g H 2 = 0. 495 mol e 68

A Problem To Consider 18.6. (Continued) b) Calculate the maximum electrical work of this cell when 0.500 g H 2 is consumed. When 0.500 g H 2 is consumed 0.495 mol of electrons are transferred, therefore: w max = n F E max = 0. 495 mol e 96500 C 0.650 V w max = 3.09 10 4 J 69

Standard Cell emf s and Standard Electrode Potentials A cell emf is a measure of the driving force of the cell reaction. The reaction at the anode has a definite oxidation potential, while the reaction at the cathode has a definite reduction potential. Thus, the overall cell emf is a combination of these two potentials: E cell = E reduction + E oxidation 70

Standard Cell emf s and Standard Electrode Potentials A cell emf is a measure of the driving force of the cell reaction. A reduction potential is a measure of the tendency to gain electrons in the reduction half-reaction. The oxidation potential for an oxidation half-reaction is the negative of the reduction potential for the reverse reaction. You can look at the oxidation half-reaction as the reverse of a corresponding reduction reaction. 71

Standard Cell emf s and Standard Electrode Potentials Consider the zinc-copper cell described earlier: 2+ Zn (s) Zn (aq) Cu 2+ (aq) Cu (s) oxidation half-reaction: Zn (s) Zn 2+ (aq) + 2e reduction half-reaction: Cu 2+ (aq) + 2e Cu (s) The copper half-reaction is a reduction. Therefore, E Cu is the reduction potential of copper. 72

Standard Cell emf s and Standard Electrode Potentials Consider the zinc-copper cell described earlier: 2+ Zn (s) Zn (aq) Cu 2+ (aq) Cu (s) oxidation half-reaction: Zn (s) Zn 2+ (aq) + 2e reduction half-reaction: Cu 2+ (aq) + 2e Cu (s) The zinc half-reaction is an oxidation. If E Zn is the reduction potential of zinc, then E Zn is the oxidation potential of zinc. 73

Standard Cell emf s and Standard Electrode Potentials Consider the zinc-copper cell described earlier: 2+ Zn (s) Zn (aq) Cu 2+ (aq) Cu (s) For this cell, the cell emf is the sum of the reduction potential for the copper half-cell and the oxidation potential for the zinc half-cell: E cell = E reduction + E oxidation = E Cu + ( E Zn ) E cell = E Cu E Zn 74

Standard Cell emf s and Standard Electrode Potentials By convention, the Standard Electrode Potentials are tabulated as reduction potentials. In general, E cell is obtained by subtracting the anode potential from the cathode potential: E cell = E cathode E anode 75

Tabulating Standard Electrode Potentials o The standard cell emf, E cell, is the emf of a cell operating under standard conditions: temperature of 25 o C concentrations of all solutes are 1 M all gases are at 1 atm Note that individual electrode potentials require that we choose a reference electrode. You arbitrarily assign this reference electrode a potential of zero and obtain the potentials of the other electrodes by measuring the cell emf. 76

Tabulating Standard Electrode Potentials By convention, the reference chosen for comparing electrode potentials is the standard hydrogen electrode (SHE). Standard electrode potentials are measured relative to this hydrogen reference. Reduction Reaction: 2 H + aq (1 M) + 2e H 2(g) (1 atm) E o = 0 V 77

Measuring Standard Electrode Potentials Emf of a cell composed of a zinc electrode connected to a hydrogen electrode, is measured to be 0.76 V. Anode (oxidation): Zn (s) Zn 2+ (aq) + 2e Cathode (reduction): + 2 H (aq) + 2e H 2(g) + Zn (s) + 2 H (aq) Cell reaction: Zn 2+ (aq) + H 2(g) Zn (s) หZn 2+ (aq) (1.0 M)ฮH + (aq) (1.0 M) H 2 g (1.0 atm) Pt (s) 78

Measuring Standard Electrode Potentials Since zinc acts as the anode (oxidation) in this cell, its reduction potential is listed as 0.76 V. o E cell o = E cathode o E anode o E cell o = E H + /H 2 o E Zn 2+ /Zn o 0. 76 V = 0 V E Zn 2+ /Zn o E Zn 2+ /Zn = 0. 76 V Zn (s) หZn 2+ (aq) (1.0 M)ฮH + (aq) (1.0 M) H 2 g (1.0 atm) Pt (s) 79

Measuring Standard Electrode Potentials Emf of a cell composed of a copper electrode connected to a hydrogen electrode, is measured to be 0.34 V. Anode (oxidation): + H 2(g) H (aq) + 2e Cathode (reduction): Cu 2+ (aq) + 2e Cu (s) Cell reaction: Cu 2+ + (aq) + H 2(g) Cu (s) + 2 H (aq) Pt (s) H 2 g (1 atm) H + aq (1 M) Cu 2+ (aq) (1 M) Cu (s) 80

Measuring Standard Electrode Potentials Since copper acts as the cathode (reduction) in this cell, its reduction potential is listed as 0.34 V. o E cell o E cell o = E cathode o = E Cu 2+ /Cu o E anode o E H + /H 2 o 0. 34 V = E Cu 2+ /Cu 0 V o E Cu 2+ /Cu = 0. 34 V Pt (s) H 2 g (1 atm) H + aq (1 M) Cu 2+ (aq) (1 M) Cu (s) 81

Standard Electrode Potentials The electrode potential is an intensive property whose value is independent of the amount of species in the reaction. Thus, the electrode potential for the half-reaction: is the same as for: Cu 2+ (aq) + 2e Cu (s) 2 Cu 2+ (aq) + 4e 2 Cu (s) 82

Standard Reduction Potentials E o is given for the reduction half-reaction (as written) The more positive E o the greater the tendency for the substance to be reduced When half-reaction is reversed, the sign of E o is reversed When a half-reaction is multiplied by an integer, the value of E o remains the same A galvanic cell runs spontaneously in the direction that gives a positive E cell o 83

Calculating Cell emf s from Standard Potentials The emf of a voltaic cell constructed from standard electrodes is easily calculated using a table of electrode potentials: o E cell o = E cathode o E anode 84

A Problem To Consider 18.7. What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Standard reduction potentials are: Cd 2+ (aq) + 2e Cd (s), E o = 0.40 V Cr 3+ (aq) + 3e Cr (s), E o = 0.74 V 85

A Problem To Consider 18.8. Calculate the standard emf of the electrochemical cell constructed from the following half-cells: Cd 2+ (aq) + 2e Cd (s), E o = 0.40 V + Ag (aq) + 1e Ag (s), E o = 0.84 V 87

A Problem To Consider 18.8. Calculate the standard emf of the electrochemical cell constructed from the following half-cells: Cd 2+ (aq) + 2e Cd (s), E o = 0.40 V (anode) + Ag (aq) + 1e Ag (s), E o = 0.84 V (cathode) o E cell o = E cathode o E anode = 0.80 V 0.40 V = 1.20 V 88

A Problem To Consider 18.9. a) Calculate the standard emf for the following voltaic cell at 25 o C: 3+ Al (s) Al aq Fe 2+ (aq) Fe (s) Standard reduction potentials are: Al 3+ (aq) + 3e Al (s), E o = 1.66 V Fe 2+ (aq) + 2e Fe (s), E o = 0.41 V b) What is the cell reaction? 89

A Problem To Consider 18.9. a) Calculate the standard emf for the following voltaic cell at 25 o C: 3+ Al (s) Al aq Fe 2+ (aq) Fe (s) Standard reduction potentials are: Al 3+ (aq) + 3e Al (s), E o = 1.66 V (anode) Fe 2+ (aq) + 2e Fe (s), E o = 0.41 V(cathode) o E cell o = E cathode o E anode = 0.41 V 1.66 V = 1.25 V 90

A Problem To Consider 18.9. b) What is the cell reaction? 3+ Al (s) Al aq Fe 2+ (aq) Fe (s) Standard reduction potentials are: Al 3+ (aq) + 3e Al (s), E o = 1.66 V (anode) Fe 2+ (aq) + 2e Fe (s), E o = 0.41 V (cathode) Al (s) Al 3+ (aq) + 3e, (oxidation) 2 Fe 2+ (aq) + 2e Fe (s), (reduction) 3 3 Fe 2+ 3+ (aq) + 2 Al (s) 3 Fe (s) + 2 Al (aq) 91

Equilibrium Constants and emf Some of the most important results from electrochemistry are the relationships among E cell, free energy, and equilibrium constant. o 92

Equilibrium Constants and emf Some of the most important results from electrochemistry are the relationships among E cell, free energy, and equilibrium constant. The measurement of cell emf s gives you yet another way of calculating equilibrium constants: o Go rxn = R T lnk o G rxn o = w cell = n F E cell 93

Equilibrium Constants and emf Combining the two equations together and rearranging: o R T lnk = n F E cell o E cell = RT nf lnk Substituting values for the constants R and F at 25 o C gives the equation: o E cell = 8.314 J 298 K mol K J n 96500 V mol lnk = 0.0257 n lnk 94

Equilibrium Constants and emf Replacing ln with log: o E cell = RT 2.303RT lnk = logk nf nf Substituting values for the constants R and F at 25 o C gives the equation: o E cell = 2.303 8.314 J 298 K mol K J n 96500 V mol o E cell = 0.0591 n logk logk 95

Spontaneity of Redox Reactions o G rxn o = R T lnk = n F E cell 96

A Problem To Consider 18.10. The standard emf for the following cell is 1.10 V: 2+ Zn (s) Zn aq Cu 2+ (aq) Cu (s) Calculate the equilibrium constant K c for the reaction: Zn (s) + Cu 2+ (aq) Zn 2+ aq + Cu (s) o E cell = 0.0591 logk n logk = ne o cell 0.0591 logk = ne o cell 0.0591 = 2 1.10 V 0.0591 = 37.2 K = antilog 37.2 = 1.6 10 37 2 10 37 98

A Problem To Consider 18.11. What is the equilibrium constant for the following reaction at 25 o C? Ag (s) + Fe 2+ + (aq) Ag aq The standard reduction potentials are: o E Fe 2+ /Fe o = 0.44 V, E Ag + /Ag = 0.80 V + Fe (s) o E cell o = E Fe 2+ /Fe o E Ag + /Ag = 0.44 V 0.80 V = 1.24 V o E cell = 0.0591 n K = antilog logk K = antilog ne cell o 0.0591 2( 1.24 V) 0.0591 = antilog 41.96 = 1 10 42 100

Dependence of emf on Concentration (Nernst Equation) Recall that the free energy change, DG, is related to the standard free energy change, DG o, by the following equation: G = G o + R T lnq where Q is the thermodynamic reaction quotient. Standard free energy change, DG o, is related to the standard emf of the cell: G o = n F E cell o And G = n F E cell 101

Dependence of emf on Concentration (Nernst Equation) G = G o + R T lnq Replacing G and G o with n F E cell and n F E o cell : o n F E cell = n F E cell Rearranging the equation: or: o E cell = E cell o E cell = E cell RT nf lnq 2.303RT nf + R T lnq logq 102

Dependence of emf on Concentration (Nernst Equation) Substituting values for the constants R and F at 25 o C gives the equations: or: o E cell = E cell 0.0257 n o E cell = E cell 0.0591 n lnq logq 103

Dependence of emf on Concentration (Nernst Equation) o E cell = E cell 0.0591 n logq The Nernst equation illustrates why cell emf decreases as the cell reaction proceeds. As reactant concentrations decrease and product concentrations increase, Q increases, thus increasing log Q which in turn decreases the cell emf. 104

A Problem To Consider 18.12. What is the emf of the following voltaic cell at 25 o C: Zn (s) Zn 2+ aq (1.00 10 5 M) Cu 2+ aq (0.100M) Cu (s) The standard emf of the cell is 1.10 V. 105

A Problem To Consider 18.12. What is the emf of the following voltaic cell at 25 o C: Zn (s) Zn 2+ aq (1.00 10 5 M) Cu 2+ aq (0.100M) Cu (s) The standard emf of the cell is 1.10 V. The cell reaction is: Zn (s) + Cu 2+ (aq) Zn 2+ aq + Cu (s) The reaction quotient is: Q = Zn2+ Cu 2+ 1.00 10 5 = 0.100 = 1.00 10 4 106

A Problem To Consider 18.12. (Continued). The cell reaction is: Zn (s) + Cu 2+ (aq) Zn 2+ aq + Cu (s) The reaction quotient is: Q = Zn2+ Cu 2+ 1.00 10 5 = 0.100 = 1.00 10 4 o E cell = E cell 0.0591 logq = 1.10 V 0.0591 log(1.00 10 4 ) n 2 E cell = 1.10 V 0.1182 V = 1.2182 V 1.22 V 107

A Problem To Consider 18.13. Will the following reaction occur spontaneously at 25 0 C if [Fe 2+ ] = 0.60 M and [Cd 2+ ] = 0.010 M? Cd (s) + Fe 2+ (aq) Cd 2+ aq + Fe (s) The standard reduction potentials are: o E Fe 2+ /Fe o = 0.44 V, E Cd 2+ /Cd = 0.40 V o E cell o = E Fe 2+ /Fe o E Cd 2+ /Cd = 0.44 V ( 0.40 V) = 0.04 V o E cell = E cell 0.0591 logq = 0.04 V 0.0591 log 0.010 n 2 0.60 E cell = 0.04 V 0.0525 V = 0.012 V 0.01 V E cell > 0, reaction is spontaneous 109

The Leclanché dry cell, or zinc-carbon dry cell, is a voltaic cell with a zinc can as the anode and a graphite rod in the center surrounded by a paste of manganese dioxide, ammonium and zinc chlorides, and carbon black, as the cathode. The electrode reactions are: anode (oxidation): Zn (s) Zn 2+ (aq) + 2e cathode (reduction): + 2 NH 4(aq) Commercial Voltaic Cells + 2 MnO 2(s) + 2e Mn 2 O 3(s) + H 2 O (l) + 2 NH 3(aq) 110

Batteries Dry cell Leclanché cell Anode: Zn (s) Zn 2+ (aq) + 2e - + Cathode: 2NH 4 (aq) + 2MnO 2 (s) + 2e - Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O (l) Zn (s) + 2NH 4 (aq) + 2MnO 2 (s) Zn 2+ (aq) + 2NH 3 (aq) + H 2 O (l) + Mn 2 O 3 (s) 111

Commercial Voltaic Cells The alkaline dry cell, is similar to the Leclanché cell, but it has potassium hydroxide in place of ammonium chloride. The electrode reactions are: anode (oxidation): Zn (s) Zn 2+ (aq) + 2e cathode (reduction): MnO 2(s) + H 2 O (l) + 2e Mn 2 O 3(s) + 2 OH (aq) 112

A small alkaline dry cell. 113

Commercial Voltaic Cells The nickel-cadmium cell (nicad cell) consists of an anode of cadmium and a cathode of hydrated nickel oxide on nickel; the electrolyte is potassium hydroxide. The electrode reactions are: anode (oxidation): Cd (s) + 2 OH (aq) Cd(OH) 2(s) + 2e cathode (reduction): NiOOH (s) + H 2 O (l) + 1e Ni(OH) 2(s) + OH (aq) 114

Nicad storage batteries. Photo courtesy of American Color. 115

Batteries Mercury Battery Anode: Zn(Hg) + 2OH - (aq) ZnO (s) + H 2 O (l) + 2e - Cathode: HgO (s) + H 2 O (l) + 2e - Hg (l) + 2OH - (aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) 116

Commercial Voltaic Cells The lead storage cell (a rechargeable cell) consists of electrodes of lead alloy grids; one electrode is packed with a spongy lead to form the anode, and the other electrode is packed with lead dioxide to form the cathode. The electrode reactions are: anode (oxidation): Pb (s) + HSO 4(aq) + PbSO 4(s) + H (aq) + 2e cathode (reduction): + PbO 2(s) + 3 H (aq) + HSO 4(aq) + 2e PbSO 4(s) + 2 H 2 O (l) 117

Batteries Lead storage battery Anode: Pb (s) + SO 2- (aq) PbSO 4 (s) + 2e - 4 Cathode: PbO 2 (s) + 4H + (aq) + SO 2- (aq) + 2e - Pb (s) + PbO 2 (s) + 4H + (aq) + 2SO 2- (aq) 4 4 PbSO 4 (s) + 2H 2 O (l) 118 2PbSO 4 (s) + 2H 2 O (l)

Some Commercial Voltaic Cells The lithium-iodine battery is a solid state battery in which the anode is lithium metal and the cathode is an I 2 complex. The solid state electrodes are separated by a thin crystalline layer of lithium iodide. Although it produces a low current, it is very reliable and is used to power pacemakers. 119

A solid-state lithium-iodide battery. 120

Commercial Voltaic Cells A fuel cell is essentially a battery, but differs by operating with a continuous supply of energetic reactants, or fuel. The electrode reactions are: anode (oxidation): 2 H 2(g) + 4 OH (aq) 4 H 2 O (l) + 4e cathode (reduction): O 2(g) + 2 H 2 O (l) + 4e 4 OH (aq) 121

A hydrogen-oxygen fuel cell. 122

Batteries A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: 2H 2 (g) + 4OH - (aq) 4H 2 O (l) + 4e - Cathode: O 2 (g) + 2H 2 O (l) + 4e - 4OH - (aq) 2H 2 (g) + O 2 (g) 2H 2 O (l) 123

Chemistry In Action: Bacteria Power CH 3 COO - + 2O 2 + H + 2CO 2 + 2H 2 O 124

Corrosion Corrosion means the breaking down of essential properties in a material due to chemical reactions with its surroundings. In the most common use of the word, this means a loss of electrons of metals reacting with water and oxygen. For example, iron corrosion: + 2 Fe (s) + O 2(g) + 4 H (aq) 2 Fe 2+ aq + 2 H 2 O (l) 125

Cathodic Protection Cathodic protection (CP) is a technique to control the corrosion of a metal surface by making it work as a cathode of an electrochemical cell. This is achieved by connecting the metal to be protected to a more easily corroded "sacrificial metal" to act as the anode. Cathodic protection systems are most commonly used to protect steel, water or fuel pipelines and storage tanks, steel pier piles, ships, offshore oil platforms and onshore oil well casings. 126

Cathodic Protection of an Iron Storage Tank 127

For larger structures, galvanic anodes cannot economically deliver enough current to provide complete protection. Impressed current cathodic protection (ICCP) systems use anodes connected to a DC power source Anodes for ICCP systems are tubular and solid rod shapes or contiguous ribbons of various specialized materials. These include high silicone cast iron, graphite, mixed metal oxides, platinum and niobium coated wire and others. A cathodic protection retifier connected to a pipeline. 128

A typical ICCP system for a pipeline would include an AC powered rectifier with a maximum rated DC output of between 10 and 50 amperes and 50 volts. The positive DC output terminal is connected via cables to the array of anodes buried in the ground. For many applications the anodes are installed in a 60 m (200 ft) deep, 25 cm (10-inch) diameter vertical hole and backfilled with conductive coke (a material that improves the performance and life of the anodes). A cable rated for the expected current output connects the negative terminal of the rectifier to the pipeline. The operating output of the rectifier is adjusted to the optimum level after conducting various tests including measurements of electrochemical potential. A cathodic protection retifier connected to a pipeline. 129

Electrolytic Cells An electrolytic cell is an electrochemical cell in which an electric current drives an otherwise nonspontaneous redox reaction. The process of producing a chemical change in an electrolytic cell is called electrolysis. Anions are oxidized at the anode. Cations are reduced at the cathode. Many important substances, such as aluminum metal and chlorine gas are produced commercially by electrolysis. 130

Electrolysis of Water 131

Electrolysis of Molten Salts A Downs cell is a commercial electrochemical cell used to obtain sodium metal by electrolysis of molten NaCl. The half-reactions are: 2 Na + + 2e 2 Na (l) reduction 2 Cl Cl 2(g) + 2e (oxidation) Lithium, magnesium, and calcium metals are all obtained by the electrolysis of their molten salts, usually chlorides. 132

A Downs cell for the preparation of sodium metal. 133

Predicting Products of Electrolysis Electrode potentials can be used to predict the products of electrolysis. More easily oxidized species is oxidized at the anode. More easily reduced species is reduced at the cathode. We will look at three different types of electrolysis: Electrolysis of pure molten salts. Electrolysis of mixed molten salts. Electrolysis of aqueous ionic solutions. 134

Predicting Products of Electrolysis Electrolysis of pure molten salts The cation of the salt (metal ion) is reduced, producing a pure metal. The anion of the salt (usually halide) is oxidized, producing a non-metal element. For example, electrolysis of molten CaCl 2. The half-reactions are: Ca 2+ + 2e Ca (l) cathode, reduction 2 Cl Cl 2(g) + 2e (anode, oxidation) The electrolysis products are Ca metal and Cl 2 gas. 135

Predicting Products of Electrolysis Electrolysis of mixed molten salts Standard potentials (given for aqueous reactions) cannot be used. However, knowledge of periodic atomic trends (electronegativity, etc.) is helpful. Metal with a higher electronegativity is less likely to give up its electrons. That metal cation gains electrons more easily (stronger oxidizing agent). Non-metal with a lower electronegativity is more likely to give up its electrons. That non-metal anion loses electrons more easily (stronger reducing agent). 136

Predicting Products of Electrolysis Electrolysis of mixed molten salts For example, electrolysis of molten NaBr and MgCl 2. The possible oxidizing species are Na + and Mg 2+. Mg is to the right of Na in the periodic table. Electronegativity of Mg is therefore higher. So Mg 2+ has a greater ability to attract electrons (stronger oxidizing agent) than Na +. Therefore Mg metal (not Na) will form on the cathode: Mg 2+ + 2e Mg (l) The possible reducing species are Br and Cl. Br is lower than Cl in the periodic table. Electronegativity of Br is therefore lower. So Br has a greater ability to lose electrons (stronger reducing agent) than Cl. Therefore Br 2 (not Cl 2 ) will form on the anode: 2 Br Br 2(l) + 2e The electrolysis products are Mg metal and Br 2 gas. 137

Predicting Products of Electrolysis Electrolysis of aqueous ionic solutions Aqueous salt solutions are mixtures of ions and water. Electrolysis of water can occur instead of salt electrolysis. Electrolysis of water: Oxidation at the anode: + 2 H 2 O (l) O 2(g) + 4 H (aq) + 4e ; E o = 1.23 V, E = 0.82 V Reduction at the cathode: 2 H 2 O (l) + 2e H 2(g) + 2 OH (aq) ; E o = 0.83 V, E = 0.42 V Overvoltage: In order to produce H 2 and O 2 additional voltage (0.4-0.6 V) is required. + 2 H 2 O (l) O 2(g) + 4 H (aq) + 4e ; E 1.4 V (with overvoltage) 2 H 2 O (l) + 2e H 2(g) + 2 OH (aq) ; E 1.0 V (with overvoltage) 138

Predicting Products of Electrolysis Electrolysis of aqueous ionic solutions Competing reactions: Oxidation at the anode: + oxidation of water: 2 H 2 O (l) O 2(g) + 4 H (aq) + 4e ; E = 0.82 V or oxidation of the anion present in the solution can occur. Cl, Br, I are oxidized. F and common oxyanions (NO 3, SO 4 2, CO 3 2, etc.) are NOT oxidized, water is oxidized to O 2 and H + instead. Reduction at the cathode: reduction of water: 2 H 2 O (l) + 2e H 2(g) + 2 OH (aq) ; E = 0.42 V or reduction of the cation present in the solution can occur. Cations of less active metals (E o > 0.00 V) are reduced. Cations of more active metals (E o < 0.00 V) are NOT reduced, water is reduced to H 2 and OH instead. 139

Predicting Products of Electrolysis Electrolysis of aqueous ionic solutions Commercial production of chlorine and sodium hydroxide from concentrated aqueous NaCl solution: Oxidation at the anode: 2 Cl (aq) Cl 2(g) + 2e ; E = 1.36 V + or: 2 H 2 O (l) O 2(g) + 4 H (aq) + 4e ; E = 0.82 V (1.4 V w/overvoltage). Cl is oxidized instead of water at the anode. Reduction at the cathode: + Na (aq) + 1e Na (l) ; E = 2.71 V or: 2 H 2 O (l) + 2e H 2(g) + 2 OH (aq) ; E = 0.42 V ( 1 V w/overvoltage). Water is reduced to H 2 and OH instead of Na + at the cathode. Products: H 2 forms on the cathode, Cl 2 forms on the anode, NaOH solution forms at the cathode. 140

Electroplating Electroplating is a process that uses electric current to reduce dissolved metal cations so that they form a metal coating on an object. The object to be plated is the cathode of the circuit. The anode is usually made of the metal to be plated on the object. Both electrodes are immersed in a solution, containing one or more dissolved metal salts as well as other ions. 141

Stoichiometry of Electrolysis Electric current is measured in amperes. Ampere (A) is the base SI unit of current equivalent to 1 coulomb/second. The quantity of electric charge passing through a circuit in a given amount of time is given by the following equation: q = I t Electric (C) = electric (A or C/sec) time (sec) charge current Faraday constant = 96,500 C per 1 mole e - 142

A Problem To Consider 18.14. How many grams of Ca metal will be produced in an electrolytic cell of molten CaCl 2 if a current of 0.452 A is passed through the cell for 1.5 hours? CaCl 2(l) electrolysis Ca(l) + Cl 2(g) q = I t = 0.452 A 1.5 h 3600 sec 1 h = 2440.8 C Note: 2 moles of electrons must be transferred to form 1 mole of calcium metal!? g Ca = 2440.8 C 1 mol e 96500 C 1 mol Ca 40.08 g Ca 2 mol e 1 mol Ca = 0.50 g Ca 144

A Problem To Consider 18.15. When an aqueous solution of potassium iodide is electrolyzed using platinum electrodes, the half-reactions are: 2 I (aq) I 2(aq) + 2 e 2 H 2 O (l) + 2 e H 2(g) + 2 OH (aq) How many grams of iodine are produced when a current of 8.52 ma flows through the cell for 10.0 min? 145

A Problem To Consider 18.15. (Continued). q = I t = 8.52 10 3 A 10.0 min 60 sec 1 min = 5.112 C 2 I (aq) I 2(aq) + 2 e Note: 2 moles of electrons must be transferred to form 1 mole of I 2!? g I 2 = 5.112 C 1 mol e 96500 C 1 mol I 2 2 mol e 254 g I 2 1 mol I 2 = 0.00673 g I 2 146