Chapter 19: Redox & Electrochemistry
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1 Chapter 19: Redox & Electrochemistry 1. Oxidation-Reduction Reactions Definitions Oxidation - refers to the of electrons by a molecule, atom or ion Reduction - refers to the of electrons by an molecule, atom or ion Chemical reactions in which the oxidation state of one or more substances changes are called oxidation-reduction reactions (or reactions) L E O goes G E R or R O L R The # of electrons lost the # of electrons gained (they are transferred!) Oxidation - a means of determining whether the atom is neutral, electron-rich, or electron-poor. Does not necessarily imply ionic charges. 1. Elements =. 2. All oxidation numbers for atoms in compounds add to. All oxidation numbers for atoms in ions add to. 3. A fixed charge cation retains its charge. (i.e Ca2+ = 4. Hydrogen attached to a nonmetal (oxid # = ) Hydrogen attached to a metal (oxid # = ) 5. Oxygen (usually has oxid # = ) (unless it violates #2; e.g. peroxides) 6. Halide (F always -1; others usually has oxid # = ) (unless it violates #2) 7. Determine remaining element oxidation # by following rule #2. Example: Find the oxidation number of Cr in K2Cr2O7. K 2 Cr 2 O 7 Note, multiply when going to upper row; divide when going to lower row. Answer is always in bottom row. 19-1
2 LP#1. Determine the oxidation number of phosphorus in PO4 3-. Recognizing Redox Reactions - keeping track of oxidation #s Zn(s) + 2H + (aq) Zn 2+ (aq) + H2(g) The oxidation numbers for the above elements and ions are: Zn(s) = Zn 2+ = H + = H2 = For a reaction to be a redox reaction, oxidation numbers must change! Note: All single replacement reactions are redox reactions, but not all redox reactions are single replacement. In the above reaction of hydrogen and oxygen: Zn is ; It is the H + is ; It is the Oxidizing Agent: causes oxidation undergoes reduction gains electrons becomes more (-) (or less +) Reducing Agent: causes reduction undergoes oxidation looses electrons becomes more (+) (or less -) 19-2
3 LP#2. Identify the oxidized reactant, the reduced reactant, the oxidizing agent and the reducing agent in the following reaction: 2 Al (s) + Cr 2 O 3(s) 2 Cr (s) + Al 2 O 3(s) Note: We can say that Cr is reduced, but we must include the entire compound and say that Cr2O3 is the oxidizing agent. We must include the entire compound for solids and covalent compounds and entire ions for polyatomic ions in agents. Balancing Oxidation-Reduction Reactions In balancing redox reactions: the number of electrons gained must equal the number of electrons lost. Some can be balanced by inspection. For example: Zn(s) + 2H + (aq) Zn 2+ (aq) + H2(g) However, some require a more sophisticated process. Half-Reactions Although oxidation and reduction must take place simultaneously (i.e. for something to be reduced, something else must be oxidized) it is often convenient to consider them as separate processes Equations that show only oxidation, or reduction, are called half-reactions The following is the oxidation of Tin(II) by Iron(III) Sn 2+ (aq) + 2 Fe 3+ (aq) Sn 4+ (aq) + 2 Fe 2+ (aq) Oxid. ½ Rxn: Red. ½ Rxn: Oxidation ½ reactions have electrons as. Reduction ½ reactions have electrons as. 19-3
4 Balancing Equations By The Method Of Half-Reactions The use of half-reactions provides a general way to balance redox reactions The following is a general set of steps that can be followed to balance almost all redox equations: Steps to Balancing Redox Reactions 1. Write the unbalanced net ionic equation and check to see if the final solution is acidic or basic. 2. Identify and label the oxidation state of each atom in every species. 3. Identify the oxidation and the reduction half reactions. Make a note of how many e- are lost or gained by each species. 4. Write the separate half-cell reaction equations. To balance the atom number: a) Balance all atoms on each side (except O and H). b) Balance the number of oxygen atoms by adding water to the opposite side. c) Balance the number of hydrogen atoms by adding H+ to the opposite side. To balance the charge: d) Make the charges on each side of the equation the same by adding e-. (NOTE: this is a good place to check; the number of e- lost or gained in step 3 per atom should be the same as in the half-reactions.) e) If the number of electrons in each half-reaction is not the same, multiply by a factor so that both are the same. (remember, e- lost must = e- gained) 5. Combine half-reactions to make a full equation. Simplify by combining like terms. 6. Check to make sure that the atom number and the total charge on each side is the same. 7. If the solution is basic, add enough OH - to neutralize all the extra H +. These turn into waters on that side. Also add the same number of OH - to the other side of the equation. (Just like algebra; what you do to one side must be done to the other.) 19-4
5 Balancing Redox Reactions That Occur in Basic Solution The procedure for balancing in this case is actually the same as for balancing in an acidic solution, with one small twist. Even though the previous reaction does not occur in basic solution, we are going to pretend that it does for the sake of demonstrating the procedures for balancing redox reactions under basic conditions. Start with the balanced reaction in acidic solution. Neutralize any H + by adding OH - to both sides of the equation. Combine H + and OH - to form water. Reduce any replicate water molecules. Finding Oxidation Numbers Oxidation Number - a means of determining whether the atom is neutral, electron-rich, or electron-poor. This does not imply ionic charges. 8. Elements = 0. K, N2, Xe (oxid # = 0) 9. All oxidation numbers add to zero (or to ion charge for an ion). 10. A fixed charge cation retains its charge. Group #1 (oxid # = +1) Group #2 (oxid. # = +2) Ag ion = +1; Cd ion = +2; Zn ion = +2; Al ion = Hydrogen attached to a nonmetal (oxid # = +1) Hydrogen attached to a metal (oxid # = -1) 12. Oxygen (usually has oxidation # = -2) (unless it violates #2; e.g. peroxides) 13. Halide (F always -1; others usually has oxidation # = -1) (unless it violates #2) 14. Determine remaining element oxidation # by following rule #
6 Example: The reaction between MnO4 - and H2C2O4 in acidic solution: Here is the unbalanced equation that describes this reaction: MnO4 - + H2C2O4 Mn 2+ (aq) + CO2(g) To balance this redox reaction using the method of half-reactions, begin by writing the incomplete oxidation and reduction half-reactions: Oxid: Red: Balance everything except H and O. Oxid: H2C2O4 CO2 Red: MnO4 - Mn 2+ Add H2O to balance any O Oxid: H2C2O4 2CO2 Red: MnO4 - Mn 2+ Add H + to balance any H Oxid: H2C2O4 2CO2 Red: MnO4 - Mn H2O Balance charge with electrons Oxid: H2C2O4 2CO2 + 2H + Red: 8H + + MnO4 - Mn H2O Check to see if electrons agree with original predictions. Make #e- lost = # e- gained. (H2C2O4 2CO2 + 2 H + + 2e - ) (5e - + 8H + + MnO4 - Mn H2O) H2C2O4 CO2 + H + + e - e - + H + + MnO4 - Mn 2+ + H2O Add half reactions. Reduce any occurrences of H + or H2O on both sides of the equation: none here. 19-6
7 Balancing Redox Reactions That Occur in Basic Solution The procedure for balancing in this case is actually the same as for balancing in an acidic solution, with one small twist. Now let s pretend that this reaction occurred in basic solution for demo of process. Start with the balanced reaction in acidic solution. 5H 2 C 2 O 4 + 6H + + 2MnO 4-10CO 2 + 2Mn H 2 O Neutralize any H + by adding OH- to both sides of the equation. 5H 2 C 2 O 4 + 6H + + 2MnO 4-10CO 2 + 2Mn H 2 O Combine H + and OH - to form water. 5H 2 C 2 O 4 + 2MnO 4-10CO 2 + 2Mn H 2 O + 6 OH - Reduce any replicate water molecules. 5H 2 C 2 O 4 + 2MnO 4-10CO 2 + 2Mn OH
8 2. Electrochemical Cells Definition: 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. Electrochemical cells are based on redox chemistry. electrochemical cells in which a spontaneous reaction generates an electric current are known as: A voltaic cell is named after Count Alessandro Volta, , (an Italian physicist) Electrochemical cells in which an electric current drives an otherwise non-spontaneous reaction are known as. Voltaic Cells Suppose we created a cell using the reaction between zinc and copper. The reaction is spontaneous. Zn(s) is oxidized and Cu2+(aq) is reduced. The redox half-reactions for the above reaction would be: Oxidation: Reduction: 2 e- transferred If zinc metal is put directly into the solution of copper ions: We can t capture the flow of electrons (or the energy of the cell.) If we separate the Cu(s)/Cu 2+ from the Zn(s)/Zn 2+? 19-8
9 What if we provide a path (wire) for electron flow? However, we now have another problem... We start with a neutral soln. ([cations] = [anions]) In the zinc solution there is a build up of: In the copper solution there is a build up of: This will cause the flow of electron to We need a way to neutralize the charge build-up in the solutions due to the change in soluble ion concentrations! What if we had a tube filled with an electrolyte that connected the two redox reactions? This would allow: - cations to move towards the: - anions to move towards the: - keeping the charges in solution neutral In turn, this would now allow the electrons to flow The connecting tube of solution is called a and is usually filled with a gel or solution of a strong electrolyte such as KCl. Summary of the movement of ions, electrons and the redox half-reactions in a voltaic cell: 19-9
10 Electrodes The two solid metals in the different half-reactions are called: The electrode in the oxidation half-reaction is called the: The electrode in the reduction half-reaction is called the: Electrons always flow from:. In voltaic (aka galvanic) cells: The cathode attracts electrons and is labeled with a: The anode produces electrons and is labeled with a: Inert (aka Passive) vs Active electrodes. Inert electrodes enter into the chemical reaction, and just serve as a surface upon which electron transfers can occur (and be conducted to the other half-cell). The material of the electrode is not important as long as it does not enter into the overall reaction. Active electrodes chemically participate in the redox reaction Question: In the previous example, are the Zn and Cu active or passive electrodes? The copper electrode in the previous example is an electrode. (The Cu could deposit on any metal!) The zinc electrode in the previous example is an electrode.(the electrode must be made of Zn to create Zn 2+ ions!) Shorthand Notation for Cells It is convenient to have a shortcut way of designating particular voltaic cells. The cell just discussed can be written as; The anode (oxidation half cell) is written on the left. The cathode (reduction half-cell) is written on the right. The connecting salt bridge is denoted by two vertical bars. The cell terminals are at the extreme ends of the cell notation. A single vertical bar indicates a phase change boundary. It follows the path of the electron flow
11 When the half-cell involves a gas, an inert material such as platinum serves as a terminal and an electrode surface on which the reaction occurs. This figure shows a hydrogen electrode The cathode half-reaction is: The notation for the hydrogen electrode as a anode is: cathode is: A complete voltaic cell notation includes concentrations of solutions and pressure of gases written in parentheses. If not noted, concentrations are assumed to be standard conditions of 1.0 M concentration and pressure = 1.0 atm. i.e. Zn(s)Zn 2+ (1.0M)H + (1.0M)H2(1.0 atm)pt LP#3. Give the overall cell reaction for the voltaic cell Cd(s)Cd 2+ (1.0M)H + (1.0M)H2(1.0 atm)pt LP#4. Write the shorthand notation for a galvanic cell that uses the reaction: Fe(s) + Sn 2+ (aq) Fe 2+ (aq) + Sn(s) Where the concentration of iron (II) is 0.5M and the concentration of tin (II) is 0.1M. 3. Electromotive Force (EMF) What causes the electrons to flow from the anode to the cathode? The movement of electrons is analogous to the movement of water over a waterfall. Water moves from a point of potential energy to a point of potential energy. Thus, a PE difference is required
12 An electric charge also moves from a point of high electrical potential to one of lower electrical potential. Charge is measured in Coulombs (C) The Coulomb is the SI derived unit of electric charge. A Coulomb is the charge carried by x electrons. It is a unit of electrical charge equal to the amount of charge transferred by a current of 1 ampere (amp) in 1 second Current is the rate of flow of electrons. An alternate way of saying this is that 1 ampere (amp) = x electrons = 1 Coulomb sec sec The work that can be done by the water over a waterfall depends on the amount of water (rate x time) and the PE difference. The work that can be done by an electrical charge moving through a conductor depends on the amount of charge (e.g. # of e-) [current x time] and the potential difference. Potential difference is the difference in electric potential (PE) between two points. Measured in:. The volt, V, is the SI unit of potential difference equivalent to 1 joule (J) of energy per coulomb (C) of charge. 1V = 1J 1C OR 1 Joule = 1 Coulomb x 1 Volt Charge of 1 mole of electrons = 96,485 C = In moving 1 mole of electrons through a circuit, the numerical value of the work done by a voltaic cell is the product of the Faraday constant (F) times the potential difference between the electrodes
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