Oxidation-Reduction Reactions

Oxidation-Reduction Reactions “LEO says GER” L E lose electrons, oxidize O says G E gain electrons, reduce R

What is a Redox Reaction? Redox is a term for reactions in which the acceptance of an electron (reduction) by a material is matched with the donation of an electron (oxidation). Possible Redox Reactions: Synthesis Reactions Decomposition Reactions Single Replacement Reactions

How Do We Know if a Reaction is a Redox Reaction? Check oxidation number of each element in a reaction to determine if there is both oxidation and reduction. Oxidation number: the charge, or apparent charge, an atom has when bonded to another atom. There are rules for determining oxidation numbers!

Redox Example 2Ca (s) + O2 (g)  2CaO (s) Synthesis reaction: Ca loses electrons; it is oxidized. O gains electrons; it is reduced. Ca is reducer or reducing agent. O is oxidizer or oxidizing agent.

Mg(s) + 2HCl(ag)  MgCl2(aq) + H2(g) Another Redox Example Single Replacement reaction: Mg(s) + 2HCl(ag)  MgCl2(aq) + H2(g) Mg loses electrons = oxidized H gains electron = reduced

Oxidation Number Rules Any atom of an element has an oxidation number of zero. Any monoatomic metallic ion has an oxidation number equal to its charge. (Transition metals have multiple oxidation numbers.) Oxygen has an oxidation number of -2. Oxygen exceptions: With fluorine, oxygen is +2 In peroxide, oxygen is -1

Oxidation Number Rules, Cont. Hydrogen has an oxidation number of +1. Exception: In a metal hydride, hydrogen is -1. Nonmetals may have multiple oxidation numbers, both positive and negative. The sum of the oxidation numbers in a compound equals zero.

Assigning Oxidation Numbers Assign oxidation numbers to all the atoms in each of the following: HNO3 CuCl2 O2 H2O2 MgSO4 CO2 PbSO4 PbO2 Ag Na2C2O4 Cr2O3 (NH4)2Ce(SO4)3

Balancing Redox Reactions Using the Ion Electron Method

Acidic or Neutral Equations Write an ionic equation in which all soluble species are dissociated. HNO3(aq) + H2S(g)  NO (g) + S8(s) + H2O(l) Becomes H+(aq) + NO3-(aq) + H2S(g) 

Acidic or Neutral Equations Write a Skeletal Equation (one that includes only substances involved in redox reaction). NO3- + H2S  NO + S8

Acidic or Neutral Equations Write equations for the half reactions. Balance the mass in each except for oxygen and hydrogen. Red: NO3-  NO Ox: 8H2S  S8

Acidic or Neutral Equations Add oxygen in the form of water to the oxygen-deficient side. Red: NO3-  2H2O + NO Ox: 8H2S  S8

Acidic or Neutral Equations Add hydrogen in the form of H+ to the hydrogen-deficient side Red: NO3- + 4H+  2H2O + NO Ox: 8H2S  16H+ + S8

Acidic or Neutral Equations Add electrons as e- to the more positive side. Red: 3e- +NO3- + 4H+  2H2O + NO Ox: 8H2S  16H+ + S8 + 16e-

Acidic or Neutral equations Find L.C.M. of electrons and multiply as needed. Red: (3e- +NO3- + 4H+  2H2O + NO)16 Ox: 3(8H2S  16H+ + S8 + 16e- )

Acidic or Neutral equations Add the half cells. Cancel out items common to both half cells. Red: 48e- +16NO3- + 64H+  32H2O + 16NO Ox: 24H2S  48H+ + 3S8 + 48e- 16H++ 16NO3- + 24H2S16NO + 32H2O + 3S8

Acidic or Neutral equations If possible, place these coefficients in the original equation (primarily done if equation has neutral species). 16HNO3(aq) + 24H2S(g)  16NO (g) + 3S8(s) + 32H2O(l)

Basic Equations Write an ionic equation in which all soluble species are dissociated. What dissociates? Strong electrolytes only! NH3(g) + O2(g)  NO(g) + H2O(l) Remains the same (nothing dissociates).

Basic Equations Write a Skeletal Equation (one that includes only substances involved in redox reaction). NH3(g) + O2(g)  NO(g) + H2O(l)

Basic Equations Write equations for the half reactions. Balance the mass for each element except oxygen and hydrogen. Ox: NH3  NO Red: O2  H2O

Basic Equations To balance oxygen, add oxygen in the form of water to the oxygen-deficient side. Ox: H2O + NH3  NO Red: O2  2H2O

Basic Equations To balance hydrogen, add hydrogen in the form of H+ to the hydrogen-deficient side. Ox: H2O + NH3  NO + 5H+ Red: O2 + 4H+  2H2O

Basic Equations For each H+ added, add an OH- to both sides. Ox: 5OH- + H2O + NH3  NO + 5H++ 5OH- Red: O2 + 4H+ + 4OH-  4OH- + 2H2O

Basic Equations Combine H+ and OH- to form water, and cancel out water molecules on both sides of half reactions. Ox: 5OH- + H2O + NH3  NO + 5H2O Red: O2 + 4H2O  4OH- + 2H2O Ox: 5OH- + NH3  NO + 4H2O Red: O2 + 2H2O  4OH-

Add electrons as e- to the more positive side. Basic Equations Add electrons as e- to the more positive side. Ox: 5OH- + NH3  NO + 4H2O + 5e- Red: 4e- + O2 + 2H2O  4OH-

Find L.C.M. of electrons and multiply as needed. Basic Equations Find L.C.M. of electrons and multiply as needed. Ox: (5OH- + NH3  NO + 4H2O + 5e-)4 Red: 5(4e- + O2 + 2H2O  4OH-)

Add the half cells. Cancel out items common to both half cells. Basic Equations Add the half cells. Cancel out items common to both half cells. Ox: 20 OH- + 4 NH3  4NO + 16H2O + 20e- Red: 20 e- + 5 O2 + 10 H2O  20 OH- 4NH3 + 5 O2  4NO + 6H2O

Basic Equations If possible, place these coefficients in the original equation (primarily done if equation has neutral species). 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)

Galvanic (Voltaic) Cells The energy released in a spontaneous redox reaction is used to perform electrical work. Galvanic cells are devices in which chemical energy is changed to electrical energy. Electricity = movement of charged particles Electrochemical cells = Galvanic cell = battery Galvanic cells are spontaneous.

Galvanic Cells If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+. Zn is spontaneously oxidized to Zn2+ by Cu2+. The Cu2+ is spontaneously reduced to Cu0 by Zn. The entire process is spontaneous.

Galvanic Cell Description Galvanic cells consist of Two solid metals for the electrodes Anode (oxidation): Zn(s)  Zn2+(aq) + 2e- AN OX Cathode (reduction): Cu2+(aq) + 2e-  Cu(s) RED CAT Electrolytic solution in each compartment Salt bridge (electrolyte + jello-like material): connects both compartments so ions flow to keep net charge in each compartment zero. Wire connecting each electrode

Salt Bridge or Porous Disk

Galvanic Cell Description As oxidation occurs, Zn is converted to Zn2+ and 2e-. The electrons flow towards the anode where they are used in reduction reaction. Flow of electrons = electricity = work Zn electrode loses mass and Cu electrode gains mass. Galvanic cell eventually “dies.”

“Rules” of Galvanic Cells 1. At the anode (oxidation) electrons are products. 2. At the cathode (reduction) electrons are reactants. 3. Electrons cannot swim.

Example Packet p 7 The following redox reaction is spontaneous: Cr2O72-(aq) + 14H+(aq) + 6I-(aq) → 2Cr3+(aq) + 3I2(s) + 7H2O(l) A voltaic cell is created using potassium dichromate and sulfuric acid in one beaker and a solution of KI in another. A salt bridge connects the two beakers and platinum is used for the electrodes. Indicate the reactions at the anode and cathode, the direction of e- flow, and the signs at the electrodes.

Example Packet p 8 The two half-reactions in a voltaic cell are Zn(s) → Zn2+(aq) + 2e- ClO3-(aq) + 6H+(aq) + 6e- → Cl-(aq) + 3H2O(l) a) Which reaction occurs at anode, which at cathode? b) Which electrode is consumed in the cell reaction? c) Which electrode is positive?

Cell Potential The flow of e-’s from anode to cathode is spontaneous. e-’s flow from anode to cathode because the cathode has a lower electrical potential energy than the anode. Potential difference = difference in electrical potential. Measured in volts.

Cell Electromotive Force One volt is the potential difference required to impart one joule of energy to a charge of one coulomb: Electromotive force (emf): force required to push electrons through external circuit. Cell potential (Ecell) = emf of a cell. Ecell : standard emf (standard cell potential) at standard conditions (1M solutions at 25 C)

Standard Reduction (Half-Cell) Potentials We have convenient tables of standard reduction potentials in aqueous solutions. Note: oxidation will NOT be on table, just use reduction value. Standard reduction potentials, Ered are measured relative to the standard hydrogen electrode (SHE).

SHE is the cathode. It consists of a Pt electrode in a tube placed in 1 M H+ solution. H2 is bubbled through the tube. For the SHE, we assign 2H+(aq, 1M) + 2e-  H2(g, 1 atm) Ered of zero.

Calculating Standard Reduction Potentials emf calculated from standard reduction potentials: Consider Zn(s)  Zn2+(aq) + 2e-. Ecell = Ered(cathode) - Ered(anode) Ecell = 0 V - Ered = -0.76 V. Standard reduction potentials must be written as reduction reactions: Zn2+(aq) + 2e-  Zn(s), Ered = -0.76 V.

Standard Reduction Potentials Reactions with Ered > 0 are spontaneous reductions relative to the SHE. Since Ered = -0.76 V, reduction of Zn2+ in the presence of the SHE is not spontaneous but the oxidation of Zn with the SHE is spontaneous. Changing the stoichiometric coefficient does not affect Ered. Therefore, 2Zn2+(aq) + 4e-  2Zn(s), Ered = -0.76 V.

Calculating E for a Cell Recall Combining two half-reactions often requires two manipulations: Half-reaction with largest potential will run as written (reduction) and other half-reaction will run in reverse. Change sign of anode reaction and add it to reduction reaction. Balance half-reactions but DO NOT change E!

Example Packet p 10 We have a Zn-Cu2+ voltaic cell: Zn(s) + Cu2+(aq, 1 M) → Zn2+(aq, 1 M) + Cu(s) Ecell = 1.10 V Given that the standard reduction potential of zinc ion is -0.76 V, calculate the Ered for the reduction of copper (II) ion to solid copper.

Example Packet p 11 Using the standard reduction potentials listed in Table 201., calculate the standard emf of a cell with the following reaction: Cr2O72-(aq) + 14H+(aq) + 6I-(aq) → 2Cr3+(aq) + 3I2(s) + 7H2O(l)

Example Packet p 12 A voltaic cell is based on the following two standard half reactions: Cd2+(aq) + 2e- → Cd(s) Sn2+(aq) + 2e- → Sn(s) Determine the reactions that occur at the cathode and the anode. Determine the cell potential.

Standard Reduction Potentials Reactions with Ered < 0 are spontaneous oxidations relative to the SHE. The larger the difference between Ered values, the larger Ecell. In a Galvanic cell (spontaneous) Ered(cathode) is more positive than Ered(anode). A negative E indicates a nonspontaneous process.

Example Packet p 14 Using the table of standard reduction potentials, determine whether the following reactions are spontaneous under standard conditions. a. Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) b. Cl2(g) + 2I-(aq) → 2Cl-(aq) + I2(s)

Oxidizing and Reducing Agents The more positive Ered the stronger the oxidizing agent on the left. The more negative Ered the stronger the reducing agent on the right. A species on the higher to the left of the table of standard reduction potentials will spontaneously oxidize a species that is lower to the right in the table. That is, F2 will oxidize H2 or Li; Ni2+ will oxidize Al(s).

Example Packet p 13 Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents. Nitrate, silver ion, dichromate

Line Notation for a Galvanic Cell Need handy notation for describing cells. Anode on left, cathode on right separated by double vertical lines for salt bridge. Phase difference in compartments shown by single vertical line. Ex: Mg(s)Mg2+(aq)  Al3+(aq)Al(s) anode cathode

Description of a Galvanic Cell Complete description includes 4 items: The cell potential (always positive) and the overall balanced cell reaction. The direction of e- flow given by half-reactions to obtain positive cell potential. Designation of anode and cathode. Notation with electrode/ions present in each compartment.

Zumdahl Example pp 832-833 Describe completely the galvanic cell based on the following half-reactions under standard conditions. Ag+ + e- → Ag E = 0.80 V Fe3+ + e- → Fe2+ E = 0.77 V

∆G and Cell Potential 1F = 96,500 C/mol = 96,500 J/V mol We can show that G = -nFE G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell. We define Since n and F are positive, if G < 0 then E  0 for a spontaneous reaction. 1F = 96,500 C/mol = 96,500 J/V mol

Example Packet p 15 Use the standard reduction potentials to calculate the standard free-energy change for the following reactions. Use G = -nFE and 1F = 96,500 J/V mol 4Ag(s) + O2(g) + 4H+(aq) → 4Ag+(aq) + 2H2O(l) What are the values of E and ∆G for: 2Ag(s) + 1/2O2(g) + 2H+(aq) → 2Ag+(aq) + H2O(l)

Nernst Equation A Galvanic cell is functional until E = 0 at which point equilibrium has been reached. The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. The Nernst equation relates emf to concentration using and noting that

Workable Nernst Equation This rearranges to give the Nernst equation: The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K and the base 10 logarithm: E = E - (0.0592 V/ n) log Q Remember that n is number of moles of electrons.

Example Packet p 17 (top) Calculate the emf at 298 K generated by the cell involving the following reaction: Cr2O72-(aq) + 14H+(aq) + 6I-(aq) → 2Cr3+(aq) + 3I2(s) + 7H2O(l) When [Cr2O72-] = 2.0 M, [H+] = 1.0 M , and [Cr3+] = 1.0 x 10-5 M, and [I-] = 1.0 M.

Example Packet p 17 (bottom) If the voltage of the Zn-H+ cell is 0.45 V at 298 K when [Zn2+] = 1.0 M and PH2 = 1.0 atm, what is the concentration of hydrogen ion? Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

Concentration Cells Nernst equation can be used to generate a cell that has an emf based solely on difference in concentration. One compartment will consist of a concentrated solution, while the other has a dilute solution. Example: 1.00 M Ni2+(aq) and 1.00 10-3 M Ni2+(aq). The cell tends to equalize the concentrations of Ni2+(aq) in each compartment. The concentrated solution has to reduce the amount of Ni2+(aq) [to Ni(s)], so must be the cathode.

Concentration Cells A concentration cell will function until the concentration of ions in each cell is equal.

Cell emf and Equilibrium A system is at equilibrium when G = 0. From the Nernst equation, at equilibrium and 298 K (E = 0.00 V and Q = Keq): So, if we know cell emf, we can calculate K.

Example Packet p 18 A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has P = 1.00 atm and an unknown [H+]. Electrode 2 is a standard hydrogen electrode (P = 1.00 atm, [H+] = 1.00 M.) At 298 K the measured cell voltage is 0.211 V, and the electrical current is observed to flow from Electrode 1 through the external circuit to Electrode 2. Use E = E - (0.0592 V/ n) log Q; Ecell = 0.00 V a) Calculate [H+] for the solution at Electrode 1. b) What is the pH?

Example Packet p 19 Using standard electrode reduction potentials , calculate the equilibrium constant for the oxidation of iron (II) ion by oxygen in acidic solution: O2(g) + 4H+(aq) + 4Fe2+(aq) → 4Fe3+(aq) + 2H2O(l)

Batteries A battery is a self-contained electrochemical power source with one or more voltaic cell. When the cells are connected in series, greater emfs can be achieved.

Lead-Acid Battery A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V. Cathode: PbO2; Anode: Pb Wood or glass-fiber spacers used to prevent electrodes form touching. Battery condition monitored by measuring density of acid solution. 3-5 yr life.

Alkaline Battery emf = 1.55 V at room temp. Anode: Zn cap (Zn powder mixed in a gel) Cathode: MnO2, NH4Cl and C paste; reduction of MnO2 Graphite rod in center is inert cathode. Alkaline battery: NH4Cl replaced with KOH.

Fuel Cells Direct production of electricity from fuels. Galvanic cell with continuously-supplied reactants. H2-O2 fuel cell primary source of electricity on Apollo moon flights. By-products are steam and hydroxide. Fuel cells being used in cars.

Corrosion Oxidation of metals Oxidation of most metals by oxygen is spontaneous. Metals such as Cu, Ag, Au, and Pt called noble metals- resistant to oxidation. Most metals develop a layer a thin oxide coating that protects inner atoms.

Corrosion of Iron Corrosion of iron is electrochemical process. Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen. Water acts as salt bridge (steel does not rust in dry air). Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3.xH2O(s).

Galvanized iron coated with a thin layer of zinc. Corrosion can be prevented by coating iron with paint or another metal. Galvanized iron coated with a thin layer of zinc.

Example Packet p 22 Predict the nature of the corrosion that would take place if an iron gutter were nailed to a house using aluminum nails.

Electrolysis Nonspontaneous reactions require an external current in order to force the reaction to proceed. Electrolysis reactions are nonspontaneous. In voltaic and electrolytic cells: reduction at cathode; oxidation at anode. However, in electrolytic cells, electrons are forced to flow from the anode to cathode.

In electrolytic cells the anode is positive and the cathode is negative. In galvanic cells the anode is negative and the cathode is positive.

Example Packet p 24 Electrolysis of AgF(aq) in an acidic solution leads to the formation of silver metal and oxygen gas. a. Write the half-reaction that occurs at each electrode. b. Calculate the minimum emf for this process under standard conditions.

Electroplating Ni plates on the inert electrode. Electroplating is important in protecting objects from corrosion.

Quantitative Aspects of Electrolyis How much material can we obtain with electrolysis? Consider Cu2+(aq) + 2e-  Cu(s). 2 mol of electrons will plate 1 mol of Cu. The charge of 1 mol of electrons is 96,500 C (1 F). Since Q = It, the amount of Cu can be calculated from the current (I) and time (t) taken to plate.

Example Packet p 25 Calculate the number of grams of aluminum produced in 1.00 hr by the electrolysis of molten AlCl3 if the electrical current is 10.0 amperes (A). Use Q = I t, where Q = charge (Coulomb = C); I = current (A); and t = time (sec)

Electrical Work Free-energy is a measure of the maximum amount of useful work that can be obtained from a system. We know If work is negative, then work is performed by the system and E is positive.

Electrical Work emf can be thought about as a measure of the driving force for a redox process. In an electrolytic cell an external source of energy is needed to force the reaction to proceed. To drive the nonspontaneous reaction the external emf must be greater than Ecell. From physics: work measured in watts (W): 1 W = 1 J/s.

Last Example Packet p 26 Calculate the number of kilowatt-hours of electricity required to produce 1.0 x 103 kg of aluminum by the electrolysis of Al3+ if the applied emf is 4.50 V. Remember: 1 J = 1 C-V 1 mol e-’s = 96,500 C 1 W = 1 J/s