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Oxidation - Reduction Reactions Oxidation - Reduction Chemistry Oxidation - reduction (redox) reactions are chemical processes that involve a transfer of electrons between substances -- this can be a complete transfer to form ionic bonds or a partial transfer to form covalent bonds In all redox reactions: • one substance loses electrons -- this substance is oxidized • one substance gains electrons -- this substance is reduced There are lots of processes in the natural world (and in the laboratory) that involve redox reactions • e.g., corrosion, batteries, photosynthesis/respiration, etc. Reaction between zinc and sulfuric acid Sulfuric acid (solution of H+ and SO42- ions) Zn strip Reaction between zinc and sulfuric acid Overall equation (molecular equation) H2 bubbles Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g) Zinc loses electrons Overall ionic equation • zinc is oxidized Zn2+(aq) + 2e- Zn(s) -- all dissolved ions are explicitly shown Zn(s) + 2 H+(aq) + SO42–(aq) Zn2+(aq) + SO42–(aq) + H2(g) Hydrogen gains electrons Net ionic equation • hydrogen is reduced 2 H+(aq) + 2e- H2(g) -- includes only substances that undergo change -- ions that are present but do not react (spectator ions) are not shown Zn(s) + 2 H+(aq) Electrons are transferred from zinc to hydrogen Reaction between Cu and AgNO3 Cu(s) Ag+(aq) Ag(s) initial final NO3–(aq) Cu2+(aq) NO3–(aq) Oxidation of Cu: Cu(s) → Cu2+(aq) + 2eReduction of Ag+: 2 Ag+(aq) + 2e- → 2 Ag(s) Zn2+(aq) + H2(g) Oxidation - Reduction Reactions Oxidation - reduction (redox) reactions are chemical processes that involve a transfer of electrons between substances • Oxidation occurs when a substance loses electrons • Reduction occurs when a substance gains electrons In a redox reaction, oxidation and reduction occur simultaneously -- one cannot occur in the absence of the other Electrons are transferred from Cu atoms to Ag+ ions in solution Cu loses electrons (oxidation) Overall Equation: Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq) Overall Ionic Equation: Cu(s) + 2 Ag+(aq) + 2 NO3–(aq) → 2 Ag(s) + Cu2+(aq) + 2 NO3–(aq) Net Ionic Equation: Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq) initial Cu(s) Ag+(aq) final Ag(s) Cu2+(aq) Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) Ag+ gains electrons (reduction) Oxidation number Oxidation / reduction memory aid The concept of oxidation numbers (also called oxidation states) was devised as a bookkeeping system for keeping track of electrons during redox reactions. L ose E lectrons O xidation -- oxidation of an element results in an increase in its oxidation number -- reduction of an element results in a decrease in its oxidation number G ain E lectrons R eduction Oxidation numbers are assigned according to a specific set of rules Rules for assigning oxidation numbers Rules for assigning oxidation numbers 1. An element in its elemental state is assigned an oxidation number of zero 2. For any monoatomic ion, the oxidation number is equal to the charge on the ion. Ba K S 0 0 0 Au 0 barium potassium sulfur gold This includes elements that exist as a diatomic molecules H2 Cl2 Cl- Ba2+ +2 barium ion (oxidation number: +2) chloride ion (oxidation number: –1) O2 0 0 0 hydrogen chlorine oxygen BaCl2 -1 +2 -1 In an ionic compound, the ions retain their oxidation number ionic compound free ions Rules for assigning oxidation numbers Rules for assigning oxidation numbers 3. Nonmetals usually have negative oxidation numbers (but sometime they can be positive) 3. Nonmetals usually have negative oxidation numbers (but sometime they can be positive) a. The oxidation number of oxygen is usually –2 b. The oxidation number of hydrogen is usually +1 when bonded to nonmetals and –1 when bonded to metals. Exceptions: Peroxide ion (O22-): oxidation number = -1 Oxygen difluoride (OF2): oxidation number = +2 H 2O H 2O -2 water MgO -2 magnesium oxide H2O2 -1 hydrogen peroxide OF2 +2 oxygen difluoride +1 water CH4 +1 methane NaH -1 sodium hydride Rules for assigning oxidation numbers 3. Nonmetals usually have negative oxidation numbers (but sometime they can be positive) c. The oxidation number of fluorine is –1 in all compounds. Other halogens have an oxidation number of -1 in most binary compounds. Exceptions: Rules for assigning oxidation numbers 4. The sum of the oxidation numbers of all atoms in a neutral compound is zero. 5. The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion. Halogens (except fluorine) combined with oxygen have positive oxidation numbers) HF NaCl -1 -1 hydrogen fluoride sodium chloride MgBr2 -1 magnesium bromide ClO4-1 +7 perchlorate ion Many elements have multiple oxidation numbers Many elements have multiple oxidation numbers It depends on the types of compounds they form It depends on the types of compounds they form N oxidation number N2 N2O NO N2O3 NO2 N2O5 NO3– 0 +1 +2 +3 +4 +5 +5 Note: The oxidation number of oxygen is –2 in all of these compounds Elemental copper (Cu) Copper (II) sulfate (CuSO4) Cu oxidation state: 0 Cu oxidation state: +2 Rules for determining the oxidation number of an element within a compound Step 1: Write the oxidation number of each known atom below the atom in the formula Step 2: Multiply each oxidation number by the number of atoms of that element in the compound Step 3: Assign oxidation numbers for the other atoms in the compound in order to make the sum of the oxidation numbers equal to zero Example: Determine the oxidation number of carbon in carbon dioxide CO2 -2 2(-2) = -4 C + (-4) = 0 C = +4 (oxidation number for carbon) Step 1: Write the oxidation number of each known atom below the atom in the formula Step 2: Multiply each oxidation number by the number of atoms of that element in the compound Step 3: Assign oxidation numbers for the other atoms in the compound in order to make the sum of the oxidation numbers equal to zero Example: Determine the oxidation number of sulfur in sulfuric acid Example: Determine the oxidation number of chromium in Cr2O72- H2SO4 +1 Cr2O72- -2 2(+1) = +2 -2 4(-2) = -8 7(-2) = -14 +2 + S + (-8) = 0 2Cr + (-14) = -2 (the charge on the ion) S = +6 (oxidation number for sulfur) Cr = +6 (oxidation number for chromium) Step 1: Write the oxidation number of each known atom below the atom in the formula Step 1: Write the oxidation number of each known atom below the atom in the formula Step 2: Multiply each oxidation number by the number of atoms of that element in the compound Step 2: Multiply each oxidation number by the number of atoms of that element in the compound Step 3: Assign oxidation numbers for the other atoms in the compound in order to make the sum of the oxidation numbers equal to zero Step 3: Assign oxidation numbers for the other atoms in the compound in order to make the sum of the oxidation numbers equal to zero Example: Determine the oxidation number of potassium and nitrogen in KNO3 KNO3 K+ Oxidation - reduction (redox) reactions are chemical processes that involve a transfer of electrons between substances NO3– Recognize that this is an ionic compound between K+ and NO3The oxidation number of potassium in For nitrogen: K+ Oxidation - Reduction Reactions is +1 (the charge on the ion) -- this can be a complete transfer to form ionic bonds or a partial transfer to form covalent bonds In redox reactions, the oxidation numbers of the elements involved in the reaction change NO3– • one substance is oxidized (loses electrons) -2 -- oxidation number increases 3(-2) = -6 • one substance is reduced (gains electrons) N + (-6) = -1 (the charge on the ion) -- oxidation number decreases N = +5 (oxidation number for nitrogen) Reaction can be rewritten to emphasize electron transfer Reaction between zinc and sulfuric acid Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g) Zn + 2 H+ + SO42- Zn2+ + SO42- + H2 0 +2 0 +2 +1 0 +1 0 •zinc loses electrons •the oxidation number of Zn increases •zinc is oxidized •zinc loses electrons •the oxidation number of Zn increases •zinc is oxidized •hydrogen gains electrons •the oxidation number of H decreases •hydrogen is reduced •hydrogen gains electrons •the oxidation number of H decreases •hydrogen is reduced Electrons are transferred from zinc to hydrogen Electrons are transferred from zinc to hydrogen Reaction between zinc and sulfuric acid Oxidizing and reducing agents Oxidizing agent: The reactant that causes another substance to be oxidized – i.e., the reactant that causes an increase in the oxidation state of another substance Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g) 0 +2 +1 The oxidizing agent is reduced in a redox reaction Reducing agent: The reactant that causes another substance to be reduced – i.e., the reactant that causes a decrease in the oxidation state of another substance The reducing agent is oxidized in a redox reaction 0 • • • • zinc loses electrons the oxidation number of Zn increases zinc is oxidized zinc is the reducing agent (it causes hydrogen to be reduced) • • • • hydrogen gains electrons the oxidation number of H decreases hydrogen is reduced hydrogen is the oxidizing agent (it causes zinc to be oxidized) Example: Is the following a redox reaction? Example: Is the following a redox reaction? Neutralization reaction between hydrochloric acid and potassium hydroxide Thermite reaction HCl (aq) + KOH (aq) +1 -1 Element +1 H2O (l) + KCl (aq) -2 +1 +1 -2 +1 -1 Oxidation number Oxidation number before reaction after reaction H +1 +1 O –2 –2 K +1 +1 Cl –1 –1 Balancing redox equations Balancing any chemical equation is based on the law of conservation of mass • the number of atoms of each element must be the same on both sides of the equation 2 Al (s) + Fe2O3 (s) 0 Element +3 Al2O3 (l) + 2 Fe (l) -2 +3 -2 Oxidation number Oxidation number before reaction after reaction Al 0 +3 Fe +3 0 O –2 –2 0 Which element is oxidized? Which element is reduced? Balancing redox equations For many simple chemical reactions, balancing the overall equation “automatically” balances the gain/loss of electrons • i.e., the equation can be balanced without explicitly considering the transfer of electrons Example: Thermite reaction When balancing redox equations, there is an additional requirement • the total gains and losses of electrons must balance each other The increase in oxidation number for the oxidized substance must be equal to the decrease in oxidation number for the reduced substance Al is oxidized (oxidation number: 0 to +3) Each Al atom loses 3 electrons Total change in oxidation number of Al: +3 x 2 atoms of Al = +6 2 Al (s) + Fe2O3 (s) Al2O3 (l) + 2 Fe (l) Fe is reduced (oxidation number: +3 to 0) Each Fe atom gains 3 electrons Total change in oxidation number of Fe: -3 x 2 atoms of Fe = -6 Balancing redox equations Method for balancing complex redox equations For complex redox equations, it can be difficult to balance the equation without explicitly accounting for the numbers of electrons lost and gained Example: Example: +6 2 Fe+2(aq) + 2 Fe+3(aq) Sn+4(aq) + Sn+2(aq) +2 +3 Cr2O7-2(aq) + Fe+2(aq) +3 Cr+3(aq) + Fe+3(aq) -2 The numbers of atoms of each element are the same on each side Step 1: Identify atoms undergoing oxidation and reduction But the total ionic charge is not balanced • +6 on the left • +5 on the right Fe undergoes oxidation ( oxidation number: +1) Electron gains/losses are also not balanced • 1 electron lost from Fe (oxidation) • 2 electrons gained by Sn (reduction) Cr undergoes reduction ( oxidation number: -3) Method for balancing complex redox equations Method for balancing complex redox equations Example: Example: +1 x 1 Fe atom = +1 +6 +2 +3 Cr2O7-2(aq) + Fe+2(aq) +3 2 Cr+3(aq) + Fe+3(aq) Cr2O7-2(aq) + Fe+2(aq) 2 Cr+3(aq) + Fe+3(aq) -3 x 2 Cr atoms = -6 Step 2: Adjust coefficients to balance atoms undergoing oxidation/reduction Step 3: Write arrows connecting the atoms being oxidized and reduced • indicate the change in oxidation indicated above the arrow Be sure to multiply the change in oxidation number for each atom by the number of atoms undergoing the change Method for balancing complex redox equations Example: Method for balancing complex redox equations Example: +1 x 1 6 Fe atom atoms==+1 +6 Cr2O7-2(aq) + 6 Fe+2(aq) 2 Cr+3(aq) + 6 Fe+3(aq) -3 x 2 Cr atoms = -6 Step 4: Check to see if the oxidation balances reduction • i.e., total change in oxidation number for oxidized substance equals total change in oxidation number for reduced substance If not, adjust the reaction coefficients again to balance oxidation and reduction Total charge on left side: -2 + +2(6) = +10 -2(aq) +2(aq) Cr2O7-2Cr (aq) Fe+2(aq) + 6 Fe +14 H+(aq) 2O+ 7 6 +3(aq) 2 Cr Cr+3 (aq) + +6 6Fe Fe+3+3(aq) (aq)+ 7 H2O(l) Total charge on right side: +3(2) + +3(6) = +24 Step 5: Balance elements and charge by adding available chemical species to the reactants or products sides of the equation a. Acid solutions: Use H2O and H+ to balance the equation It is best to add H+ ions first to obtain charge balance, and then add H2O to the equation to obtain material balance Add 14 H+ to left side to obtain charge balance Add 7 H2O to right side to obtain material balance Method for balancing complex redox equations Example: Method for balancing complex redox equations Note: For redox reactions taking place in basic solutions, replace Step 5a with Step 5b. Cr2O7-2(aq) + 6 Fe+2(aq) + 14 H+(aq) 2 Cr+3(aq) + 6 Fe+3(aq) + 7 H2O(l) Step 6: Check your final equation to verify charge and material balance Step 5b. Basic solutions: Use H2O and OH– to balance the equation It is best to add OH– ions first to obtain charge balance, and then add H2O to the equation to obtain material balance Material Balance Element Left Right Cr 2 2 O 7 7 Fe 6 6 H 14 14 Charge Balance Net Charge on Left = (-2) + 6(+2) + 14(+1) = +24 Net Charge on Right = 2(+3) + 6(+3) = +24 For redox reactions taking place in neutral solutions, use either Step 5a or Step 5b.