Oxidation - Reduction Chemistry

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1 Oxidation - Reduction Chemistry Oxidation - Reduction Reactions 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. Sulfuric acid (solution of H + and SO4 ions) Zn strip H2 bubbles Zinc loses electrons zinc is oxidized Zn(s) Zn 2+ (aq) + 2e - Hydrogen gains electrons hydrogen is reduced 2 H + (aq) + 2e - H 2 (g) Overall equation (molecular equation) Zn(s) + H 2 SO 4 (aq) ZnSO 4 (aq) + H 2 (g) Overall ionic equation -- all dissolved ions are explicitly shown Zn(s) + 2 H + (aq) + SO 2 4 (aq) Zn 2+ (aq) + SO 2 4 (aq) + H 2 (g) Net ionic equation -- includes only substances that undergo change -- ions that are present but do not react (spectator ions) are not shown Electrons are transferred from zinc to hydrogen Zn(s) + 2 H + (aq) Zn 2+ (aq) + H 2 (g) Reaction between Cu and AgNO 3 Oxidation - Reduction Reactions Cu(s) Ag(s) Oxidation - reduction (redox) reactions are chemical processes that involve a transfer of electrons between substances Ag + (aq) NO 3 (aq) initial final Cu 2+ (aq) NO 3 (aq) Oxidation occurs when a substance loses electrons Reduction occurs when a substance gains electrons Oxidation of Cu: Cu(s) Cu 2+ (aq) + 2e - Reduction of Ag + : 2 Ag + (aq) + 2e - 2 Ag(s) Electrons are transferred from Cu atoms to Ag + ions in solution Overall Equation: Cu(s) + 2 AgNO 3 (aq) 2 Ag(s) + Cu(NO 3 ) 2 (aq) In a redox reaction, oxidation and reduction occur simultaneously -- one cannot occur in the absence of the other Cu loses electrons (oxidation) Overall Ionic Equation: Cu(s) + 2 Ag + (aq) + 2 NO 3 (aq) 2 Ag(s) + Cu 2+ (aq) + 2 NO 3 (aq) initial final Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) Net Ionic Equation: Cu(s) + 2 Ag + (aq) 2 Ag(s) + Cu 2+ (aq) Cu(s) Ag + (aq) Ag(s) Cu 2+ (aq) Ag + gains electrons (reduction)

2 Oxidation / reduction memory aid L E O G E R ose lectrons xidation ain lectrons eduction The concept of oxidation numbers (also called oxidation states) was devised as a bookkeeping system for keeping track of electrons during redox reactions. -- oxidation of an element results in an increase in its oxidation number -- reduction of an element results in a decrease in its oxidation number s are assigned according to a specific set of rules 1. An element in its elemental state is assigned an oxidation number of zero Ba barium K potassium This includes elements that exist as a diatomic molecules S sulfur Au gold 2. For any monoatomic ion, the oxidation number is equal to the charge on the ion. Ba 2+ barium ion (oxidation number: +2) Cl chloride ion (oxidation number: 1) BaCl In an ionic compound, the ions retain their oxidation number H 2 Cl 2 O hydrogen chlorine oxygen free ions ionic compound 3. Nonmetals usually have negative oxidation numbers (but sometime they can be positive) a. The oxidation number of oxygen is usually 2 Exceptions: Peroxide ion (O2 ): oxidation number = -1 Oxygen difluoride (OF2): oxidation number = +2 H2O MgO H2O2 OF2 water magnesium oxide -1 hydrogen peroxide +2 oxygen difluoride 3. Nonmetals usually have negative oxidation numbers (but sometime they can be positive) b. The oxidation number of hydrogen is usually +1 when bonded to nonmetals and 1 when bonded to metals. H2O CH4 NaH water methane sodium hydride -1

3 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: Halogens (except fluorine) combined with oxygen have positive 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. HF NaCl MgBr2 ClO4-1 hydrogen fluoride -1-1 sodium chloride -1 magnesium bromide +7 perchlorate ion Many elements have multiple oxidation numbers It depends on the types of compounds they form Many elements have multiple oxidation numbers It depends on the types of compounds they form N 2 N 2 O NO N 2 O 3 NO 2 N 2 O 5 NO 3 N oxidation number Note: The oxidation number of oxygen is 2 in all of these compounds Elemental copper (Cu) Cu oxidation state: 0 Copper (II) sulfate (CuSO 4 ) 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 Determine the oxidation number of carbon in carbon dioxide CO 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

4 Determine the oxidation number of sulfur in sulfuric acid H 2 SO (+1) = +2 4() = S + (-8) = 0 S = +6 (oxidation number for sulfur) 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 Determine the oxidation number of chromium in Cr 2 O 7 Cr 2 O 7 7() = -14 2Cr + (-14) = (the charge on the ion) Cr = +6 (oxidation number for chromium) 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 Determine the oxidation number of potassium and nitrogen in KNO 3 KNO 3 Recognize that this is an ionic compound between K + and NO 3 - The oxidation number of potassium in K + is +1 (the charge on the ion) For nitrogen: NO 3 K + NO 3 3() = -6 N + (-6) = -1 (the charge on the ion) N = +5 (oxidation number for nitrogen) Oxidation - Reduction Reactions 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 redox reactions, the oxidation numbers of the elements involved in the reaction change one substance is oxidized (loses electrons) -- oxidation number increases one substance is reduced (gains electrons) -- oxidation number decreases Zn(s) + H 2 SO 4 (aq) ZnSO 4 (aq) + H 2 (g) zinc loses electrons the oxidation number of Zn increases zinc is oxidized hydrogen gains electrons the oxidation number of H decreases hydrogen is reduced Reaction can be rewritten to emphasize electron transfer Zn + 2 H + + SO 4 Zn 2+ + SO 4 + H zinc loses electrons the oxidation number of Zn increases zinc is oxidized 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

5 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 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 Zn(s) + H 2 SO 4 (aq) ZnSO 4 (aq) + H 2 (g) 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) Is the following a redox reaction? Is the following a redox reaction? Neutralization reaction between hydrochloric acid and potassium hydroxide Thermite reaction HCl (aq) + KOH (aq) H 2 O (l) + KCl (aq) 2 Al (s) + Fe2O3 (s) Al2O3 (l) + 2 Fe (l) Element before reaction after reaction H O 2 2 K Cl 1 1 Element before reaction after reaction Al 0 +3 Fe +3 0 O 2 2 Which element is oxidized? Which element is reduced? 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 When balancing redox equations, there is an additional requirement the total gains and losses of electrons must balance each other 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 Thermite reaction 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 The increase in oxidation number for the oxidized substance must be equal to the decrease in oxidation number for the reduced substance 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

6 Balancing 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 2 Fe +2 (aq) + Sn +4 (aq) 2Fe +3 (aq) + Sn +2 (aq) The numbers of atoms of each element are the same on each side But the total ionic charge is not balanced +6 on the left +5 on the right Electron gains/losses are also not balanced 1 electron lost from Fe (oxidation) 2 electrons gained by Sn (reduction) Cr2O7 (aq) + Fe +2 (aq) Cr +3 (aq) + Fe +3 (aq) Step 1: Identify atoms undergoing oxidation and reduction Fe undergoes oxidation ( oxidation number: +1) Cr undergoes reduction ( oxidation number: -3) +1 x 1 Fe atom = Cr2O7 (aq) + Fe +2 (aq) 2 Cr +3 (aq) + Fe +3 (aq) Cr2O7 (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 +1 x 16 Fe atoms = = +1+6 Total charge on left side: + +2(6) = +10 Cr2O7 (aq) + 6Fe +2 (aq) 2 Cr +3 (aq) + 6Fe +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 Cr2O7 Cr2O7 (aq) + 6 (aq) Fe +2 (aq) + 6Fe (aq) H + (aq) 2 Cr +3 (aq) Fe (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

7 Cr2O7 (aq) + 6 Fe +2 (aq) + 14 H + (aq) 2 Cr +3 (aq) + 6 Fe +3 (aq) + 7 H2O(l) Note: For redox reactions taking place in basic solutions, replace Step 5a with Step 5b. Step 5b. Basic solutions: Use H2O and OH to balance the equation Step 6: Check your final equation to verify charge and material balance Material Balance It is best to add OH ions first to obtain charge balance, and then add H2O to the equation to obtain material balance Element Left Right Cr 2 2 O 7 7 Fe 6 6 H Charge Balance Net Charge on Left = () + 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.

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