Electrochemistry Reduction-Oxidation

Oxidation Historically means “to combine with oxygen” Reactions of substances with oxygen, ie Combustion, Rusting (corrosion)

Oxidation Substances other than oxygen could cause reactions with characteristics of oxidation, so we say: Oxidation is the loss of electrons A reactant that oxidizes another reactant is called an oxidizing agent. Example: Zn (s) + Cu 2+ (aq) → Cu(s) + Zn 2+ (aq)

Reduction Reduction is the gain of electrons. A reactant that reduces another reactant is called a reducing agent. Example: Zn (s) + Cu 2+ (aq)  Cu(s) + Zn 2+ (aq)

Reduction-Oxidation Reaction Reduction-Oxidation reactions, or redox reactions, are reactions in which both oxidation and reduction occur.

3Fe(s) + Al 2 (SO 4 ) 3 (aq)  2Al(s) + 3FeSO 4 (aq) Write the total and net ionic equation for the reaction. Try to determine which species undergo oxidation/reduction and which are oxidizing/reducing agents.

Mnemonic Device OIL RIG Oxidation is loss Reduction is gain portal.com/academy/lesson/balancing-redox- reactions-and-identifying-oxidizing-and-reducing- agents.html#lesson

Write a balanced net ionic equation for calcium with aqueous aluminum sulfate. Identify the reactant oxidized and reduced as well as the reducing and oxidizing agent.

Practice P715 #

Half-Reactions To monitor the transfer of electrons in a redox reaction, you can represent the oxidation and reduction separately, these reactions are called half- reactions. Example: Zn(s) + Cu 2+ (aq)  Cu(s) + Zn 2+ (aq) Oxidation half-reaction: Zn(s)  Zn 2+ (aq) + 2e - Reduction half-reaction: Cu 2+ (aq) + 2e -  Cu(s)

Disproportionation A disproportionation reaction is a reaction in which a single element undergoes both oxidation and reduction in the same reaction. Example: 2Cu + (aq)  Cu(s) + Cu 2+ (aq)

Practice p 716 #5-8

Oxidation Numbers Oxidation numbers are actual or hypothetical charges, assigned using a set of rules, used to describe redox reactions with covalent compounds. They are also used to identify redox reactions, and to identify oxidizing and reducing agents.

Oxidation Numbers of an Element Some elements (transition metals) have more than one possible charge. Vanadium can form V 2+, V 3+,V 4+, V 5+. Oxidation numbers help to determine the oxidation state. Ie. Vanadium is oxidized by removing 2 e - s, it has an oxidation number of +2. V  V e - If this ion is oxidized again, it has an oxidation number of +3. V 2+  V 3+ + e - Note: every time you oxidize vanadium by removing another electron, its oxidation number increases by 1.

Oxidation Numbers of an Element If you add electrons to an element, the oxidation number will drop. Example: If you add 2 electrons to sulfur, the oxidation number is -2. S + 2e -  S 2- Rules have been established for determining oxidation numbers of elements within compounds.

Rules for Assigning Oxidation Numbers 1. A pure element has an oxidation number of The oxidation number of a monoatomic ion is equal to its charge. 3. The oxidation number of hydrogen in its compounds is +1, except in metal hydrides, where it is The oxidation number of oxygen in its compounds is usually -2. Exceptions include peroxides and OF2. 5. In covalent compounds that do not contain oxygen or hydrogen, the oxidation number of the less electronegative element is positive and the more electronegative element is negative. 6. The algebraic sum of the oxidation numbers in the formula of a compound must equal zero. 7. The algebraic sum of the oxidation numbers in the formula of a polyatomic ions must equal the charge of the ion.

Determine the oxidation number of each atom, in each of the following formulas. a) Cl 2 b) CH 4 c) NaCl d) OF 2

Using Oxidation Numbers During a redox reaction, if the oxidation number increases, there was a loss of electrons. If the oxidation number decreases, there was a gain of electrons. Determining the oxidation number of each atom in an equation can help to identify the species being oxidized/reduced and the oxidizing/reducing agent.

Use oxidation numbers to determine whether or not the reaction is a redox reaction and to identify all species. CH 4(g) + Cl 2(g)  CH 3 Cl (g) + HCl (g)

Use oxidation numbers to determine whether or not the reaction is a redox reaction and to identify all species. CaCO 3(s) + 2HCl (aq) → CaCl 2(aq) + H 2 O (l) + CO 2(g)

Practice P 726 #9-12 P 728 #13-16 Try the end of section review on page

Half-Reaction Method for Acidic Solutions 1. Step 1 Write an unbalanced half-reaction that shows the formulas of the given reactant(s) and product(s). 2. Step 2 Balance any atoms other than oxygen and hydrogen first. 3. Step 3 Balance any oxygen atoms by adding water molecules. 4. Step 4 Balance any hydrogen atoms by adding hydrogen ions. 5. Step 5 Balance the charges by adding electrons.

Write a balanced half-reaction that shows the reduction of permanganate ions, MnO 4 -, to manganese (II) ions in an acidic solution. Step 1 MnO 4 − → Mn 2+ Step 2 Balance the atoms. Here, the Mn atoms are already balanced. Step 3 Add water molecules to balance the oxygen atoms. MnO 4 − → Mn H 2 O Step 4 The reaction occurs in acidic solution, add hydrogen ions to balance the hydrogen atoms. MnO 4 − + 8H + → Mn H 2 O Step 5 The net charge on the left side is 7+, the net charge on the right side is 2+. Add five electrons to the left side to balance the charges. MnO 4 − + 8H + + 5e − → Mn H 2 O

Half-Reaction Method for Basic Solutions 1. Use the half-reaction method for acidic solutions to get the final balanced equation as if H + ions were present. 2. To both sides of the equation, add the # of OH - ions that is equal to the number of H+ ions present. 3. Combine H + and OH - ions to form H 2 O on the side containing both and eliminate the H 2 O molecules that appear on both sides of the equation. 4. Check that elements and charges are balanced.

Write a balanced half-reaction that shows the oxidation of thiosulfate ions, S 2 O 3 2−, to sulfite ions, SO 3 2−, in a basic solution. Step 1 S 2 O 3 2− → SO 3 2− Step 2 Balance the atoms, beginning with the sulfur atoms. S 2 O 3 2− → 2SO 3 2− Step 3 Balance the oxygen and hydrogen atoms as if the solution is acidic. S 2 O 3 2− + 3H 2 O → 2SO 3 2− S 2 O 3 2− + 3H 2 O → 2SO 3 2− + 6H +

Step 4 There are six hydrogen ions present, so adjust for basic conditions by adding six hydroxide ions to each side. S 2 O 3 2− + 3H 2 O + 6OH − → 2SO 3 2− + 6H + + 6OH− Step 5 Combine the hydrogen ions and hydroxide ions on the right side into water molecules. S 2 O 3 2− + 3H 2 O + 6OH − → 2SO 3 2− + 6H 2 O Step 6 Remove three water molecules from each side. S 2 O 3 2− + 6OH − → 2SO 3 2− + 3H 2 O Step 7 The net charge on the left side is 8−, the net charge on the right side is 4−. Add four electrons to the right side to balance the charges. S 2 O 3 2− + 6OH − → 2SO 3 2− + 3H 2 O + 4e−

Practice Page 732 # Page 734 # 21-24

The Half-Reaction Method for Balancing Equations The half-reaction method for balancing equations involves writing two balanced half-reactions for a given redox reaction and combining them to form one balanced net reaction. This method is especially useful for reactions that take place under acidic or basic conditions.

Half-Reaction Method 1. Write the unbalanced net ionic equation. 2. Divide the unbalanced net ionic equation into an oxidation half- reaction and a reduction half-reaction.  you may need to assign oxidation numbers to all of the elements to determine what is oxidized and what is reduced. 3. Balance the oxidation half-reaction and the reduction half- reaction independently. 4. Determine the least common multiple of the numbers of electrons in the oxidation and reduction half-reactions

Half-Reaction Method 5. Use the coefficients to write each half-reaction so that it includes the LCM of the number of electrons 6. Add the balanced half-reactions that include equal numbers of electrons. 7. Remove the electrons from both sides of the equation. 8. Remove any identical molecules or ions that are present on both sides of the equation. 9. If you require a balanced chemical equation, include any spectator ions in the chemical formulas.

Problem: Write a balanced net ionic equation to show the reaction of perchlorate ions, ClO 4 -, and nitrogen dioxide in acidic solution to produce chloride ions and nitrate ions. Step 1 ClO 4 − + NO 2 → Cl − + NO 3 − Step 2 Write two unbalanced half-reactions. Oxidation: NO 2 → NO 3 − Reduction: ClO 4 − → Cl − Step 3 Balance the two half-reactions for acidic conditions.

Step 4 The LCM of 1 and 8 is 8. Step 5 Multiply the oxidation half-reaction by 8, so that equal numbers of electrons are lost and gained. 8NO 2 + 8H 2 O → 8NO 3 − + 16H + + 8e − Step 6 Add the half reactions. 8NO 2 + 8H 2 O → 8NO 3 − + 16H + + 8e − ClO 4 − + 8H + + 8e − → Cl − + 4H 2 O________ 8NO 2 +8H 2 O +ClO 4 − +8H + +8e − →8NO H + +8e − +Cl − +4H 2 O Step 7 Simplify by removing 8 electrons from both sides. 8NO 2 +8H 2 O +ClO 4 − +8H + →8NO H + +Cl − + 4H 2 O Step 8 Simplify by removing 4 water molecules, and 8 hydrogen ions from each side. 8NO 2 + 4H 2 O + ClO 4 − → 8NO H + + Cl −

Practice Try questions on page 738 & 739.

The Oxidation Number Method for Balancing Equations The oxidation number method is a method of balancing redox equations by ensuring that the total increase in the oxidation numbers of the oxidized elements equals the total decrease in the oxidation numbers of the reduced elements.

Oxidation Number Method 1. Write an unbalanced equation, if it is not given. 2. Determine whether the reaction is a redox reaction by assigning an oxidation number to each element wherever it appears in the equation. 3. If the reaction is a redox reaction, identify the element that undergoes an increase in oxidation number and the elements the undergoes a decrease. 4. Find the numerical values of the increase and decrease. 5. Determine the smallest whole-number ratio of the oxidized and reduced elements so that the total increase in oxidation numbers equals the total decrease in oxidation numbers. 6. Use the smallest whole-number ratio to balance the numbers of atoms of the element oxidized and the element reduced. 7. Balance the other elements by inspection, if possible. 8. For reactions that occur in acidic or basic solutions, include water molecules, hydrogen ions, or hydroxide ions as needed to balance the equation.

Write a balanced net ionic equation to show the formation of iodine by bubbling oxygen gas through a basic solution that contains iodide ions.

Practice Try problems on page 749.