Mark S. Cracolice Edward I. Peters Mark S. Cracolice The University of Montana Chapter 19 Oxidation–Reduction (Redox) Reactions

Electron Transfer Reactions The reaction between zinc and copper(II) ions is spontaneous.

Electron Transfer Reactions Zinc is said to be oxidized to zinc ion and copper(II) ions is said to be reduced to copper.

Electron Transfer Reactions The same reaction between zinc and copper(II) ions can be performed using the following arrangement.

Electron Transfer Reactions When the two metallic rods are connected, zinc is oxidized to zinc ions and releases electrons at the zinc electrode and copper (II) ions collect electrons from the copper electrode to become copper.

Electron Transfer Reactions The overall oxidation-reduction reaction can be separated into two half reactions. Zn (s)  Zn 2+ (aq) + 2e - ( at the zinc rod) Cu 2+ (aq) + 2e -  Cu (s) ( at the copper rod) _________________________________________ Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s)

Electron Transfer Reactions The process of loosing electrons is called oxidation. The chemical change that occurs at the zinc electrode is oxidation, zinc is oxidized to Zn 2+. Zn(s)  Zn 2+ (aq) + 2 e – The process of gaining electrons is called reduction. Copper (II) cations are reduced to Cu at the copper rod. Cu 2+ (aq) + 2 e –  Cu(s)

Electron Transfer Reactions If half-reaction equations are added algebraically, the result is the net ionic equation for the oxidation–reduction reaction: Zn(s) Zn 2+ (aq) + 2 e – Cu 2+ (aq) + 2 e – Cu(s) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) The chemical change is an electron transfer reaction; electrons have been transferred from zinc atoms to copper(II) ions.

Voltaic (Galvanic) Cells Voltaic (Galvanic) Cell A cell in which a spontaneous chemical reaction is used to produce electricity.

Voltaic (Galvanic) Cells Anode: Electrode at which oxidation occurs. The species being oxidized releases electrons into the electrode. Cathode: Electrode at which reduction occurs. The species being reduced collect electrons from the electrode

Commercial Galvanic Cells Alkaline battery One electrode is zinc in contact with a gel of potassium hydroxide. One electrode is graphite in contact with a mixture of manganese (IV) oxide, graphite and potassium hydroxide. Reaction at the anode: Zn (s) + 2 OH - (aq)  ZnO (s) + H 2 O (l) + 2e Reaction at the cathode: 2 MnO 2 (s) + H 2 O (l) + 2e  Mn 2 O 3 (s) + 2 OH -

Commercial Galvanic Cells Lead battery One electrode is lead in contact with a sulfuric acid solution One electrode is lead in contact with lead (IV) oxide and sulfuric acid Reaction at the anode: Pb (s) + SO 4 2- (aq)  PbSO 4 (s) + 2e Reaction at the cathode: PbO 2 (s) + 4 H + + SO 4 2- (aq) + 2e  PbSO 4 (s) + 2 H 2 O Lead battery is rechargeable.

Electrolytic Cells Electrolytic Cell In an electrolytic cell, current supplied by an external source is used to drive a non-spontaneous redox reaction Components An electrolyte and two electrodes. When the electrodes are connected to an outside source of electricity, they become charged, one positively and one negatively. The ions in the electrolyte move to the oppositely charged electrode, where chemical reactions occur. The movement of ions is an electric current in solution.

Electrolytic Cells Electrolysis of liquid sodium chloride. Two electrodes are connected to an external electric source and dipped in liquid sodium chloride.

Electrolytic Cells At Negative Electrode: sodium ions Na + are attracted to the negative electrode, accept electrons from the electrode and are reduced to sodium. The reaction is reduction so the electrode is called cathode. Na + + e -  Na At Positive Electrode: Chloride ions Cl - are attracted to the positive electrode, where they are oxidized to chlorine gas. The reaction is oxidation, so the electrode is an anode. 2 Cl e -  Cl 2

Oxidation Numbers & Redox Reactions Electron-transfer reaction is called redox reaction. The reactant that gains electrons is called the oxidizing agent. The reactant that loses electrons is called the reducing agent. In the reaction, the oxidizing agent is reduced, and the reducing agent oxidized. To determine which species has gained and which has lost electrons, it is useful to determine the oxidation states, or oxidation numbers of reactants and products.

Rules for Determination of Oxidation Numbers The oxidation number of any elemental substance is 0 The oxidation number of a monatomic ion is the same as the charge on the ion The oxidation number of combined oxygen is -2, except in peroxide(-1), superoxide(-1/2), and OF 2 (+2) The oxidation number of combined hydrogen is +1(except hydride ion, H - ). In any molecular or ionic species the sum of the oxidation numbers of all atoms in a molecule or polyatomic ion is equal to the charge on the species. Oxidation is an increase in oxidation number. Reduction is a decrease in oxidation number.

Oxidizing and Reducing Agents Oxidizing Agent (Oxidizer) Species that removes electrons in a redox reaction (the species that is reduced). Reducing Agent (Reducer) Species from which electrons are removed (the species that is oxidized).

Strengths of Ox and Red Agents Strong oxidizing agent A species with a strong attraction for electrons. Weak oxidizing agent A species that attracts electrons only slightly. Strong reducing agent A species that releases electrons readily. Weak reducing agent A species that does not easily give up its electrons.

Strengths of Oxidizing and Reducing Agents

Predicting Redox Reactions

Writing Redox Equations Writing Redox Equations for Half-Reactions in Acidic Solutions 1. After identifying the element oxidized or reduced, write a skeleton half-reaction equation with the element in its original form (element, monatomic ion, or part of a polyatomic ion or compound) on the left and its final form on the right. 2. Balance the element oxidized or reduced. 3. Balance elements other than hydrogen or oxygen, if any.

Writing Redox Equations Writing Redox Equations for Half-Reactions in Acidic Solutions 4. Balance oxygen by adding water molecules where necessary. 5. Balance hydrogen by adding H + where necessary. 6. Balance charge by adding electrons to the more positive side. 7. Recheck the equation to be sure it is balanced in both number of atoms and charge.

Writing Redox Equations Example: Iron(II) ion solution and a permanganate ion (MnO 4 – ) solution are combined in an acidic solution. Manganese(II) ion and iron(III) ion are produced. Write a balanced net ionic equation for the reaction. Solution: First, we summarize the reaction description: Reactants: Fe 2+ (aq) and MnO 4 – (aq) The reaction is in acidic solution, so H + (aq) is also a reactant Products: Fe 3+ (aq) and Mn 2+ (aq)

Writing Redox Equations Reactants: Fe 2+ (aq), MnO 4 – (aq), H + (aq) Products: Fe 3+ (aq), Mn 2+ (aq) Solution: Step 1: Write skeleton half-reaction equations Fe 2+ (aq) Fe 3+ (aq) MnO 4 – (aq) Mn 2+ (aq)

Writing Redox Equations Reactants: Fe 2+ (aq), MnO 4 – (aq), H + (aq) Products: Fe 3+ (aq), Mn 2+ (aq) Solution: Step 2: Balance element oxidized/reduced Fe 2+ (aq) Fe 3+ (aq)MnO 4 – (aq) Mn 2+ (aq) Both Fe and Mn are already balanced

Writing Redox Equations Reactants: Fe 2+ (aq), MnO 4 – (aq), H + (aq) Products: Fe 3+ (aq), Mn 2+ (aq) Solution: Step 3: Balance elements other than H, O Fe 2+ (aq) Fe 3+ (aq)MnO 4 – (aq) Mn 2+ (aq) There are none

Writing Redox Equations Reactants: Fe 2+ (aq), MnO 4 – (aq), H + (aq) Products: Fe 3+ (aq), Mn 2+ (aq) Solution: Step 4: Balance O by adding H 2 O Fe 2+ (aq) Fe 3+ (aq)MnO 4 – (aq) Mn 2+ (aq) + 4 H 2 O(l)

Writing Redox Equations Reactants: Fe 2+ (aq), MnO 4 – (aq), H + (aq) Products: Fe 3+ (aq), Mn 2+ (aq) Solution: Step 5: Balance H by adding H + Fe 2+ (aq) Fe 3+ (aq) 8 H + + MnO 4 – (aq) Mn 2+ (aq) + 4 H 2 O(l)

Writing Redox Equations Reactants: Fe 2+ (aq), MnO 4 – (aq), H + (aq) Products: Fe 3+ (aq), Mn 2+ (aq) Solution: Step 6: Balance charge by adding electrons Fe 2+ (aq) Fe 3+ (aq) + e – 5 e – + 8 H + + MnO 4 – (aq) Mn 2+ (aq) + 4 H 2 O(l)

Writing Redox Equations Reactants: Fe 2+ (aq), MnO 4 – (aq), H + (aq) Products: Fe 3+ (aq), Mn 2+ (aq) Solution: Step 7: Check for charge and atom balance Fe 2+ (aq) Fe 3+ (aq) + e – 2+ each side, 1 Fe each side 5 e – + 8 H + + MnO 4 – (aq) Mn 2+ (aq) + 4 H 2 O(l) 2+ each side, 8 H, 1 Mn, 4 O each side

Writing Redox Equations Reactants: Fe 2+ (aq), MnO 4 – (aq), H + (aq) Products: Fe 3+ (aq), Mn 2+ (aq) Solution: Add the half reactions to yield the complete net ionic equation 5 × [Fe 2+ (aq) Fe 3+ (aq) + e – ] 5 e – + 8 H + + MnO 4 – (aq) Mn 2+ (aq) + 4 H 2 O(l) 5 Fe 2+ (aq) + 8 H + + MnO 4 – (aq) 5 Fe 3+ (aq) + Mn 2+ (aq) + 4 H 2 O(l)

Homework Homework: 3, 9, 11, 17, 19, 21, 23,25, 29, 49, 59 Write net equations for all reactions of the Experiment 13