Oxidation Numbers Rules for Assigning Oxidation States The oxidation state of an atom in an uncombined element is 0. The oxidation state of a monatomic ion is the same as its charge. Oxygen is assigned an oxidation state of –2 in most of its covalent compounds. Important exception: peroxides (compounds containing the O2 2- group), in which each oxygen is assigned an oxidation state of –1) In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of +1 For a compound, sum total of ON s is zero. For an ionic species (like a polyatomic ion), the sum of the oxidation states must equal the overall charge on that ion.

Redox: Reduction occurs when an atom gains one or more electrons. Ex:   Oxidation occurs when an atom or ion loses one or more electrons. LEO goes GER Copper metal reacts with silver nitrate to form silver metal and copper nitrate: Cu + 2 Ag(NO3)  2 Ag + Cu(NO3)2.

Identifying OX, RD, SI Species Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Oxidation = loss of electrons. The species becomes more positive in charge. For example, Ca0  Ca+2, so Ca0 is the species that is oxidized. Reduction = gain of electrons. The species becomes more negative in charge. For example, H+1  H0, so the H+1 is the species that is reduced. Spectator Ion = no change in charge. The species does not gain or lose any electrons. For example, Cl-1  Cl-1, so the Cl-1 is the spectator ion.

Oxidizing Agent and Reducing Agent: Oxidizing agent gets reduced itself and reducing agent gets oxidized itself, so a strong oxidizing agent should have a great tendency to accept e and a strong reducing agent should be willing to lose e easily. What are strong oxidizing agents- metals or non metals? Why? Which is the strongest oxidizing agent and which is the strongest reducing agent?

Agents Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Since Ca0 is being oxidized and H+1 is being reduced, the electrons must be going from the Ca0 to the H+1. Since Ca0 would not lose electrons (be oxidized) if H+1 weren’t there to gain them, H+1 is the cause, or agent, of Ca0’s oxidation. H+1 is the oxidizing agent. Since H+1 would not gain electrons (be reduced) if Ca0 weren’t there to lose them, Ca0 is the cause, or agent, of H+1’s reduction. Ca0 is the reducing agent.

Steps for Balancing a Redox Reaction: Half Reaction Method In half reaction method, oxidation and reduction half- reactions are written and balanced separately before combining them into a balanced redox reaction. It is a good method for balancing redox reactions because this method can be used both for reactions carried out in acidic and basic medium .

Steps for Balancing Redox Reaction Using Half Reaction Method IN ACIDIC MEDIUM: Step 1: Write unbalanced equation in ionic form. Step 2: Write separate half reactions for the oxidation and reduction processes. (Use Oxidation Numbers for identifying oxidation and reduction reactions) Step 3: Balance atoms in the half reactions First, balance all atoms except H and O Balance O by adding H2O Balance H by adding H+ Step 4: Balance Charges on each half reaction, by adding electrons. Step 5: Multiply each half reaction by an appropriate number to make the number of electrons equal in both half reactions. Step 6: Add two half reactions and simplify where possible by canceling species appearing in both sides. Step 7: Check equation for same number of atoms and charges on both sides.

Writing Half-Reactions Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Oxidation: Ca0  Ca+2 + 2e- Reduction: 2H+1 + 2e-  H20 The two electrons lost by Ca0 are gained by the two H+1 (each H+1 picks up an electron). PRACTICE SOME!

Practice Half-Reactions Don’t forget to determine the charge of each species first! 4 Li + O2  2 Li2O Oxidation Half-Reaction: Reduction Half-Reaction: Zn + Na2SO4  ZnSO4 + 2 Na

Steps for Balancing Redox Reaction Using Half Reaction Method IN BASIC MEDIUM: For balancing redox reactions in basic solutions, all the steps are the same as acidic medium balancing, except you add one more step to it. The H+ ions can then be “neutralized” by adding an equal number of OH- ions to both sides of the equation. Ex.

Standard Cell Potential Just as the water tends to flow from a higher level to a lower level, electrons also move from a higher “potential” to a lower potential. This potential difference is called the electromotive force (EMF) of cell and is written as Ecell. The standard for measuring the cell potentials is called a SHE (Standard Hydrogen Electrode). Description of SHE (Standard Hydrogen Electrode) Reaction 2H+(aq, 1M)+ 2e - H2(g, 101kPa) E0= 0.00 V

Standard Reduction Potentials Many different half cells can be paired with the SHE and the standard reduction potentials for each half cell is obtained. Check the table for values of reduction potential for various substances: Would substances with high reduction potential be strong oxidizing agents or strong reducing agents? Why?

Activity Series For metals, the higher up the chart the element is, the more likely it is to be oxidized. This is because metals like to lose electrons, and the more active a metallic element is, the more easily it can lose them. For nonmetals, the higher up the chart the element is, the more likely it is to be reduced. This is because nonmetals like to gain electrons, and the more active a nonmetallic element is, the more easily it can gain them.

Metal Activity 3 K0 + Fe+3Cl-13 REACTION Metallic elements start out with a charge of ZERO, so they can only be oxidized to form (+) ions. The higher of two metals MUST undergo oxidation in the reaction, or no reaction will happen. The reaction 3 K + FeCl3  3 KCl + Fe WILL happen, because K is being oxidized, and that is what Table J says should happen. The reaction Fe + 3 KCl  FeCl3 + 3 K will NOT happen. Fe0 + 3 K+1Cl-1 NO REACTION

Voltaic Cells (Galvanic Cells) A voltaic cell converts chemical energy from a spontaneous redox reaction into electrical energy. Ex: Cu and Zn voltaic cell (More positive reduction potential is the cathode) Key Words: Cathode Anode Salt Bridge How a Voltaic Cell Works: An Ox, Red Cat Representing Electrochemical Cells

Voltaic Cells Produce electrical current using a spontaneous redox reaction Used to make batteries! Materials needed: two beakers, piece of the metals (anode, - electrode and cathode + electrode), solution of each metal, porous material (salt bridge), solution of a salt that does not contain either metal in the reaction, wire and a load to make use of the generated current! Use Reference Table J to determine the metals to use Higher = (-) anode (lower reduction potential) Lower = (+) cathode (higher reduction potential)

Making Voltaic Cells

Electrolytic Cells Use electricity to force a nonspontaneous redox reaction to take place. Uses for Electrolytic Cells: Decomposition of Alkali Metal Compounds Decomposition of Water into Hydrogen and Oxygen Electroplating Differences between Voltaic and Electrolytic Cells: ANODE: Voltaic (-) Electrolytic (+) CATHODE: Voltaic (+) Electrolytic (-) Voltaic: 2 half-cells, a salt bridge and a load Electrolytic: 1 cell, no salt bridge, IS the load

Decomposing Alkali Metal Compounds 2 NaCl  2 Na + Cl2 The Na+1 is reduced at the (-) cathode, picking up an e- from the battery The Cl-1 is oxidized at the (+) anode, the e- being pulled off by the battery (DC)

Decomposing Water 2 H2O  2 H2 + O2 The H+ is reduced at the (-) cathode, yielding H2 (g), which is trapped in the tube. The O-2 is oxidized at the (+) anode, yielding O2 (g), which is trapped in the tube.

Electroplating The Ag0 is oxidized to Ag+1 when the (+) end of the battery strips its electrons off. The Ag+1 migrates through the solution towards the (-) charged cathode (ring), where it picks up an electron from the battery and forms Ag0, which coats on to the ring.

Spontaneity of Redox Reactions: E0 = E0red( reduction process-cathode) – E0red (oxidation process-anode) A positive value of E0 indicates a spontaneous process and a negative value of E0 indicates a nonspontaneous value. Steps for Predicting Spontaneity of Redox Reactions First write the reaction as oxidation and reduction half reactions. Then plug standard reduction potential values in the equation given above. Check for the spontaneity by a positive or a negative value of E0 Ex: