Chapter 12 Oxidation-Reduction and Electrochemistry

Chapter 12: Oxidation-Reduction and Electrochemistry - The branch of chemistry that deals w/ electricity-related applications of oxidation-reduction is called electrochemistry. Oxidation-reduction reactions involve a transfer of electrons from the substance oxidized to the substance reduced. Voltaic Cells (Galvanic Cells) The substance that is oxidized during the reaction is separated from the substance that is reduced during the reaction. The electron transfer is accompanied by a transfer of electrical energy instead of heat One means separating oxidation and reduction half-reactions is w/ a porous barrier, which prevents the metal atoms of one-half reaction from mixing w/ the ions of the other half-reaction. - Ions can move through the porous barrier. Electrons can be transferred from one side to the other through an external connecting wire. Electric current moves in a closed loop path so this movement of electrons through the wire is balanced by the movement of ions in the solution. - The zinc strip is in an aqueous solution of 𝑍𝑛𝑆𝑂4 ; the copper strip, in an aqueous solution of 𝐶𝑢𝑆𝑂4 . - An electrode is a conductor used to establish electrical contact w/ a nonmetallic part of the circuit. - A single electrode immersed in a solution of its ions is a half-cell. - The zinc strp in aqueous 𝑍𝑛𝑆𝑂4is an anode, the electrode where oxidation takes place. - The copper strip in 𝐶𝑢𝑆𝑂4 is a cathode, the electrode where reduction takes place. - The copper half-cell can be written as 𝐶𝑢2+ /𝐶𝑢, and the zinc half-cell as 𝑍𝑛2+ /𝑍𝑛. - The two half-cells together constitute an electrochemical cell. An electrochemical cell is a system of electrodes and electrolytes in which either chemical reactions produce electrical energy or an electric current produces chemical change. Half-reactions: 𝐶𝑢2+ (𝑎𝑞) + 2𝑒 − → 𝐶𝑢(𝑠) 𝑍𝑛(𝑠) → 𝑍𝑛2+ (𝑎𝑞) + 2𝑒 − 2+ −2 Oxidized Half-reaction: 𝐶𝑢(𝑠) → 𝐶𝑢 (𝑎𝑞) + 2𝑒 Reduction Half-reaction: 𝑍𝑛2+ (𝑎𝑞) + 2𝑒 − → 𝑍𝑛(𝑠) - The gain of electrons is reduction and the loss of electrons is oxidation. - “LEO the lion says GER”: “Loss of Electrons is Oxidation” and “Gain of Electrons is Reduction” - The fact that the zinc in this reaction is oxidized by giving up electrons makes it possible for electrons to be gained by the copper, which is acting as a reducing agent. - Because the copper in this reaction is being reduced by gaining electrons, electrons can be lost by the zinc, which is acting as an oxidizing agent. Electrolytic Cells - The redox reactions do not occur spontaneously but can be forced to take place by supplying energy w/ an external current, these are called electrolytic reactions. - In electroplating, where electrolysis is used to coat a material w/ a layer of metal, the object to be plated is made the cathode in the reaction. BALANCING REDOX EQUATIONS The Rules for Assigning an Oxidation State - NOTE: The sum of the oxidation states must equal zero for an electrically neutral compound. For an ion, the sum of the oxidation states must equal the charge on theion, including polyatomic ions. 1. The oxidation state of an atom in an element is zero. 2. The oxidation state of a monatomic ion is the same as its charge. 3. The oxidation state of fluorine is -1 in it's compounds. 4. The oxidation state of oxygen is usually -2 and its compounds. a. One exception to this rule occurs when oxygen is bonded to flooring in the oxidation state of chlorine takes precedence. b. Another exception occurs in peroxide compounds, where the oxidation state is assigned the value of -1. 5. The oxidation state of hydrogen in most compounds is +1. a. In hydrides, there is an exception. Oxygen is assigned the value of -1. The Electron Shift Method - The oxidation state method breaks the reaction into two half reactions, the reduction reaction and the oxidation reaction. Ex: Balance the equation 𝐻𝐶𝑙(𝑎𝑞) + 𝑀𝑛𝑂2 (𝑠) → 𝐻2 𝑂(𝑙) + 𝑀𝑛𝐶𝑙2 (𝑎𝑞) + 𝐶𝑙2 (𝑔) 1. The oxidation states are written above all the elements in the equation . +1 -1 +4 -2 +1 -2 +2 -1 0 𝐻𝐶𝑙(𝑎𝑞) + 𝑀𝑛𝑂2 (𝑠) → 𝐻2 𝑂(𝑙) + 𝑀𝑛𝐶𝑙2 (𝑎𝑞) + 𝐶𝑙2 (𝑔) 2. The oxidation state of the manganese has changed from +4 to +2, while the oxidation state of the chlorine that emerges as a gas has changed from -1 to 0. A change from +4 to +2 is a reduction in the oxidation state and a game in electrons. These are written as the half reactions of the ions involved with the electron changes. NOTE: oxidation states have a + or - on the left of the number and that ionic charges are shown on the right of the number. 𝑀𝑛4+ (𝑎𝑞) + 2𝑒 − → 𝑀𝑛2+ (𝑎𝑞) Reduction reaction 1− − 𝐶𝑙 (𝑎𝑞) → 𝐶𝑙(𝑔) + 1𝑒 Oxidation reaction 3. The next step is to assure that the number of atoms required by the formulas in the reactants equals the number of atoms in the products. Notice that the product𝐶𝑙2 in the original equation is a diatomic molecule and requires 2 atoms in the molecular formula. This means that there must be 2 𝐶𝑙 − ions provided on the left side of the equation. A total of 2𝑒 − are lost, one for each 𝐶𝑙 − . 𝑀𝑛4+ (𝑎𝑞) + 2𝑒 − → 𝑀𝑛2+ (𝑎𝑞) Reduction reaction 2𝐶𝑙 − (𝑎𝑞) → 𝐶𝑙2 (𝑔) + 2𝑒 − Oxidation reaction 4. If the number of electrons gained in reduction did not equal the number electrons lost in oxidation, we would have to multiply either one reaction line or both by an appropriate number to ensure that the number of electrons lost equaled the number of electrons gained. 5. Next add the two half reactions. 𝑀𝑛4+ (𝑎𝑞) + 2𝐶𝑙 − (𝑎𝑞) → 𝑀𝑛2+ + 𝐶𝑙2 (𝑔) 6. Now go back to the original equation and insert the coefficients. 2𝐻𝐶𝑙(𝑎𝑞) + 1𝑀𝑛𝑂2 (𝑠) →? 𝐻2 𝑂(𝑙) + 1𝑀𝑛𝐶𝑙2 (𝑎𝑞) + 𝐶𝑙2 (𝑔) 7. From the numbers thus established, the remaining coefficients can be deduced by infection. Notice that 2 more molecules of HCl are required to furnish the chlorine for the 𝑀𝑛𝐶𝑙2 , and the 2 atoms of oxygen in 𝑀𝑛𝑂2 form 2𝐻2 𝑂. 8. The final balanced equation is: 4𝐻𝐶𝑙(𝑎𝑞) + 𝑀𝑛𝑂2 (𝑠) → 2𝐻2 𝑂(𝑙) + 𝑀𝑛𝐶𝑙2 (𝑎𝑞) + 𝐶𝑙2 (𝑔)