Another substance, called a reducing agent, causes or promotes the reduction of a metal compound to an elemental compound.

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1 Electrochemistry Oxidation and Reducation The technology of metalurrgy has allowed humanity to progress from the Stone Age, through the Bronze Age and the Iron Age to modern times. Very few metals exist as pure elements, most exist in a variety of compounds mixed with other substances in rocks called ores. Although the technological process of refining vary from one metal to another, the processes of refining involve a large volume of ore, that is reduced to a smaller volume of metal. The term reduction came to be associated with producing metals from their compounds. Fe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g) SnO2(s) + C(s) --> Sn(s) + CO2(g) CuS(s) + H2(g) --> Cu(s) + H2S(g) Another substance, called a reducing agent, causes or promotes the reduction of a metal compound to an elemental compound. Even before metals, humans discovered fire. Fire has been a crucial piece of technology for the development of many cultures. But only recently (18th century) have we come to realize the role of oxygen in burning. Understanding the connection of corrosion (rusting, tarnishing, etc) and burning is an even more recent development. Reactions of substances with oxygen, whether they were the explosive combustion of gunpowder, the burning of wood, or the slow corrosion of iron came to be called oxidation. It soon became apparent that oxygen was not the only substance that could cause reactions with characteristics of oxidation. 2Mg(s) + O2(g) --> 2MgO(s) 2Al(s) + 3Cl2(g) --> 2AlCl3(s) Cu(s) + Br2(g) --> CuBr2(s) A substance that causes or promotes oxidation is called an oxidizing agent. - A piece of zinc is placed in an aqueous solution of copper(ii) sulfate. Zn(s) + CuSO4(aq) --> Cu(s) + ZnSO4(aq) single displacement Zn(s) + Cu 2+ (aq) + SO4 2- (aq) --> Cu(s) + Zn 2+ (aq) + SO4 2- (aq)

2 The sulfate ions are spectating. If we omit the spectator ions, we can obtain the net ionic equation. Zn(s) + Cu 2+ (aq) --> Cu(s) + Zn 2+ (aq) Notice what happens to the reactants. The zinc atoms lose electrons to form zinc ions. The copper ions gain electrons to form copper atoms. Oxidation is the loss of electrons. Reduction is the gain of electrons. Oxidation and reduction are two halves of a reaction and one cannot occur without the other. These are known as oxidation-reduction reactions or redox reactions. The number of electrons lost by one substance must equal the number of electrons gained by the other. In order to track electron transfer, oxidation number are used. For the above reaction: Zinc is oxidized, it donates electrons and is also the reducing agent. Copper ion is reduced, accepts electrons and is the oxidizing agent. OIL RIG - Oxidation is Loss. Reduction is Gain. Are Not Redox Double displacement reactions are not redox reactions because there is not change in electronic configuration. BaCl2 + Na2SO4 --> BaSO4 + 2NaCl Precipitation are non-redox as well. Acid - Base reactions are also non-redox.

3 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 the oxidation number of hydrogen is The oxidation number of oxygen in its compounds is usually -2, but there are exceptions. These include peroxides, such as H2O2, and the compound OF2. 5. In covalent compounds (two nonmetals) that do not contain oxygen or hydrogen, the oxidation number of the less electronegative element is positive and the more electronegative element 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 ion must equal the charge on the ion. Notice that during a redox reaction if the oxidation number increases there was a loss of electrons and if the oxidation number decreases there was a gain of electrons.

4 The Half-Reaction Method for Balancing Equations There are specific techniques for balancing redox equations. One method is especially useful for reactions that take place under acidic or basic conditions. Half-Reactions A half-reaction is a balanced equation that shows the number of electrons involved in either oxidation or reduction. Because a redox reaction involves both oxidationi and reduction, two half-reactions are needed to represent a redox reaction. One half-reaction shows oxidation and the other half-reaction shows reduction. Zn(s) + Cu 2+ (aq) --> Cu(s) + Zn 2+ (aq) As already given above You can write an oxidation half-reaction to show this change. Zn(s) --> Zn 2+ (aq) + 2e - You can write a reduction half-reaction to show this change. Cu 2+ (aq) + 2e - --> Cu(s) Steps for Balancing by the Half-Reaction Method (Given in Table) 1. Write the unbalanced net ionic equation, if it is not already given. 2. Divide the unbalanced net ionic equation into an oxidation half-reaction and a reduction half-reaction. To do this, you may need to assign oxidation to all of the elements in the net ionic equation 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 (LCM) of the numbers of electrons in the oxidation half-reaction and the reduction half-reaction. 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 the 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. 10. If necessary, include states.

5 Balance the following redox reaction that results when Mg(s) displaces aluminum from aqueous aluminum nitrate, Al(NO3)3(aq). Steps in the Half-Reaction Method for Balancing Equations For Redox Reactions Occurring in Acid Solution 1. Write separate equations for the oxidation and reduction half-reactions. 2. For each half-reaction: a) Balance all the elements except hydrogen and oxygen. b) Balance oxygen using H2O. c) Balance hydrogen using H +. d) Balance the charge using electrons. 3. If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions. 4. Add the half-reactions and cancel identical species. 5. Check that all the elements and charges are balanced. Balance the equation for the reaction between permanganate and iron (II) ions in acidic solution, Acidic MnO4 - (aq) + Fe 2+ (aq) > Fe 3+ (aq) + Mn 2+ (aq) This reaction can be used to analyze iron ore for its iron content. Potassium dichromate (K2Cr2O7) is a bright orange compound that can be reduced to a blue-violet solution of Cr 3+ ions. Under acidic conditions, K2Cr2O7 reacts with ethyl alcohol (C2H5OH) as follows: H + (aq) + Cr2O7 2- (aq) + C2H5OH(l) --> Cr 3+ (aq) + CO2(g) + H2O(l) Balance this equation using the half-reaction method.

6 The Half-Reaction Method for Balancing Equations for Redox Reactions Occurring in Basic Solutions 1. Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H + ions were present. 2. To both sides of the equation obtained above, add a number of OH - ions that is equal to the number of H + ions. (We want to eliminate H + by forming H2O.) 3. Form H2O on the side containing both H + and OH - ions, and eliminate these number of H2O molecules that appear on both sides of the equation. 4. Check that elements and charges are balanced. Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores. An aqueous solution containing cyanide ion is often used to extract the silver using the following reaction that occurs in basic solution: Basic Ag(s) + CN - (aq) + O2(g) > Ag(CN)2 - (aq) Balance this equation using the half-reaction method.

7 The Oxidation Number Method for Balancing Equations When balancing equations by the half-reaction method in the previous section, you sometimes used oxidation numbers to determine the reactants and products in each half-reaction. You can use oxidation numbers to balance a chemical equation by a new method. 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. 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 undergo an increase in oxidation number and the element that undergo a decrease in oxidation number. 4. Find the numerical values of the increase and the decrease in oxidation numbers. 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.

CHM 101 GENERAL CHEMISTRY FALL QUARTER 2008

CHM 101 GENERAL CHEMISTRY FALL QUARTER 2008 Section 2 Lecture Notes 10/29/2008 (last revised: 10/29/08, 2:00 PM) 4.9 Oxidation Reduction Reactions Introduction: Your text uses the reaction between solid