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Topic 9 Reduction and Oxidation 714 Version 1 2012 Definitions Learn these off by heart Anode: Where oxidation takes place. In electrolysis, it is the + electrode and anions are attracted here. Cathode: Where reduction takes place. In electrolysis, it is the – electrode and cations are attracted here. Electrolysis: Passage of electric current through an electrolyte. Amount of discharge is affected by: 1) current; 2) charge on ion, 3) duration of electrolysis. Electrolyte: A substance which does not conduct electricity when solid, but does when molten or in aqueous solution and is chemically decomposed in the process. Electrolytic cell: Used to make non-spontaneous redox reactions occur by providing energy in the form of electricity from an external source. Electroplating: A process of coating one metal with a thin layer of another metal, by electrolysis. Half cell: A metal in contact with an aqueous solution of its own ions. Oxidation: The loss of electrons Oxidizing agent: A substance that readily oxidizes other substances. Oxidizing agents are thus reduced. Reactivity: A measure of the readiness of a substance to gain or lose electrons. The stronger the reducing agent or oxidizing agent, the more reactive it is. Redox reaction: A reaction in which there is a transfer of electrons, i.e. reduction and oxidation occurring simultaneously. Reducing agent: A substance that readily reduces other substances. Reducing agents are thus oxidized. Reduction: The gain of electrons Salt bridge: Allows the free movement of ions in a voltaic cell. Paper dipped in a saturated solution of KNO3 is an example of a salt bridge. Shorthand notation: For a voltaic cell. Example: Cu(s)/Cu2+(aq) || H+(aq) / H2(g) Standard conditions: 298 K, 1 atm, 1.0 M. 715 Version 1 2012 Standard electrode potential: The electrode potential of one half-cell compared against another half-cell, by convention, the hydrogen half-cell, which is arbitrarily given a value of 0 V. Standard cell potential: Difference between the two standard electrode potentials of the two half cells. Standard hydrogen electrode: Arbitrarily assigned a potential of zero. Electrode consists of an inert metal such as platinum dipped into a 1 M solution of HCl, where hydrogen gas at 1 atm flows in. Voltaic cell: Two different half-cells connected together to enable to electron transferred during the redox reaction to produce energy in the form of electricity. The electrons are produced at the half-cell that is most easily oxidized. 716 Version 1 2012 Introduction to Oxidation and Reduction Oxidation----Loss of electrons Reduction ---- Gain of Electrons LEO says "GER" Loss of Electrons = Oxidation Gain of Electrons = Reduction OIL RIG Oxidation is LOSS, Reduction is Gain 717 Version 1 2012 3. Rules for Assigning Oxidation Numbers Oxidation Numbers always refer to single atoms The oxidation number of an uncombined element i.e is always 0 The oxidation number of Hydrogen is usually +1 i.e 4. The oxidation number of Oxygen is usually -2 5. The oxidation numbers of Alkali metals are +1, The Oxidation numbers of the Alkaline earth metals are +2 Oxidation numbers on monatomic ions carry the charge of the ion The sum of the oxidation numbers in a neutral compound is 0. In a polyatomic ion it equals the charge of the ion 1. 2. 6. 7. i.e H2, Ne, Zn NH3, HCl H2O, SO2 KCl, CaBr2 K+, Cl-, O2i.e. H2SO4 H=+1, S= +6, O = -2 2(+1) + (+6) + 4(-2)=0 i.e. SO42S= +6, O = -2 (+6) +4(-2) = -2 Practice: Assign the oxidation numbers for each element in the following: 1. NO2 2. N2O5 3. HClO3 4. HNO3 5. Ca(NO3)2 6. KMnO4 7. Fe(OH)3 8. K2Cr2O7 . N= +4 O = -2 718 Version 1 2012 9. CO32- 10. CN- 11. K3Fe(CN)6 Use oxidation numbers to determine if a substance has been oxidized or reduced. An increase in the oxidation number indicates that an atom has lost electrons and therefore oxidized. A decrease in the oxidation number indicates that an atom has gained electrons and therefore reduced Example Zn + CuSO4 ZnSO4 + Cu 0 +2 +6-2 +2+6-2 0 Zn: 0 + 2 -- Oxidized Cu: +2 0 -- Reduced Practice: For each of the following reactions find the element oxidized and the element reduced 1. Cl2 + 2. Cu 3. HNO3 + + KBr HNO3 I2 KCl + Cu(NO3)2 + Br2 NO2 + HIO3 + 719 H2 O NO2 Version 1 2012 Oxidation States Practice 720 Version 1 2012 721 Version 1 2012 Oxidation-Reduction Reactions Oxidation-reduction (redox) reactions are reactions in which oxidation numbers change. Oxidation numbers are either real charges or formal charges which help chemists keep track of electron transfer. In practice, oxidation numbers are best viewed as a bookkeeping device. Oxidation cannot occur without reduction. In a redox reaction, the substance oxidized contains atoms which increase in oxidation number. Oxidation is associated with electron loss (helpful mnemonic: LEO = Loss of Electrons, Oxidation). The substance reduced contains atoms which decrease in oxidation number during the reaction. Reduction is associated with electron gain (helpful mnemonic: GER = Gain of Electrons, Reduction). An oxidizing agent is a substance which oxidizes something else: it itself is reduced! Also, a reducing agent is a substance that reduces another reactant: it itself is oxidized. A disproportionation reaction is a reaction in which the same element is both oxidized and reduced. How to Assign Oxidation Numbers: The Fundamental Rules The oxidation number of any pure element is zero. Thus the oxidation number of H in H2 is zero. The oxidation number of a monatomic ion is equal to its charge. Thus the oxidation number of Cl in the Cl- ion is -1, that for Mg in the Mg+2 ion is +2, and that for oxygen in O2- ion is -2. The sum of the oxidation numbers in a compound is zero if neutral, or equal to the charge if an ion. The oxidation number of alkali metals in compounds is +1, and that of alkaline earths in compounds is +2. The oxidation number of F is -1 in all its compounds. The oxidation number of H is +1 in most compounds. Exceptions are H2 (where H = 0) and the ionic hydrides, such as NaH (where H = -1). The oxidation number of oxygen (O) is -2 in most compounds. Exceptions are O2 (where O = 0) and peroxides, such as H2O2 or Na2O2, where O = -1. For other elements, you can usually use the sum rule above to solve for the unknown oxidation number. Examples: NO(g) has O = -2, so N = +2. Thus N + 2(-2) = 0, so N = +4. equation gives S = -2 + 8 = +6. 2) = 0; 2 Cr = 12; Cr = +6. NO2 (g) has two oxygen atoms and each has O = -2. SO42- has O = -2. Thus S + 4(-2) = -2. Solving the K2Cr2O7 has K = +1 and O = -2. Thus 2(+1) + 2 Cr + 7(- 722 Version 1 2012 Recognizing Oxidation-Reduction Reactions Oxidation-reduction reactions are reactions in which one type of atom increases in oxidation number (is oxidized) and another type of atom decreases in oxidation number (is reduced). Thus to show that a reaction is a redox reaction, you need to calculate oxidation numbers for the atoms in the reactants and products, and document that changes are taking place. There are, however, a few useful generalizations. A large number (but not all!) of oxidation-reduction reactions contain one or more reactants or products which are pure elements. Why is this true? Also, all electrochemical reactions are redox reactions. Most acid-base reactions and most precipitation reactions are not redox reactions. Why? Give some examples! 723 Version 1 2012 Oxidation-Reduction Reactions Oxidation-reduction (redox) reactions are reactions in which oxidation numbers change. Oxidation numbers are either real charges or formal charges which help chemists keep track of electron transfer. In practice, oxidation numbers are best viewed as a bookkeeping device. Oxidation cannot occur without reduction. In a redox reaction, the substance oxidized contains atoms which increase in oxidation number. Oxidation is associated with electron loss (helpful mnemonic: LEO = Loss of Electrons, Oxidation). The substance reduced contains atoms which decrease in oxidation number during the reaction. Reduction is associated with electron gain (helpful mnemonic: GER = Gain of Electrons, Reduction). An oxidizing agent is a substance which oxidizes something else: it itself is reduced! Also, a reducing agent is a substance that reduces another reactant: it itself is oxidized. A disproportionation reaction is a reaction in which the same element is both oxidized and reduced. How to Assign Oxidation Numbers: The Fundamental Rules The oxidation number of any pure element is zero. Thus the oxidation number of H in H2 is zero. The oxidation number of a monatomic ion is equal to its charge. Thus the oxidation number of Cl in the Cl- ion is -1, that for Mg in the Mg+2 ion is +2, and that for oxygen in O2- ion is -2. The sum of the oxidation numbers in a compound is zero if neutral, or equal to the charge if an ion. The oxidation number of alkali metals in compounds is +1, and that of alkaline earths in compounds is +2. The oxidation number of F is -1 in all its compounds. The oxidation number of H is +1 in most compounds. Exceptions are H2 (where H = 0) and the ionic hydrides, such as NaH (where H = -1). The oxidation number of oxygen (O) is -2 in most compounds. Exceptions are O2 (where O = 0) and peroxides, such as H2O2 or Na2O2, where O = -1. For other elements, you can usually use the sum rule above to solve for the unknown oxidation number. Examples: NO(g) has O = -2, so N = +2. Thus N + 2(-2) = 0, so N = +4. equation gives S = -2 + 8 = +6. 2) = 0; 2 Cr = 12; Cr = +6. NO2 (g) has two oxygen atoms and each has O = -2. SO42- has O = -2. Thus S + 4(-2) = -2. Solving the K2Cr2O7 has K = +1 and O = -2. Thus 2(+1) + 2 Cr + 7(- 724 Version 1 2012 Recognizing Oxidation-Reduction Reactions Oxidation-reduction reactions are reactions in which one type of atom increases in oxidation number (is oxidized) and another type of atom decreases in oxidation number (is reduced). Thus to show that a reaction is a redox reaction, you need to calculate oxidation numbers for the atoms in the reactants and products, and document that changes are taking place. There are, however, a few useful generalizations. A large number (but not all!) of oxidation-reduction reactions contain one or more reactants or products which are pure elements. Why is this true? Also, all electrochemical reactions are redox reactions. Most acid-base reactions and most precipitation reactions are not redox reactions. Why? Give some examples! 725 Version 1 2012 Intro to Redox 1. 9.1.1 Define oxidation and reduction in terms of electron loss and gain. (1) a. How was oxidation originally defined: b. How do we now define oxidation and reduction? c. How can you remember these simple definitions? 2. 9.1.2 Deduce the oxidation number of an element in a compound. (3) a. Record the basic rules for assigning oxidation numbers: i. ii. iii. iv. v. vi. 726 Version 1 2012 b. Determine the oxidation number for each element in the following reaction, then show the oxidation and reduction half-reactions: S(s) + O2(g) SO2(g) 3. 9.1.3 State the names of compounds using oxidation numbers. (1) 4. 9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers. (3) a. How can it be determined whether a reaction is considered to be a Redox equation? NO2 N2O5 HClO3 HNO3 Ca(NO3)2 KMnO4 Fe(OH)3 K2Cr2O7 CO32K3Fe(CN)6 b. For each of the following reactions, first determine the oxidation state, then write the oxidation and reduction half reactions for each: i. Cl2 + KBr KCl + Br2 ii. Cu + HNO3 Cu(NO3)2 + NO2 + H2O iii. HNO3 + I2 HIO3 + NO2 c. Provide an example for an element which is both oxidized and reduced. What is this known as? 727 Version 1 2012 Oxidation State Worksheet In each of the following chemicals, determine the oxidation states of each element: 1) sodium nitrate ____________________________________ 2) ammonia ____________________________________ 3) zinc oxide ____________________________________ 4) water ____________________________________ 5) calcium hydride ____________________________________ 6) carbon dioxide ____________________________________ 7) nitrogen ____________________________________ 8) sodium sulfate ____________________________________ 9) aluminum hydroxide ____________________________________ 10) magnesium phosphate ____________________________________ In each of the following reactions, determine what was oxidized and what was reduced. 11) Ca + H2O CaO + H2 Element oxidized: ____________________________________ Element reduced: ____________________________________ 12) 2 H2 + O2 2 H2O Element oxidized: ____________________________________ Element reduced: ____________________________________ 728 Version 1 2012 Balancing Oxidation-Reduction Equations A trial-and-error approach to balancing chemical equations involves playing with the equation adjusting the ratio of the reactants and products until the following goals have been achieved. Goals for Balancing Chemical Equations 1. The number of atoms of each element on both sides of the equation is the same and therefore mass is conserved. 2. The sum of the positive and negative charges is the same on both sides of the equation and therefore charge is conserved. (Charge is conserved because electrons are neither created nor destroyed in a chemical reaction.) There are two situations in which relying on trial and error can get you into trouble. Sometimes the equation is too complex to be solved by trial and error within a reasonable amount of time. Consider the following reaction, for example. 3 Cu(s) + 8 HNO3(aq) 3 Cu2+(aq) + 2 NO(g) + 6 NO3-(aq) + 4 H2O(l) Other times, more than one equation can be written that seems to be balanced. The following are just a few of the balanced equations that can be written for the reaction between the permanganate ion and hydrogen peroxide, for example. 2 MnO4-(aq) + H2O2(aq) + 6 H+(aq) 2 Mn2+(aq) + 3 O2(g) + 4 H2O(l) 2 MnO4-(aq) + 3 H2O2(aq) + 6 H+(aq) 2 Mn2+(aq) + 4 O2(g) + 6 H2O(l) 2 MnO4-(aq) + 5 H2O2(aq) + 6 H+(aq) 2 Mn2+(aq) + 5 O2(g) + 8 H2O(l) 2 MnO4-(aq) + 7 H2O2(aq) + 6 H+(aq) 2 Mn2+(aq) + 6 O2(g) + 10 H2O(l) Equations such as these have to be balanced by a more systematic approach than trial and error. The Half-Reaction Method of Balancing Redox Equations A powerful technique for balancing oxidation-reduction equations involves dividing these reactions into separate oxidation and reduction half-reactions. We then balance the halfreactions, one at a time, and combine them so that electrons are neither created nor destroyed in the reaction. The steps involved in the half-reaction method for balancing equations can be illustrated by considering the reaction used to determine the amount of the triiodide ion (I3-) in a solution by titration with the thiosulfate (S2O32-) ion. STEP 1: Write a skeleton equation for the reaction. The skeleton equation for the reaction on which this titration is based can be written as follows. I3- + S2O32- I- + S4O62- 729 Version 1 2012 STEP 2: Assign oxidation numbers to atoms on both sides of the equation. The negative charge in the I3- ion is formally distributed over the three iodine atoms, which means that the average oxidation state of the iodine atoms in this ion is -1/3. In the S4O62- ion, the total oxidation state of the sulfur atoms is +10. The average oxidation state of the sulfur atoms is therefore +21/2. I3- + S2O32- I- + S4O62- -1/3 -1 +2 -2 +21/2 -2 STEP 3: Determine which atoms are oxidized and which are reduced. STEP 4: Divide the reaction into oxidation and reduction half-reactions and balance these half-reactions one at a time. This reaction can be arbitrarily divided into two half-reactions. One half-reaction describes what happens during oxidation. Oxidation: S2O32- S4O62+21/2 +2 The other describes the reduction half of the reaction. Reduction: I3-1/3 I-1 It doesn't matter which half-reaction we balance first, so let's start with the reduction halfreaction. Our goal is to balance this half-reaction in terms of both charge and mass. It seems reasonable to start by balancing the number of iodine atoms on both sides of the equation. Reduction: I3- 3 I- We then balance the charge by noting that two electrons must be added to an I 3- ion to produce 3 I- ions, Reduction: I3- + 2 e- 3 I- as can be seen from the Lewis structures of these ions shown in the figure below. We now turn to the oxidation half-reaction. The Lewis structures of the starting material and the product of this half-reaction suggest that we can get an S4O62- ion by removing two electrons from a pair of S2O32- ions, as shown in the figure below. 730 Version 1 2012 Oxidation: 2 S2O32- S4O62- + 2 e- STEP 5: Combine these half-reactions so that electrons are neither created nor destroyed. Two electrons are given off in the oxidation half-reaction and two electrons are picked up in the reduction half-reaction. We can therefore obtain a balanced chemical equation by simply combining these half-reactions. (2 S2O32- S4O62- + 2 e-) + (I3- + 2 e- 3 I-) I3- + 2 S2O32- 3 I- + S4O62- STEP 6: Balance the remainder of the equation by inspection, if necessary. Since the overall equation is already balanced in terms of both charge and mass, we simply introduce the symbols describing the states of the reactants and products. I3-(aq) + 2 S2O32-(aq) 3 I-(aq) + S4O62-(aq) Redox Reactions In Acidic Solutions Some might argue that we don't need to use half-reactions to balance equations because they can be balanced by trial and error. The half-reaction technique becomes indispensable, however, in balancing reactions such as the oxidation of sulfur dioxide by the dichromate ion in acidic solution. H+ SO2(aq) + Cr2O72-(aq) SO42-(aq) + Cr3+(aq) The reason why this equation is inherently more difficult to balance has nothing to do with the ratio of moles of SO2 to moles of Cr2O72-; it results from the fact that the solvent takes an active role in both half-reactions. Practice Problem 3: Use half-reactions to balance the equation for the reaction between sulfur dioxide and the dichromate ion in acidic solution. 731 Version 1 2012 The reaction between oxalic acid and potassium permanganate in acidic solution is a classical technique for standardizing solutions of the MnO4- ion. These solutions need to be standardized before they can be used because it is difficult to obtain pure potassium permanganate. There are three sources of error. Samples of KMnO4 are usually contaminated by MnO2. Some of the KMnO4 reacts with trace contaminants when it dissolves in water, even when distilled water is used as the solvent. The presence of traces of MnO2 in this system catalyzes the decomposition of MnO4- ion on standing. Solutions of this ion therefore have to be standardized by titration just before they are used. A sample of reagent grade sodium oxalate (Na2C2O4) is weighed out, dissolved in distilled water, acidified with sulfuric acid, and then stirred until the oxalate dissolves. The resulting oxalic acid solution is then used to titrate MnO4- to the endpoint of the titration, which is the point at which the last drop of MnO4- ion is decolorized and a faint pink color persists for 30 seconds. Practice Problem 4: We can determine the concentration of an acidic permanganate ion solution by titrating this solution with a known amount of oxalic acid until the charactistic purple color of the MnO4- ion disappears. H2C2O4(aq) + MnO4-(aq) CO2(g) + Mn2+(aq) Use the half-reaction method to write a balanced equation for this reaction. Solutions of the MnO4- ion that have been standardized against oxalic acid, using the equation balanced in the previous practice problem, can be used to determine the concentration of aqueous solutions of hydrogen peroxide, using the equation balanced in the following practice problem. 732 Version 1 2012 Practice Problem 5: An endless number of balanced equations can be written for the reaction between the permanganate ion and hydrogen peroxide in acidic solution to form the manganese (II) ion and oxygen: MnO4-(aq) + H2O2(aq) Mn2+(aq) + O2(g) Use the half-reaction method to determine the correct stoichiometry for this reaction. Redox Reactions in Basic Solutions Half-reactions are also valuable for balancing equations in basic solutions. The key to success with these reactions is recognizing that basic solutions contain H2O molecules and OH- ions. We can therefore add water molecules or hydroxide ions to either side of the equation, as needed. The following equation describes the reaction between the permanganate ion and hydrogen peroxide in an acidic solution. 2 MnO4-(aq) + 5 H2O2(aq) + 6 H+(aq) 2 Mn2+(aq) + 5 O2(g) + 8 H2O(l) It might be interesting to see what happens when this reaction occurs in a basic solution. Practice Problem 6: Write a balanced equation for the reaction between the permanganate ion and hydrogen peroxide in a basic solution to form manganese dioxide and oxygen. MnO4-(aq) + H2O2(aq) 733 MnO2(s) + O2(g) Version 1 2012 Reactions in which a single reagent undergoes both oxidation and reduction are called disproportionation reactions. Bromine, for example, disproportionates to form bromide and bromate ions when a strong base is added to an aqueous bromine solution. OHBr - + BrO3- Br2 Practice Problem 7: Write a balanced equation for the disproportionation of bromine in the presence of a strong base. Molecular Redox Reactions Lewis structures can play a vital role in understanding oxidation-reduction reactions with complex molecules. Consider the following reaction, for example, which is used in the Breathalyzer to determine the amount of ethyl alcohol or ethanol on the breath of individuals who are suspected of driving while under the influence. 3 CH3CH2OH(g) + 2 Cr2O72-(aq) + 16 H+(aq) 734 3 CH3CO2H(aq) + 4 Cr3+(aq) + 11 H2O(l) Version 1 2012 We could balance the oxidation half-reaction in terms of the molecular formulas of the starting material and the product of this half-reaction. Oxidation: C2H6O C2H4O2 It is easier to understand what happens in this reaction, however, if we assign oxidation numbers to each of the carbon atoms in the Lewis structures of the components of this reaction, as shown in the figure below. The carbon atom in the CH3 group in ethanol is assigned an oxidation state of -3 so that it can balance the oxidation states of the three H atoms it carries. Applying the same technique to the CH2OH group in the starting material gives an oxidation state of -1. The carbon in the CH3 group in the acetic acid formed in this reaction has the same oxidation state as it did in the starting material: -3. There is a change in the oxidation number of the other carbon atom, however, from -1 to +3. The oxidation half-reaction therefore formally corresponds to the loss of four electrons by one of the carbon atoms. CH3CO2H + 4 e- Oxidation: CH3CH2OH Because this reaction is run in acidic solution, we can add H+ and H2O molecules as needed to balance the equation. CH3CO2H + 4 e- + 4 H+ Oxidation: CH3CH2OH + H2O The other half of this reaction involves a six-electron reduction of the Cr2O72- ion in acidic solution to form a pair of Cr3+ ions. Reduction: Cr2O72- + 6 e- 2 Cr3+ Adding H+ ions and H2O molecules as needed gives the following balanced equation for this halfreaction. Reduction: Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O We are now ready to combine the two half-reactions by assuming that electrons are neither created nor destroyed in this reaction. CH3CO2H + 4 e- + 4 H+) 3(CH3CH2OH + H2O 2(Cr2O72- + 14 H+ + 6 e- 735 2 Cr3+ + 7 H2O) Version 1 2012 3 CH3CH2OH + 2 Cr2O72- + 28 H+ + 3 H2O 3 CH3CO2H + 4 Cr3+ + 12 H+ + 14 H2O Simplifying this equation by removing 3 H2O molecules and 12 H+ ions from both sides of the equation gives the balanced equation for this reaction. 3 CH3CH2OH(g) + 2 Cr2O72-(aq) + 16 H+(aq) 3 CH3CO2H(aq) + 4 Cr3+(aq) + 11 H2O(l) Practice Problem 8: Methyllithium (CH3Li) can be used to form bonds between carbon and either main-group metals or transition metals: HgCl2(s) + 2 CH3Li(l) Hg(CH3)2(l) + 2 LiCl(s) WCl6(s) + 6 CH3Li(l) W(CH3)6(l) + 6 LiCl(s) It can be used also to form bonds between carbon and other nonmetals: PCl3(s) + 3 CH3Li(l) P(CH3)3(l) + 3 LiCl(s) or between carbon atoms: CH3Li(l) + H2CO(g) [CH3CH2OLi] CH3CH2OH(l) Use Lewis structures to explain the stoichiometry of the following oxidation-reaction, which is used to synthesize methyllithium: CH3Br(l) + 2 Li(s) 736 CH3Li(l) + LiBr(s) Version 1 2012 How to Balance Equations for Oxidation-Reduction Reactions Oxidation-reduction (redox) reactions are reactions in which oxidation numbers change. Oxidation numbers are either real charges or formal charges which help chemists keep track of electron transfer. In practice, oxidation numbers are best viewed as a bookkeeping device. Oxidation cannot occur without reduction. In a redox reaction the substance which is oxidized contains atoms which increase in oxidation number. Oxidation is associated with electron loss (helpful mnemonic: LEO = Loss of Electrons, Oxidation). Conversely, the substance which is reduced contains atoms which decrease in oxidation number during the reaction. Reduction is associated with electron gain (helpful mnemonic: GER = Gain of Electrons, Reduction). Chemists often talk about oxidizing and reducing agents. Be careful with these terms! An oxidizing agent is a substance which oxidizes something else: it itself is reduced! Also, a reducing agent is a substance that reduces another reactant: it itself is oxidized. A disproportionation reaction is a reaction in which the same element is both oxidized and reduced. How to Assign Oxidation Numbers: The Fundamental Rules Rules for assigning oxidation numbers are as follows: • The oxidation number of any pure element is zero. Thus the oxidation number of H in H2 is zero. • The oxidation number of a monatomic ion is equal to its charge. Thus the oxidation number of Cl in the Cl- ion is -1, that for Mg in the Mg+2 ion is +2, and that for oxygen in O2- ion is -2. • The sum of the oxidation numbers in a compound is zero if neutral, or equal to the charge if an ion. • The oxidation number of alkali metals in compounds is +1, and that of alkaline earths in compounds is +2. The oxidation number of F is -1 in all its compounds. • The oxidation number of H is +1 in most compounds. Exceptions are H2 (where H = 0) and the ionic hydrides, such as NaH (where H = -1). • The oxidation number of oxygen (O) is -2 in most compounds. Exceptions are O2 (where O = 0) and peroxides, such as H2O2 or Na2O2, where O = -1. • For other elements, you can usually use rule (3) to solve for the unknown oxidation number. 737 Version 1 2012 Examples: NO(g) has O = -2, so N = +2. NO2(g) has O = -2, so N = +4. SO42- has O = -2. Thus S + 4(-2) = -2. Solving the equation gives S = -2 + 8 = +6. K2Cr2O7 has K = +1 and O = -2. Thus 2(+1) + 2 Cr + 7(- 2) = 0; 2 Cr = 12; Cr = +6. How to Balance Redox Reactions Using the Method of Half-Reactions Oxidation-reduction reactions are often tricky to balance without using a systematic method. We shall use the method of half-reactions which is outlined in detail below. Method in Acidic (or Neutral) Solution Suppose you are asked to balance the equation below: NO2– + MnO4– NO3– + Mn+2 (in acid solution) Begin by writing the unbalanced oxidation and reduction half-reactions (you do not need to know which is which): NO2– NO3– MnO4– Mn+2 Next, balance for atoms. First do this for atoms other than O and H. (Both equations above are already balanced for N and Mn, so no change is needed in this example.) Then balance for O atoms by adding H2O to the reaction side deficient in O: H2O + NO2– NO3– MnO4– 738 Mn+2 + 4 H2O Version 1 2012 This leaves H atoms unbalanced. In acidic (or neutral) solution, balance for H atoms by adding H+ to the side deficient in H: H2O + NO2– NO3– + 2 H+ 8 H+ + MnO4– Mn+2 + 4 H2O The next step is to balance for charge. To do this, add electrons (e-) to the more positive side: H2O + NO2- NO3- + 2 H+ + 2 e- 5 e- + 8 H+ + MnO4- Mn+2 + 4 H2O Now you need to multiply the equations by appropriate factors so that the number of electrons lost in the oxidation half-reaction (LEO) is equal to the number of electrons gained in the reduction half-reaction (GER): 5 x [ H2O + NO2- NO3- + 2 H+ + 2 e- ] 2 x [ 5 e- + 8 H+ + MnO4- Mn+2 + 4 H2O ] Then, sum the above equations to obtain 5H2O + 5NO2- + 10 e- + 16H+ + 2MnO4- 5NO3- + 10H+ + 10 e- + 2Mn+2 + 8H2O Finally, simplify by subtracting out species that are identical on both sides. Our final balanced redox equation is 5 NO2- + 6 H+ + 2 MnO4- 5 NO3- + 2 Mn+2 + 3 H2O Check this equation to confirm that it is balanced for atoms and balanced for charge. 739 Version 1 2012 Method in Basic Solution Suppose you are asked to balance the equation below: I- + MnO4- I2 + MnO2 (in basic solution) Begin by writing the unbalanced oxidation and reduction half-reactions (you do not need to know which is which): I- I2 MnO4- MnO2 Next, balance for atoms. First do this for atoms other than O and H: 2 I- I2 MnO4- MnO2 Then balance for O atoms by adding H2O to the reaction side deficient in O: 2 I- I2 MnO4- MnO2 + 2 H2O This leaves H atoms unbalanced. In basic solution (just as in acidic or neutral solution) first balance for H atoms by adding H+ to the side deficient in H: 4 H+ + MnO4- MnO2 + 2 H2O In basic solution, follow this step by neutralizing the H+; do this by adding an equivalent amount of OH- to both sides of the equation. 4 OH- + 4 H+ + MnO4- 740 MnO2 + 2 H2O + 4 OH- Version 1 2012 Then form water on the side which has both H+ and OH- (recall that H+ + OH- H2O): in this case we form 4 H2O on the left: 4 H2O + MnO4- MnO2 + 2 H2O + 4 OH- Next simplify the water by subtracting 2 H20 from both sides. The half-reactions are now: 2 I– I2 2 H2O + MnO4– MnO2 + 4 OH– At this point the equations should be balanced for atoms. The next step is to balance for charge. To do this, add electrons (e–) to the more positive side: 2 I– 3 e– + 2 H2O + MnO4– 2 + 2 e– MnO2 + 4 OH– Now you need to multiply the equations by appropriate factors so that the number of electrons lost in the oxidation half-reaction (LEO) is equal to the number of electrons gained in the reduction half-reaction (GER): 3 x [ 2 I- I2 + 2 e- ] 2 x [ 3 e- + 2 H2O + MnO4- MnO2 + 4 OH- ] Sum the equations to obtain 6 I- + 6 e- + 4 H2O + 2 MnO4- 2 + 6 e- 2 + 8 OH- Finally, simplify by subtracting out species that are identical on both sides: 6 I- + 4 H2O + 2 MnO4- 741 I2 + 2 MnO2 + 8 OH- Version 1 2012 Check our final equation above to confirm that it is balanced for atoms and balanced for charge. Exercises: Balance the following redox reactions. In each case • (a) give the balanced half-reactions; identify the oxidation half-reaction and the reduction half-reaction. • (b) give the balanced net reaction. • (c) identify the oxidizing agent and the reducing agent. _______________________________________________________________________ 1. Cl2(g) + S2O32-(aq) Cl-(aq) + SO42-(aq) in acid solution. 742 Version 1 2012 Answers: (a) S2O32-(aq) + 5 H20 2 SO42-(aq) + 10 H+(aq) + 8 e- (oxidation half-reaction – LEO); Cl2(g) + 2 e- 2 Cl-(aq) (reduction half-reaction – GER). (b) S2O32-(aq) + 5 H20 + 4 Cl2(g) 2 SO42-(aq) + 10 H+(aq) + 8 Cl-(aq) (c) S2O32-(aq) is the reducing agent; Cl2(g) is the oxidizing agent. _______________________________________________________________________ 2. O3(g) + Br-(aq) O2(g) + BrO-(aq) in basic solution. Answers: (a) Br - (aq) + H20 + 2 OH-(aq) BrO-(aq) + 2H2O + 2 e- or, after simplifying, Br - (aq) + 2 OH-(aq) BrO-(aq) + H2O + 2 e- (oxidation half-reaction – LEO); O3(g) + 2 H2O + 2 e- O2(g) + H2O + 2 OH-(aq) or, after simplifying, O3(g) + H2O + 2 e- O2(g) + 2 OH-(aq) (reduction half-reaction – GER). (b) Br - (aq) + O3(g) BrO-(aq) + O2(g) 743 Version 1 2012 (c) Br - (aq) is the reducing agent; O3(g) is the oxidizing agent. _______________________________________________________________________ 3. Balance the reaction, Br2(l) Br-(aq) + BrO3-(aq) in basic solution. Hint: this is a disproportionation reaction! Answer: 6 Br2(l) + 12 OH- -(aq) + 2 BrO3-(aq) + 6 H2O 744 Version 1 2012 How to Balance Redox Equations 1) Above each atom in the original equation write the atoms oxidation number. 2) Identify which atoms have changed their charge. 3) Write the half-reactions and make the number of electrons gained EQUAL the number of electrons lost. 4) Rewrite the original equation with the new coefficients. 5) Add H+ and H2O (acidic) or H2O and OH– (basic) as needed to balance the number of oxygen atoms and the charges. 6) Check that the sum of the charges on both sides of the equation are equal. For Example: Balance the following equation: Cr2+ + I2 ----> Cr3+ + I1Step 1 Cr2+ + I20 ----> Cr3+ + I1Steps 2 & 3 2 ( Cr2+ – 1 e- ---> Cr3+ ) Cr became more positive, it lost electrons ( I2 + 2e ---> 2 I1- ) I became more negative, it gained electrons Step 4 2 Cr2+ + I20 ----> 2 Cr3+ + 2I1Step 5 Not needed Step 6 (4 +) + ( 0 ) (6 +) + (2 -) 4+=4+ Equation is balanced! 745 Version 1 2012 Even More Redox Equations to Balance 1) 2) Fe2+ + NO3 - + 3) H2O2 + 4) C2H4 5) 6) MnO4 I2 I- H2O MnO4- + PbO2(s) As2S3 + + Cl- NO3- Fe3+ + Mn2+ IO3- + NO2 + I2 Mn2+ + Pb2+ + AsO43- CO2 Cl2 + SO42- 746 + NO Version 1 2012 Redox and Electrochemical Cells 1. 2. Assign oxidation numbers to all elements in the following componds and ions: a. NH4+ b. CuCl2 f. NO3- g. MnO2 c. H2O h. PO43- d. SO2 e. Fe2O3 i. K2Cr2O7 j. MnO4- Use oxidation numbers to deduce which species is oxidised and which is reduced in the following reactions: 3. a. Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq) b. Cl2(aq) + 2NaBr(aq) → Br2(aq) + 2NaCl(aq) c. 2FeCl2(aq) + Cl2(aq) → 2FeCl3(aq) d. H2O(l) + 2F2(aq) → 4HF(aq) + O2(g) e. I2(aq) + SO3-(aq) + H2O(l) → 2I-(aq) + SO42-(aq) + 2H+(aq) Identify the oxidizing and reducing agents in each of the following reactions: a. H2(g) + Cl2(g) → 2HCl(g) 747 Version 1 2012 4. b. 2Al(s) + 3PbCl2(s) → 2AlCl3(s) + 3Pb(s) c. Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(aq) d. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Deduce the half equations of oxidation and reduction for the following reactions: a. Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g) b. 2Fe2+(aq) + Cl2(aq) → 2Fe3+(aq) + 2Cl-(aq) c. Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq) d. Cl2(aq) + 2Br-(aq) → 2Cl-(aq) + Br2(aq) 748 Version 1 2012 5. Use the half-reaction method to write balanced equations for the following reactions: a. Magnesium reducing lead ions to lead metal b. Sulfur dioxide being oxidised to sulfate through reducing iodine to iodide ions c. Hydrogen peroxide oxidising iron(II) to iron(III) in acidic solution d. Zinc reducing acidified dichromate(VI) ions to chromium(III) e. Acidified permanganate(VII) ions oxidising methanol to CO2 and H2O 749 Version 1 2012 6. If wine is left in an open bottle it often tastes ‘vinegary’ a few days later. This is because of the oxidation of the ethanol to ethanoic acid (remember that?) by atmospheric oxygen. a. Write the half-equation for the oxidation of ethanol to ethanoic acid. b. What is the intial and final average oxidation number for the carbons in this change? c. Write the half-equation for the reduction of oxygen in acidic solution. d. Combine these to produce an overall equation for the reaction. 750 Version 1 2012 Here is a small part of the reactivity series of metals: Mg strongest reducing agent Al Zn Fe Pb Cu Ag weakest reducing agent 7. Refer to the reactivity series given above to predict whether the following reactions will occur: a. ZnCl2(aq) + 2Ag(s) → 2AgCl(s) + Zn(s) b. 2FeCl3(aq) + 3Mg(s) → MgCl2(aq) + 2Fe(s) We learned earlier that the tendency for the halogens to act as oxidizing agents decreases down a group: F2 strongest oxidizing agent Cl2 Br2 I2 weakest oxidizing agent 8. Use the two reactivity series given to predict whether reactions will occur between the following reactants and write equations where relevant. a. CuCl2(aq) + Ag(s) 751 Version 1 2012 9. b. Fe(NO3)2 + Al(s) c. NaI(aq) + Br2(aq) d. KCl + I2 a. Use the following reactions to deduce the order of reactivity of the elements w, x, y, z putting the most reactive first. w + x+ → w+ + x; b. y+ + z → no rxn; x + z+ → x+ + z; x + y+ → x+ + y Which of the following reactions would you expect to occur according to the reactivity series you established in a? i. w + + y → w + y+ ii. 752 w + + z → w + z+ Version 1 2012 10. The following reactions are spontaneous as written. Fe(s) + Cd2+(aq) → Fe2+(aq) + Cd(s) Cd(s) + Sn2+(aq) → Cd2+(aq) + Sn(s) Sn(s) + Pb2+(aq) → Sn2+(aq) + Pb(s) a. State and explain which is the strongest oxidizing agent in the examples above. b. State and explain which is the strongest reducing agent in the examples above. c. State with a reason which of the following pairs will react spontaneously: Sn(s) + Fe2+; Cd(s) + Pb2+; or Fe(s) + Pb2+. 753 Version 1 2012 11. Draw a voltaic cell with one half cell consisting of Mg and a solution of Mg2+ ions and the other consisting of Zn and a solution of Zn2+ ions. Label the electrodes with name and charge, the direction of electron and ion movement and write equations for the reactions occurring at each electrode. 12. Predict what would happen if an iron spatula was left in a solution of copper(II) sulfate overnight. 754 Version 1 2012 13. You are intending to construct an electrochemical cell using two metals chosen from copper, iron, magnesium and zinc, in aqueous solutions of their salts. Explain how you would use a reactivity series to choose the pair of metals that would produce the greatest potential difference for the cell and predict which metal will be the anode and which the cathode. 14. In each of the following half-reactions, give the species being reduced and the number of electrons needed to balance the half-reactions: a. AgBrO3 + ?e- → Ag + BrO3- b. HCrO4- + 7H+ + ?e- → Cr3+ + 4H2O c. WO3 + 6H+ + ?e- → W + 3H2O 755 Version 1 2012 15. Identify the species in each of the following reactions that would receive electrons from the cathode and that would lose electrons at the anode in each of the following galvanic cells: 16. a. Au3+(aq) + Zn(s) Au+(aq) + Zn2+(aq) b. 3 Pu6+(aq) + 2 Cr3+(aq) 2 Cr6+(aq) + 3 Pu4+(aq) In electrolysis as well as in a battery, oxidation occurs at the _______________ and reduction 17. occurs at the _______________. a. Write the two half-reactions for the electrolysis of molten NaCl and indicate which reaction occurs at each electrode. b. Write the overall equation for the electrolysis of NaCl(l). 756 Version 1 2012 Common Oxidizing Agents and Reducing Agents In looking at oxidation-reduction reactions, we can focus on the role played by a particular reactant in a chemical reaction. What is the role of the permanganate ion in the following reaction, for example? 2 MnO4-(aq) + 5 H2C2O4(aq) + 6 H+(aq) 10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l) Oxalic acid is oxidized to carbon dioxide in this reaction and the permanganate ion is reduced to the Mn2+ ion. Oxidation: H2C2O4 Reduction: CO2 +3 +4 MnO4- Mn2+ +7 +2 The permanganate ion removes electrons from oxalic acid molecules and thereby oxidizes the oxalic acid. Thus, the MnO4- ion acts as an oxidizing agent in this reaction. Oxalic acid, on the other hand, is a reducing agent in this reaction. By giving up electrons, it reduces the MnO4- ion to Mn2+. Atoms, ions, and molecules that have an unusually large affinity for electrons tend to be good oxidizing agents. Elemental fluorine, for example, is the strongest common oxidizing agent. F2 is such a good oxidizing agent that metals, quartz, asbestos, and even water burst into flame in its presence. Other good oxidizing agents include O2, O3, and Cl2, which are the elemental forms of the second and third most electronegative elements, respectively. Another place to look for good oxidizing agents is among compounds with unusually large oxidation states, such as the permanganate (MnO4-), chromate (CrO42-), and dichromate (Cr2O72) ions, as well as nitric acid (HNO3), perchloric acid (HClO4), and sulfuric acid (H2SO4). These compounds are strong oxidizing agents because elements become more electronegative as the oxidation states of their atoms increase. Good reducing agents include the active metals, such as sodium, magnesium, aluminum, and zinc, which have relatively small ionization energies and low electro-negativities. Metal hydrides, such as NaH, CaH2, and LiAlH4, which formally contain the H- ion, are also good reducing agents. Some compounds can act as either oxidizing agents or reducing agents. One example is hydrogen gas, which acts as an oxidizing agent when it combines with metals and as a reducing agent when it reacts with nonmetals. 2 Na(s) + H2(g) H2(g) + Cl2(g) 2 NaH(s) 2 HCl(g) Another example is hydrogen peroxide, in which the oxygen atom is in the -1 oxidation state. Because this oxidation state lies between the extremes of the more common 0 and -2 oxidation states of oxygen, H2O2 can act as either an oxidizing agent or a reducing agent. 757 Version 1 2012 The Relative Strengths of Oxidizing and Reducing Agents Spontaneous oxidation-reduction reactions convert the stronger of a pair of oxidizing agents and the stronger of a pair of reducing agents into a weaker oxidizing agent and a weaker reducing agent. The fact that the following reaction occurs, for example, suggests that copper metal is a stronger reducing agent than silver metal and that the Ag+ ion is a stronger oxidizing agent than the Cu2+ ion. Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) stronger stronger reducing oxidizing agent agent weaker weaker oxidizing reducing agent agent On the basis of many such experiments, the common oxidation-reduction half-reactions have been organized into a table in which the strongest reducing agents are at one end and the strongest oxidizing agents are at the other, as shown in the table below. By convention, all of the half-reactions are written in the direction of reduction. Furthermore, by convention, the strongest reducing agents are usually found at the top of the table. The Relative Strengths of Common Oxidizing Agents and Reducing Agents K+ + e- K Best 2+ - Ba reducing 2+ - Ca agents Ba + 2 e Ca + 2 e + - Na + e Na 2+ - Mg + 2 e Mg H2 + 2 e- 2 H- Al3+ + 3 e- Al Mn2+ + 2 e- Mn 2+ - Zn 3+ - Cr Zn + 2 e Cr + 3 e S + 2 e- S2- 2 CO2 + 2 H+ + 2 eCr3+ + e- Cr2+ Fe2+ + 2 e- Fe 2+ - Co 2+ - Ni 2+ - Sn + 2 e Sn Pb2+ + 2 e- Pb Co + 2 e Ni + 2 e 3+ - Fe + 3 e Fe 2 H+ + 2 e2- H2C2O4 H2 - S4O6 + 2 e 2 S2O32- 758 Version 1 2012 Sn4+ + 2 e- Sn2+ Cu2+ + e- Cu+ O2 + 2 H2O + 4 eCu+ + e- 4 OH- Cu I2 + 2 e- 2 I- oxidizing MnO4- + 2 H2O + 3 epower O2 + 2 H+ + 2 e- increases Fe3+ + eHg22+ H2O2 +2e Ag+ + e- power 2 Hg increases Ag - Hg + 2 e Hg - 2 OH- H2O2 + 2 e HNO3 + 3 H+ + 3 eBr2(aq) + 2 e - NO + 2 H2O 2 Br - 2 IO3- + 12 H+ + 10 eCrO42- + - +8H +3e Pt2+ + 2 e- I2 + 6 H2O Cr3+ + 4 H2O Pt + MnO2 + 4 H + 2 eO2 + 4 H+ + 4 e- Mn2+ + 2 H2O 2 H2O Cr2O72- + 14 H+ + 6 eCl2(g) + 2 e- MnO4- + 8 H+ + 5 eH2O2 + 2 H + 2 e- Co + e Best Mn2+ + 4 H2O 2 H2O 2+ Co S2O82- + 2 e- 2 SO42- oxidizing O3(g) + 2 H+ + 2 eagents Pb2+ + 2 H2O Au + 3+ 2 Cr3+ + 7 H2O 2 Cl- PbO2 + 4 H+ + 2 eAu+ + e- Reducing Fe2+ - 2+ MnO2 + 4 OH- F2(g) + 2 H+ + 2 e- O2(g) + H2O 2 HF(aq) Fortunately, you don't have to memorize these conventions. All you have to do is remember that the active metals, such as sodium and potassium, are excellent reducing agents and look for these entries in the table. The strongest reducing agents will be found at the corner of the table where sodium and potassium metal are listed. 759 Version 1 2012 Practice Problem 9: Arrange the following oxidizing and reducing agents in order of increasing strength: Reducing agents: Cl-, Cu, H2, H-, HF, Pb, and Zn Oxidizing agents: Cr3+, Cr2O72-, Cu2+, H+, O2, O3, and Na+ Practice Problem 10: Predict whether the following oxidation-reduction reactions should occur as written: (a) 2 Ag(s) + S(s) Ag2S(s) (b) 2 Ag(s) + Cu2+(aq) 2 Ag+(aq) + Cu(s) (c) MnO4-(aq) + 3 Fe2+(aq) + 2 H2O(l) MnO2(s) + 3 Fe3+(aq) + 4 OH-(aq) (d) MnO4-(aq) + 5 Fe2+(aq)+ 8 H+(aq) Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l) 760 Version 1 2012 Practice Problem 11: Which of the following pairs of ions cannot exist simultaneously in aqueous solutions? (a) Cu+ and Fe3+ (b) Fe3+ and I- c) Al3+ and Co2+ Displacement Reactions For each of the following, either complete the word equation or write ‘no reaction’ potassium + copper sulphate iron + zinc sulphate sodium + iron chloride zinc + silver nitrate lithium + sodium chloride calcium + iron nitrate sodium + potassium sulphate 761 Version 1 2012 More Redox Equations to Balance 1) NO3-(aq) Zn(s) + 2) IO3-(aq) 3) Ce4+(aq) + 4) PH3(g) 5) F2(g) 6) H2O2(aq) + I-(aq) Cl-(aq) + + Cr(OH)3(s) N2(g) I2(aq) Cl2(aq) CrO42-(aq) H2O(l) + Zn2+(aq) + Ce3+(aq) + CrO2-(aq) + F- + P4(s) O2(g) CrO42-(aq) 762 Version 1 2012 Oxidation and Reduction Practice In each of the following equations, indicate the element that has been oxidized and the one that has been reduced. You should also label the oxidation state of each before and after the process: 1) 2 Na + FeCl2 2 NaCl + Fe 2) 2 C2H2 + 5 O2 4 CO2 + 2 H2O 3) 2 PbS + 3 O2 2 SO2 + 2 PbO 4) 2 H2 + O2 2 H2O 5) Cu + HNO3 CuNO3 + H2 6) AgNO3 + Cu CuNO3 + Ag 763 Version 1 2012 Redox Equations 1. 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. (3) a. Using Cr2O72- as an example, demonstrate the steps for writing the proper half-reaction for oxidation: i. ii. iii. iv. v. vi. b. Using HNO2 as an example, demonstrate the steps for writing the proper half-reaction for reduction: i. ii. iii. iv. v. vi. 764 Version 1 2012 2. 9.2.2 Deduce redox equations using half equations. (3) a. If given both the reduction and oxidation half equations, you should be able to balance and simplify the equations. A common example is the oxidation of a primary or secondary alcohol (via dichromate or manganate). Following Topic 10 again, if the half-equations for the oxidation of methanol (and reduction of manganate) 2H2O + CH3OH CO2 + H2O + 6H+ + 6eMnO4- + 8H+ + 5e- Mn2+ + 4H2O 3. 9.2.3 Define the terms oxidizing agent and reducing agent. (1) 4. 9.2.4 Identify the oxidizing and reducing agents in redox equations. (2) Reducing Agents (oxidized) Oxidizing Agents (reduced) Hydrogen Oxygen Carbon Ozone Carbon Monoxide Chlorine Metals Acidified KMnO4 Acidified K2Cr2O7 Acidified H2O2 Metal Ions Hydrogen Ions MnO2 765 Version 1 2012 5. Often, chemical equations are difficult to balance based due to large and uneven numbers of reacting and produced species. A systematic approach using the principles of redox may allow you to balance such equations. Balance each of the following equations using this method: a. Cu + HNO3 Cu(NO3)2 + NO + H2O b. HNO3 + I2 HIO3 + NO2 + H2O 766 Version 1 2012 Redox Equations 1) Complete the table below by filling in the charges on the ions. The first one has been done for you. 2) Identify the charge on each species. ZnSO4 PbSO4 HNO3 KBr FeSO4 MgSO4 NaCl Zn= +2 SO4= -2 3) Redox equations tell us what species have gained electrons, and which species have lost electrons. The information below will help you to write redox equations. 1) Word equation 2) Symbol equation 3) Ionic equation 4) Write down the charges on each species 5) Identify the electron transfer (OILRIG). If a species becomes more positive (0 +2) it has lost electrons (oxidation). If a species has become more negative (+2 0) it has gained electrons (reduction). Complete the missing information in the table for the reaction of Magnesium with Zinc Sulphate. Magnesium + Zinc Sulphate Magnesium Sulphate + Mg + + Zn + Zn2+ + SO42- + Zn 0 +2 -2 Mg: 0 +2 lost electrons, become oxidised. Zn: gained electrons, become 767 Version 1 2012 4) Identify what has been oxidised and reduced in the following reactions using the same procedure as above. a) Zinc reacting with Tin(II) Sulphate. b) Magnesium reacting with Iron(II) Sulphate. 768 Version 1 2012 Working with Oxidation & Reduction 1) What is the difference between oxidation and reduction? Include an example to show what you mean. 2) How can you differentiate between an oxidizing agent and a reducing agent ? 3) In the reaction: 2 K + I2 2 KI , which chemical is the oxidizing agent and which chemical is oxidized ? 4) For each of the following reactions, indicate whether the reaction is a REDOX or a NONREDOX reaction. a) Ba(NO3)2 + H2SO4 BaSO4 + 2 HNO3 b) H2O2 + MnO4- O2 + Mn2+ c) HNO3 + I2 HIO3 + NO2 + H2O d) H2CO3 H2O + CO2 5) A "redox" reaction is different from a "nonredox" reaction because: a) only redox reactions are balanced by using H+ and water. b) redox reactions are the more common reaction type. c) in redox reactions the electrical charges of the atoms change. d) in nonredox reactions the charges of the atoms change. 6) What is the oxidation number of carbon in each of the following substances? a) C b) CaC2 c) CO2 d) CO e) CO3-2 7) Explain what happens to manganese in this equation: I-1 + MnO4 -1 I2 + MnO2 769 Version 1 2012 8) In the following redox reaction which element is oxidized and which is reduced? 4NH3 + 3Ca(ClO)2 2N2 + 6H2O + 3CaCl2 A. B. C. D. E. H is oxidized and N is reduced Cl is oxidized and O is reduced N is oxidized and Cl is reduced Cl is oxidized and N is reduced N is oxidized and O is reduced 9) Which one of the following is a redox reaction? A) 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g) B) 2KBr(aq) + Pb(NO3)2(aq) 2KNO3(aq) + PbBr2(s) C) CaBr2(aq) + H2SO4(aq) CaSO4(s) + 2HBr(g) D) H+(aq) + OH–(aq) H2O(l) E) CO32–(aq) + HSO4–(aq) HCO3–(aq) + SO42–(aq) 10) Which of the following equations does not represent an oxidation-reduction reaction? A) 3Al + 6HCl 3H2 + AlCl3 B) 2H2O 2H2 + O2 C) 2NaCl + Pb(NO3)2 PbCl2 + 3NaNO3 D) 2NaI + Br2 2NaBr + I2 E) Cu(NO3)2 + Zn Zn(NO3)2 + Cu 11) Identify the element being oxidized in the following reaction. 4Al + 3O2 2Al2O3 12) Identify the oxidizing agent in the following reaction. 4Al + 3O2 2Al2O3 13) Identify the element being reduced in the following reaction. 2KBr + F2 Br2 + 2KF 14) Determine the oxidation number of each of the elements in Cs2Cr2O7? 15) Determine the oxidation number of each of the elements in K2TaF7? 770 Version 1 2012 REDOX WORD FRAME Oxidation-reduction, or _______________________, reactions are an important category of chemical reactions because they are what makes batteries work, they provide a way for plants and animals to transfer and store energy in cells, they allow police officers to measure the blood alcohol level of a driver, they explain why silver tarnishes, why iron rusts, why the Statue of Liberty has turned green, and a myriad of other useful things. Oxidation was originally only defined as the chemical combination of a substance with ________________________ but is now defined as any chemical change in which electrons are _________________ by an atom. Reduction was originally defined as only the chemical __________________ of oxygen but is now defined as any chemical change in which electrons are ________________________ by an atom. An oxidation reaction is always accompanied by a _________________________ reaction. The substance that undergoes oxidation is also called the _________________ agent. This is because it causes the reduction of the other element. The substance that undergoes reduction is also called the ___________________________ agent. This is because it causes the ______________________ of the other element. All _______________________ replacement reactions are oxidation-reduction reactions because one pure element replaces another atom in a compound. Combustion, doubledisplacement, synthesis, and decomposition reactions can also be oxidation–reduction reactions. A change in _______________________________ determines whether or not an oxidationreduction reactions has occurred. An oxidation number can be assigned to an element in a substance according to a set of rules. The oxidation number of a pure element is ______________. The oxidation number of a monatomic cations can be determined from the ________________________ __________________. The oxidation state of oxygen is usually ______________. The charge on a hydrogen ion is _________________. The oxidation states of all of the atoms in a compound must add up to be ____________. 771 Version 1 2012 The oxidation states of all of the atoms in an ion must add up to be the same as _______________________________________. From the above rules, the oxidation states of elements such as sulfur, nitrogen, and chlorine in covalent compounds can be determined. One way to describe how __________________________ are transferred between elements in a redox reaction is to write _______________ reactions. The half reaction in which electrons are a reactant is called _____________________________. The half reaction in which electrons are a product is called ___________________________. Word Bank: -2 +1 electrons gained half loss lost oxidation (2x) oxidation state oxidizing oxygen periodic table redox reducing reduction (2x) single the ion’s charge zero (2x) 772 Version 1 2012 Hi y’all ahm Erin Brockovich and ah want to talk to y’all about chromium and ahxidation’. Well now, I managed to use a relationship between humans and chemicals to get the biggest settlement ever for a civil class action lawsuit. You see, I could not understand why medical records were being used by PG&E* in a real estate lawsuit in the town of Hinkley, California. When I went and spoke to the residents of Hinkley, they presented a variety of conditions (including throat cancer, respiratory, liver and kidney problems, reproductory problems etc.) that can be caused by the chromium 6 ion (which was present in unlined ponds of wastewater.) When I went to see the scientists, they told me that chromium is found naturally in the Earth, plants and animals (including us humans), but not in its element form. However, chromium can have a variable oxidation state in compounds (between -2 and +6) but most commonly it exists as chromium 3 (Cr3+, in vegetables and meat) and chromium 6 (Cr6+, in industry.) Chromium 6 is used to stop rust forming, and Insert picture of Erin Brockovich when inhaled causes the health problems mentioned above. However, PG&E argued that chromium 6 is here turned into chromium 3 in the stomach, so the Hinkley residents were actually ingesting the non-toxic chromium 3 form. What do you think? Can you help me write a letter to the residents of Hinkley explaining the science behind their problems? Thank y’all very much! *Pacific Gas and Electric You will be graded according to the following stair of success: A B C D E Uses reactivity series to suggest why chromium prevents rust; constructs general statement about oxidising ability and reactivity. Independently creates a half equation for the process that PG&E claim happens in the stomach and explain whether it is redox or not, and why. Describes the reaction in the stomach as oxidation, reduction or redox, using different definitions of oxidation and reduction in answer; uses books to describe redox (reduction and oxidation) reactions Suggests where chromium is found on the periodic table, giving a reason from text; suggests whether chromium 3 or 6 is toxic. Defines the words underlined correctly, using ideas about particles: gives own opinion about the case. Suggests how the people ended up ingesting chromium. 773 Version 1 2012 Random Questions about Electrochemistry 1) What is a reducing agent ? What is an oxidizing agent ? 2) Which substance is the best reducing agent? Which is the best oxidizing agent? 3) When Ni and Fe are added to a solution that contains both Ni2+ and Fe2+, which will be reduced and which will be oxidized? 4) Write the half-reactions that will occur if Cl2 and Br2 are added to a solution containing Cl- and Br-. Is this reaction spontaneous? 5) 6) What reaction happens at the anode ? At the cathode ? Give an example of each. In a certain zinc-copper cell, the concentrations are [Cu2+] = 0.0100M and [Zn2+]= 1.0 M. Zn(s) + Cu2+ Zn2+ + Cu(s) If the standard cell potential is 1.10 V, what is the cell potential? 774 Version 1 2012 7) In the analysis of two water samples, cell potentials of 0.57 V and 0.82 were obtained. Calculate the Cu2+ ion concentration for both samples. E° = 0.46V and the overall reaction is Cu(s) + 2 Ag+ Cu2+ + 2 Ag(s) 775 Version 1 2012 Reactivity 1. 9.3.1 Deduce a reactivity series based on the chemical behavior of a group of oxidizing and reducing agents. (3) a. Why do metals generally behave as reducing agents? b. Provide an example of a replacement reaction where Mg takes the place of Cu2+ in the compound CuSO4: i. Write the Total Ionic Equation: ii. Write the Net Ionic Equation: iii. Provide each of the ½ reactions: iv. Which metal is more reactive? c. Describe and diagram a thermite reaction: d. What happens when common metals are added to water? Use Na and H2O as an example: 2. 9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series. (3) 776 Version 1 2012 Reactivity Series 1. (a) The table shows what happens when four metals are added to the same volume of dilute hydrochloric acid in a test tube. Use the information in the table to help you decide which of these metals is (i) (ii) the most reactive. the least reactive. (b) Zinc reacts with iron(II) sulphate solution in a displacement reaction. (i) Why does this reaction occur? (ii) Complete the word equation for the reaction. zinc + iron(II) sulphate → .............................. 777 + .............................. Version 1 2012 2. The list gives the order of reactivity of some metals. (a) Iron is sometimes coated with zinc to prevent the iron rusting. The iron does not rust even if the coating of zinc becomes damaged. (i) What is the name given to this method of rust prevention? (ii) Give one example where this method of rust prevention is used. (iii) Explain how this method of rust prevention works. 778 Version 1 2012 (b) A student is given some solid nickel nitrate and several small pieces of magnesium, zinc, iron, copper and silver. Describe and explain how he can find the position of nickel in the reactivity series given above. 779 Version 1 2012 3. Use information from the table to answer this question. (a) When zinc is added to magnesium sulphate solution, no reaction occurs. Explain why. (b) When iron filings are added to copper(II) sulphate solution, a reaction takes place. (i) (ii) Write a chemical equation for this reaction. Describe the colour changes during this reaction. colour change of solid colour change of solution 780 Version 1 2012 (c) When copper is added to dilute sulphuric acid, no reaction occurs. When iron is added to dilute sulphuric acid, hydrogen gas and iron(II) sulphate solution are formed. What does this show about the reactivity of hydrogen compared to the reactivity of copper and the reactivity of iron? 4. This question is about the reactions of the metals calcium, iron and zinc. (a) Samples of each of the powdered metals were placed in separate beakers of water. Only calcium reacted immediately. Describe two observations that could be made during the reaction of calcium with water. Write a chemical equation for the reaction. (b) A reaction occurred when powdered zinc was heated in steam. Name the zinc compound formed. Write a chemical equation for the reaction. 781 Version 1 2012 (c) Some powdered zinc was added to a solution of iron(II) sulphate. (i) Write an ionic equation to show the reaction that occurs. (ii) State the type of reaction occurring. (d) Iron rusts slowly in the presence of water. Name one other substance that must be present for iron to rust. (e) Galvanising is one method used to prevent iron from rusting. (i) Describe how a sheet of iron is galvanised. (ii) A sheet of galvanised iron was scratched and left in the rain. The exposed iron did not rust. Explain why. 782 Version 1 2012 Electrochemical Cells and the Activity Series Write the appropriate half reactions and calculate the cell potential for each of the following electrochemical cells 1. Copper and aluminum 2. Silver and zinc 3. Copper and nickel 4. Nickel and cadmium 5. Silver and aluminum 6. Magnesium and zinc 7. Lead and iron 8. Copper and magnesium 9. Magnesium and iron 783 Version 1 2012 10. Copper and iron Use the electrochemical series to determine which of the following reactions can occur. those that occur write an appropriate balanced chemical reaction. 11. For Zn + H2SO4 12. Mg + FeCl2 13. Pb + SnCl2 14. Mg + Fe SO4 15. Ag + Cu(NO3)2 16. A1 + CuSO4 17. Zn + AgNO3 784 Version 1 2012 18. Mg + Fe(NO3)2 19. Mg + NaCl 20. FeSO4 + Cu 785 Version 1 2012 Daniel’s Cell 1) If this is a standard cell, what is the concentration of the zinc sulphate solution? 2) What is the pressure and temperature in this standard cell? 3) Write down the 2 half equations for Zn2+ and Cu2+ from the data booklet. Include their voltages. 4) How do you know the zinc electrode is the negative electrode? 5) If it is negative, electrons are being produced here. Does your equation in 3) show this? If not reverse it and replace the ↔ with a . NB: You never have to reverse the sign for the voltage. 6) How do you know the copper electrode is positive? 786 Version 1 2012 7) As it is positive electrons are attracted here. Does your equation in 3) show this? If not reverse it and replace the <--> with a . If it does show the addition of electrons then the forward reaction is correct; replace the ↔ with a . 8) Add the 2 half equations to get the equation for the cell. 9) Find the DIFFERENCE in the 2 voltages. This number must be positive in Chemistry. ___________V 10) Oxidised = _________ Oxidizing Agent: _________ Reduced:_______ Reducing Agent: ___________ 11) Why does the zinc electrode get lighter? 12) Why does the concentration of the copper ions decrease? 13) What are the charge carriers in a) The wires _______________ b) the salt bridge __________________ 14) Which direction do the potassium ions in the salt bridge move and why? 15) If 2 electrons go around the circuit: a) How many zinc ions are produced? _________ b) How many copper atoms are produced? _______ 16) If 2 MOLES of electrons go around the circuit: a) How many zinc ions are produced? _______ b) How many moles of copper atoms are produced? _________ 787 Version 1 2012 17) Which electrode’s mass would change the most? Explain. 19 ) 1) HL: Electrochemical cells can be written in the following shorthand notation: Cu/Cu2+║Zn2+/Zn Most+ Most – What does the ║ mean? ___________________ 788 Version 1 2012 Activity Series Simulation **DATA NOTEBOOK WILL BE ASSESSED** (you can get to this link from my website, unit 8: Redox, first link under Labs) http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redox/home.html Purpose: To determine an activity series of various metals. Procedures: Describe briefly how you will determine the activity series. Include important details like the state of matter of the reactants and how you will know that a reaction has taken place. Data: A large grid is the best with 8 rows (for each metal you are testing) and 9 columns for each solution you will put the solid in. Some boxes of the grid will not be filled in. Record not only if there is a reaction but before and after observations. Click on microscopic views when available and describe what is happening. Analysis 1. 2. 3. 4. 5. 6. 7. Give an order for your activity series and EXPLAIN. USE DATA in your explanation. Compare to the actual activity series Write half reactions for Fe (s) reactions Write the net ionic reaction for iron reactions For each half reaction state whether oxidation or reduction is occurring. For each reaction state how many moles of electrons are transferred per mole of metal atoms. If 2.00 g of aluminum completely reacts with a copper nitrate solution, what mass of copper would be formed? 8. Give at least 3 changes you would see for the Aluminum + Copper nitrate reaction. 9. Using the activity series from your data book a. Which of the following is the most powerful reducing agent? i. Copper ii. Magnesium iii. Iron iv. Zinc b. Zinc reacts with dilute acids to liberate hydrogen. This is because i. The zinc ion is a more powerful oxidizing agent than the hydrogen ion. ii. They hydrogen ion is a more powerful oxidizing agent than the zinc ion. iii. The zinc ion is a more powerful reducing agent than the hydrogen ion. iv. The hydrogen ion is a more powerful reducing agent than the zinc ion. c. Write the order of reactivity for halogens. d. Why is the order for halogens different from alkali metals? e. Br2 is brown, I2 is Pink, and Cl2 has a slightly yellowish green color. State and explain what would you see when i. Chlorine gas is bubbled through aqueous sodium bromide ii. Bromine is added to sodium chloride iii. Chlorine is bubbled through sodium iodide 789 Version 1 2012 Voltaic Cells Notes 1. 9.4.1 Explain how a redox reaction is used to produce electricity in a voltaic cell. (3) a. For those of you not in physics, discuss a brief intro to charges: b. Define Amperes and what it’s a measure of: i. A large current could be produced from (two scenarios): 1. 2. c. Define potential difference and what it means for the activity of a cell: 2. 9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode). (1) a. Draw and Diagram a Voltaic (Daniell) Cell: b. What purpose does the salt bridge serve? 790 Version 1 2012 c. A voltaic cell will continue until: d. How can we remember which electrode is the Cathode and which is the Anode? e. What does the voltage of a cell depend on? f. Provide the shorthand notation for the Daniell cell and describe how to name them. 791 Version 1 2012 Electrolytic Cells Notes 1. 9.5.1 Describe, using a diagram, the essential components of an electrolytic cell. (2) a. Diagram a simple Electrolytic Cell 2. 9.5.2 State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode). (1) a. Define an electrolyte b. During electrolysis, we still follow RedCat and AnOx, what does this mean? 3. 9.5.3 Describe how current is conducted in an electrolytic cell. (2) a. How do we test conductivity? b. What does it mean for a substance to be a conductor or an insulator? 792 Version 1 2012 4. 9.5.4 Deduce the products of the electrolysis of a molten salt. (3) a. What happens when an electrical current passes through an ionic substance (molten or solution)? Diagram, and provide Redox half equations as Diagram, and provide Redox half equations as well as the full equation, for the decomposition well as the full equation, for the decomposition of PbBr2 of NaCl Diagram, and provide Redox half equations as well as the full equation, for the decomposition of H2O 793 Diagram, and provide Redox half equations as well as the full equation, for electroplating a material with Cu Version 1 2012 Electrochemical Cell Tutorial Open http://www.wwnorton.com/college/chemistry/gilbert2/contents/ch18/studyplan.asp in a browser and click on Zinc-Copper Cell **If the address above doesn’t work you can get to the site from my website. In the Redox unit under assignments, click on Electrochemcial Cell tutorial. Directions: Read through the entire tutorial and take notes. Some important notes to take are listed below (110). You should read through all of the tutorial including the parts on the Nernst equation and answer the questions at the end of the tutorial. However, you will not have to perform calculations using the Nernst equation so you can skip these questions. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. Sketch and label the electrochemical cell. Which electrode (anode or cathode) is the negative terminal? Which electrode is the positive terminal? What process (oxidation or reduction) happens at the anode? What process (oxidation or reduction) happens at the cathode? What is current? In what direction do the electrons flow? What is a salt bridge used for? What do you think would happen to the cell if there was no salt bridge? Do you think the cell could continue to function forever? Explain. Read the Chem 1 text book pg 663 to 666 and 673 to 682 and answer the questions on the worksheets provided. Then come back to this worksheet. In the electrochemical cell from the tutorial, a cell was made with copper and zinc. Which metal is more active? In the electrochemical cell from the worksheet, a cell was made with Platinum and Nickel. Which metal is more active? What is the relationship between how active a metal is and whether it is at the anode or cathode? Sketch an electrochemical cell for Magnesium and Iron. You will need to use the activity series. The sketch should identify a. Positive and negative electrode b. Cathode and anode c. ½ reactions taking place at each electrode d. the direction of electron flow e. the direction of flow through the salt bridge 794 Version 1 2012 Voltaic Versus Electrolytic Cells Do some research to answer the following questions. We will take up the answers as a class. 1. Valtaic cells have several other names, list two: 2. Define Anode: 3. Define Cathode: 4. Complete the following table comparing general properties of voltaic and electrolytic cells. Voltaic Electrolytic Spontaneous? (Yes/No) Anode (negative or positive) Cathode (negative or positive) Electrons flow from which terminal?(negative/positive) (anode./cathode) Electrons flow towards which terminal? (negative/positive) (anode/cathode) 795 Version 1 2012 5. How is spontaneity (or lack therefore) explained by the direction of flow of electrons? 6. Below is a diagram of a Voltaic Cell. Label the following components: a. Anode + charge b. Cathode + charge c. Salt Bridge i. What ions are found within the salt bridge? d. Metal Ion Solutions i. Which ions are present in solution at each terminual? e. Zinc Metal, Copper Metal, Zinc Ion, Copper Ion f. Direction of electron flow g. Direction of Na+ flow h. Direction of Cl- flow 6. Write the half reaction occuring at each terminal 7. What purpose does the salt bridge serve? 796 Version 1 2012 8. Below is a diagram of an Electrolytic Cell. Label the following components: a. Anode + charge b. Cathode + charge c. Power Source d. Direction of electron flow e. Deduce and label the products of electrolysis of NaCl 9. State and explain one practical application of an electrolytic cell. 10. Redraw the voltaic cell from question 5 as an electrolytic cell. 797 Version 1 2012 Sodium Chemistry 798 Version 1 2012 799 Version 1 2012 Topic 9 Oxidation and Reduction IB Exam Questions 1. What happens at the negative electrode in a voltaic cell and in an electrolytic cell? Voltaic cell Electrolytic cell A. oxidation reduction B. reduction oxidation C. oxidation oxidation D. reduction reduction (Total 1 mark) 2. Consider how current is conducted in an electrolytic cell. Which statement is correct? A. Electrons move through the electrolyte and the external circuit. B. Ions move through the electrolyte and the external circuit. C. Electrons move through the external circuit and ions move through the electrolyte. D. Electrons move through the electrolyte and ions move through the external circuit. (Total 1 mark) 3. Which species is oxidized in the following reaction? 2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq) A. Ag+ B. Cu C. Ag D. Cu2+ (Total 1 mark) 4. Which list represents the halogens in increasing order of oxidizing strength (weakest oxidizing agent first)? A. Cl2 I2 Br2 B. I2 Br2 Cl2 C. I2 Cl2 Br2 D. Cl2 Br2 I2 (Total 1 mark) 800 Version 1 2012 5. Which are redox reactions? I. 2FeCl2 + Cl2 → 2FeCl3 II. Mg + 2HNO3 → Mg(NO3)2 + H2 III. H2O + SO3 → H2SO4 A. I and II only B. I and III only C. II and III only D. I, II and III (Total 1 mark) 6. Which compound contains nitrogen with an oxidation number of +3? A. NH4Cl B. HNO3 C. N2O4 D. KNO2 (Total 1 mark) 7. In the electrolytic cell shown, at which electrode will chlorine form, and what is the process taking place there? Electrode Process A. P reduction B. Q reduction C. P oxidation D. Q oxidation (Total 1 mark) 801 Version 1 2012 8. Magnesium is higher in the reactivity series than zinc. In the cell shown, in which direction do the electrons flow in wire X and which metal is oxidized? Electron flow Oxidized A. Zn to Mg Zn B. Mg to Zn Zn C. Zn to Mg Mg D. Mg to Zn Mg (Total 1 mark) 9. Outline two differences between an electrolytic cell and a voltaic cell. ............................................................................................................................................... . ............................................................................................................................................... . ............................................................................................................................................... . ............................................................................................................................................... . (Total 2 marks) 10. (i) Define oxidation in terms of oxidation numbers. (1) 802 Version 1 2012 (ii) Describe using a labelled diagram, the essential components of an electrolytic cell. (3) (iii) Explain why solid sodium chloride does not conduct electricity but molten sodium chloride does. (2) (iv) Molten sodium chloride undergoes electrolysis in an electrolytic cell. For each electrode deduce the half-equation and state whether oxidation or reduction takes place. Deduce the equation of the overall cell reaction including state symbols. (5) 803 Version 1 2012 (v) Electrolysis has made it possible to obtain reactive metals such as aluminium from their ores, which has resulted in significant developments in engineering and technology. State one reason why aluminium is preferred to iron in many uses. (1) (vi) Outline two differences between an electrolytic cell and a voltaic cell. (2) (Total 14 marks) 11. (a) Define oxidation in terms of electron transfer. ...................................................................................................................................... ...................................................................................................................................... (1) (b) Chlorine can be made by reacting concentrated hydrochloric acid with potassium manganate(VII), KMnO4. 2KMnO4(aq) + 16HCl(aq) → 2MnCl2(aq) + 2KCl(aq) + 5Cl2(aq) + 8H2O(aq) (i) State the oxidation number of manganese in KMnO4 and in MnCl2. KMnO4 .............................................................................................................. MnCl2 ................................................................................................................ (2) (ii) Deduce which species has been oxidized in this reaction and state the change in oxidation number that it has undergone. ........................................................................................................................... ........................................................................................................................... (2) (Total 5 marks) 804 Version 1 2012 Topic 9 IB Exam Question Markacheme 1. A [1] 2. C [1] 3. B [1] 4. B [1] 5. A [1] 6. D [1] 7. D [1] 8. D [1] 9. electrolytic cell converts electrical energy to chemical energy and voltaic cell converts chemical energy to electrical energy / electrolytic cell uses electricity to carry out a (redox) chemical reaction and voltaic cell uses a (redox) chemical reaction to produce electricity / electrolytic cell requires a power supply and voltaic cell does not; electrolytic cell involves a non-spontaneous (redox) reaction and voltaic cell involves a spontaneous (redox) reaction; in an electrolytic cell, cathode is negative and anode is positive and vice-versa for a voltaic cell / electrolytic cell, anode is positive and voltaic cell, anode is negative / electrolytic cell, cathode is negative and voltaic cell, cathode is positive; voltaic cell has two separate solutions and electrolytic cell has one solution / voltaic cell has salt bridge and electrolytic cell has no salt bridge; electrolytic cell, oxidation occurs at the positive electrode/anode and voltaic cell, oxidation occurs at the negative electrode/anode/vice-versa; If descriptions are reversed for electrolytic and voltaic cell, penalize first marking point but award second marking point as ECF. 2 max [2] 10. (i) increase in the oxidation number; (ii) Annotated diagram of cell showing: 1 805 Version 1 2012 power supply/battery; electrolyte; cathode/negative electrode and anode/positive electrode; 3 (iii) (iv) (v) (vi) (solid) ions in a lattice / ions cannot move; (molten) ions mobile / ions free to move; 2 reduction occurs at the cathode/negative electrode and oxidation occurs at the anode/positive electrode; Cathode/negative electrode: Na+ + e– → Na; Anode/positive electrode: 2Cl– → Cl2 + 2e– / Cl– → 12 Cl2 + e–; Award [1 max] if the two electrodes are not labelled/labelled incorrectly for the two half-equations. Overall cell reaction: Na+(l) + Cl–(l) → Na(l) + 12 Cl2(g) Award [1] for correct equation and [1] for correct state symbols. Allow NaCl(l) instead of Na+(l) and Cl–(l). 5 Al does not corrode/rust / Al is less dense/better conductor/more malleable; Accept Al is a lighter (metal compared to Fe). Accept converse argument. 1 electrolytic cell converts electrical energy to chemical energy and voltaic cell converts chemical energy to electrical energy / electrolytic cell uses electricity to carry out a (redox) chemical reaction and voltaic cell uses a (redox) chemical reaction to produce electricity / electrolytic cell requires a power supply and voltaic cell does not; electrolytic cell involves a non-spontaneous (redox) reaction and voltaic cell involves a spontaneous (redox) reaction; in an electrolytic cell, cathode is negative and anode is positive and vice-versa for a voltaic cell / electrolytic cell, anode is positive and voltaic cell, anode is negative / electrolytic cell, cathode is negative and voltaic cell, cathode is positive; 806 Version 1 2012 voltaic cell has two separate solutions and electrolytic cell has one solution / voltaic cell has salt bridge and electrolytic cell has no salt bridge; electrolytic cell, oxidation occurs at the positive electrode/anode and voltaic cell, oxidation occurs at the negative electrode/anode and vice-versa; 2 max [14] 11. (a) Loss of (one or more) electrons; (b) (i) (ii) 1 (KMnO4) + 7; (MnCl2) + 2; Must have + sign for mark. [1 max] if roman numerals or 7+ or 2+ used or if + signs are missing. Cl– / chloride / chlorine / Cl (has been oxidized) / HCl; oxidation number from –1 to 0 / has increased by one; If HCl is given for first mark, it must be clear that it is the Cl that has the change of oxidation number. 2 2 [5] 807 Version 1 2012