Week 10 Group Seatwork Answers and Notes PDF
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This document provides notes and answers for a week 10 group chemistry exercise. It covers oxidation-reduction reactions, redox balancing, and discusses how to balance equations in acidic and basic solutions. The document includes examples of half-reactions and balanced redox reactions.
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WEEK 10 GROUP SEATWORKANSWERSANDNOTES PART I. Review: Calculating the oxidation state of an element in a compound. OXIDATION-REDUCTION REACTIONS xidation-reduction reactions, orredoxreactions,are reactions in which one or more O elements change their oxidation state. ...
WEEK 10 GROUP SEATWORKANSWERSANDNOTES PART I. Review: Calculating the oxidation state of an element in a compound. OXIDATION-REDUCTION REACTIONS xidation-reduction reactions, orredoxreactions,are reactions in which one or more O elements change their oxidation state. he term “oxidation” originated from the reaction of oxygen gas with metals, in which a T metal, for example Fe, was converted to its oxide: 2 Fe(s) + 3/2 O2 (g) → Fe2O 3 (s) In this case, the Fe was converted from an elemental form, Fe, to the ions Fe2+ and Fe3+. Today, we define the wordoxidationas theincreaseof the oxidation numberof an atom. he term “reduction”, in turn, originated from the reaction of a metal oxide to form the metal. T This reaction resulted in a reduction of mass of the solid as water was being produced, hence the term “reduction”. For example, heating copper(I) oxide in hydrogen gas formed copper metal and water: Cu2O(s) + H2 (g) → 2 Cu(s) + H2O (g) Today, we define the wordreductionas thedecreaseof the oxidation numberof an atom. reaction is called anoxidation-reductionreactionbecause oxidation and reduction A cannot happen separately. A chemical species that undergoes oxidation always has a partner species that undergoes reduction in the same equation. For example, consider the reaction of copper(II) ions and zinc metal: Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) In this reaction, copper(II) ions are converted to copper metal. Since copper(II) ion has an oxidation number of +2, and copper metal an oxidation number of 0, the oxidation number of Cu isdecreased, and the copper(II) ions are saidto have beenreducedto copper metal. Likewise, zinc metal is converted to zinc(II) ions in the same reaction. Since zinc metal has an oxidation number of 0, and zinc(II) ion has an oxidation number of +2, the oxidation number of Zn isincreased, and the zinc metal is saidto have beenoxidizedto zinc(II) ions. he reason for these changes happening together in the same reaction is because there is T anexchange of electrons. Copper(II) iongainselectrons,while zinc metalloseselectrons: u2+ +2 e- → Cu C ( reduction) Zn → Zn2+ +2 e- (oxidation) e refer to these two halves of the reaction ashalf-reactions.Note that the number of W electrons lost in the oxidation half-reaction are gained in the reduction half-reaction. The number of electrons gained in one and lost in the othermust be equal. e can see that areductionreaction involves thegainof electrons, and anoxidation W reaction involves thelossof electrons. This is thebasis for theOIL RIGmnemonic for remembering which half-reactions are oxidation half-reactions and which are reduction half-reactions: “Oxidationis lo ss,reductionis ga in” 1) Classify the following half-reactions as oxidation or reduction half-reactions: a) Cr → Cr3+ + 3 e- oxidation b) Al3+ + 3 e- → Alreduction c) F2 + 2 e- → 2 F- reduction + - d) H2O → O2 + 4 H + 4 e oxidation BALANCING OXIDATION-REDUCTION REACTIONS BY THE ION-ELECTRON METHOD ost oxidation-reduction reactions happen in the presence of water, most of them in M + - aqueous solution. In aqueous solution there exist H2O molecules, H3O ions, and OH ions. - These species can participate in an oxidation-reduction reaction. H2O molecules and OH + + ions can participate directly, while H3O can givea proton, H : + + H3O (aq) → H2O(l) + H (aq) his is the basis for the balancing technique known as theion-electron method. This works T by recognizing two basic principles: The totalnumbers of atomsmust be the same on bothsides. The totalchargesmust be the same on both sides. o start, divide the oxidation-reduction reactions into half-reactions. Then balance each one T separately, and then combine the balanced half-reactions together. Balancing Half-Reactions or balancing the half-reactions, there are special steps for balancing hydrogen and oxygen F atoms: AddH+, to balancehydrogenon either side. AddH2O , to balanceoxygenon either side. Follow the steps below for balancing half-reactions: tep 1:For each half-reaction, determine which elementis undergoing a change in oxidation S state. Step 2:Balance the element determined in Step 1 (ifnot hydrogen or oxygen). Step 3:Balance oxygen by adding H2O on either side. Step 4:Balance hydrogen by adding H+on either side. Step 5:Balance the charges by adding the electronsto the side with more positive charge. 2) Balance the following half-reactions: a) Cu → Cu2+Cu → Cu2+ + 2 e- - Balance the Cu on both sides. - There are no hydrogen or oxygen atoms to balance, so all atoms in the half-reaction are balanced. - The charge on the left side is 0, and the charge on the right side is +2, so add 2 electrons to the right side: Cu → Cu2+ + 2 e- b) MnO4- → Mn2+ MnO4- + 8 H++ 5 e- → Mn2+ + 4 H2O - Balance the Mn on both sides. - There are 4 O atoms on the left side, and 0 O atoms on the right side, so add 4 H2O molecules to the right side: MnO4- → Mn2+ +4 H2O - Now there are 0 H atoms on the left side, but 8 H atoms on the right side. Add 8 H+ ions to the left side: MnO4- +8 H+→ Mn2+ + 4 H2O - Check that each side has the same number of Mn, H, and O atoms. - The charge on the left side is +7 and the charge on the right side is +2, so add 5 electrons to the left side: MnO4- + 8 H++5 e- → Mn2+ + 4 H2O 2+ 3+ 2+ + - 3+ c) VO → V VO + 2 H + e → V + H2O (solveon your own) d) W → WO42 - W + 4 H2O → WO 2- 4 + 8 + H + 6 e- (solveon your own) e) SO32 - → HS- SO32 - + 7 H+ + 6 e- → HS-+ 3 H2O (solveon your own) Combining Balanced Half-Reactions fter the half-reactions are combined, it is time to combine them. Let’s take as an example A the unbalanced equation Cu + MnO4- → Cu2+ + Mn2+ hich we separate into two half-reactions and identify as either the oxidation or reduction w half-reaction based on the change in oxidation state: u → Cu2+ C ( oxidation) MnO4- → Mn2+ (reduction) and, balancing them both, we obtain u → Cu2+ +2 e- C nO4- + 8 H++5 e- → Mn2+ + 4 H2O M e align the arrows to clearly see that the electrons in the top half-reaction are on the right, W while the electrons in the bottom half-reaction are on the left. To combine these two, multiply both equations to get theleast common multiple (LCM)of the two numbers of electrons. For this case, there are 2 electrons in the top equation and 5 electrons in the bottom equation. The LCM of 2 and 5 is 10, so each equation should have 10 electrons. The only way to do this is to multiply the top half-reaction by 5: 5 Cu → 5 Cu2+ +10 e- and to multiply the bottom half-reaction by 2: 2 MnO4- + 16 H++10 e- → 2 Mn2+ + 8 H2O Now the two can be combined. The electrons will be canceled out completely: Cu → 5 Cu2+ +10 e- 5 2 MnO4- + 16 H++10 e- → 2 Mn2+ + 8 H2O 5 Cu + 2 MnO4- + 16 H+→ 5 Cu2+ +2 Mn2+ + 8 H2O The resulting equation is balanced both in mass and charge. (Please check.) Reactions in Acidic or Basic Medium The form of a redox reaction differs if the pH of the aqueous solution is acidic or basic. + - In acidic medium, H2O molecules and H3O ions areplentiful, while OH ions are scarce. So + we expect the equation to haveH2O moleculesandH ions. Therefore, the above method for balancing equations is sufficient. In basic medium, however, H2O molecules and OH- ionsare plentiful, while H3O + ions are - scarce. So we expect the equation to haveH2 O moleculesandOH ions. After combining the equations in the above method, we can transform the equation byadding as many OH- ions as there are H+ ionsto both sides of the equation. Consider the balanced equation we derived above: 5 Cu + 2 MnO4- + 16 H+→ 5 Cu2+ + 2 Mn2+ + 8 H2O If the reaction occurs in acidic medium, this is ok. But if it occurs in basic medium, we need to add OH- ions to both sides. Since the equationhas 16 H+ ions, we should add16 OH- ionsto both sides: 5 Cu + 2 MnO4- +16 H++16 OH- → 5 Cu2+ + 2 Mn2+ + 8 H2O - +16 OH The H+ and OH- ions should then be combined to form16 H2O molecules: 5 Cu + 2 MnO4- +16 H2O→ 5 Cu2+ + 2 Mn2+ + 8 H2O - + 16 OH And, because H2O is now found on both sides, someH2 O molecules should be cancelled: 5 Cu + 2 MnO4- + 168 H2O→ 5 Cu2+ + 2 Mn2+ +8 H2O - + 16 OH The final balanced equation is 5 Cu + 2 MnO4- + 8 H2O 2+ 2+ - → 5 Cu + 2 Mn + 16 OH - nd we can see that it contains H2O a molecules andOH ions, as expected for a reaction occurring in basic solution. Actually this reaction simplifies to 5 Cu + 2 MnO4- + 8 H2O → 5 Cu(OH)2 + 2 Mn(OH)2 + 2OH- ecause Cu2+ and Mn2+ form their hydroxides in basicsolution, both of which precipitate as b solids. 3) Balance the following equations: a) Cr2+ + MnO2 → Cr3+ + Mn(OH)2 (basic solution) b) MnO4- + C2O 2- → Mn2+ + CO2 (acidic solution) 4 c) AsO3 + I2 → AsO43- + I- (basic solution) 3- d) VO2+ + HClO → V2+ + HClO2 (acidic solution) PART II. THE VOLTAIC CELL AND BATTERIES s we saw in the previous section, redox reactions involve anexchange of electrons A during the reaction. But this was not known to early scientists in the early 1800s, because the electron particle had not been discovered then. Scientists at that time thought that electricity was a fluid, with a current (its “speed”), that flowed from object to object. he scientist Luigi Galvani was the first to discover electricity flow in biological tissues. When T two dissimilar metals touched each other, and both of them touched the same frog leg, the frog leg twitched. Galvani thought this was because animals had electricity. However, Alessandro Volta challenged this thinking, because he thought that the electricity flow had something to do with the fact that the metals were dissimilar, and the frog leg merely contained a substance that allowed the flow. o prove this, he made what is now recognized as the first battery, called avoltaic pile, to T demonstrate that electricity flows when two dissimilar metals are connected by a cardboard soaked in brine (salt solution). ome twenty years later, Michael Faraday was able to use such a battery to generate S electricity for electrolysis. Faraday also first used the wordelectrodeto describe the metals in the battery, and the wordelectrolyteto describethe solution used to wet the cardboard or paper in between the metals. The electrode that receives the flow of electricity from the electrolyte he called theanode(Greek for “upwardpath”), and the electrode that gives the flow of electricity to the electrolyte he called thecathode(Greek for “downward path”). Since the whole assembly is a self-sustaining system producing electricity, this was also called a cell, after the already-discovered unit of biologicallife. oday, with the discovery of the electron as a particle, we know that it is the electrons that T travel through the electrodes. One half-reaction supplies the electrons, which travel from one end of the cell, through the anode and the cathode, to the other end of the cell, where the electrons are received by the reactants of the other half-reaction. To complete the circuit, a salt bridge containing the electrolyte is added. his type of cell is called avoltaic cell(for AlessandroVolta). It produces anelectric T current, and therefore generates anelectric potential.This potential (also known as voltage) is measured in volts. Each of the two compartmentsof this cell is called ahalf-cell. CELL NOTATION e do not need to draw a voltaic cell to refer to it each time. Rather we usecell notationto W describe a specific cell and write acell diagram. onsider the cell in the picture in the previous section, and assume that the concentrations C of Zn2+ in the left half-cell and Cu2+ in the righthalf-cell are both 1.0 mol/L, or 1.0 M. The cell diagram for this cell can be written as Zn(s) | Zn2+ (1.0 M) || Cu2+ (1.0 M) | Cu(s) he phase boundaries (between solid and solution) are marked with a single vertical line (|), T and the boundaries between the half-cell solutions are marked with two vertical lines (||), representing the salt bridge, because there are two phase boundaries in this area (one between the Zn2+ solution and the salt bridge, andone between the salt bridge and the Cu2+ solution). In a cell diagram, the solid in theleft half-cellis theanode, while the solid in theright half cellis thecathode. A half-cell can be referred toas itself, like: Zn(s) | Zn2+ (1.0 M) the Zn metal here clearly being the anode. Or: Cu2+ (1.0 M) | Cu(s) the Cu metal here clearly being the cathode. REACTIONS AT THE ELECTRODES The half-reactions happening at the electrodes depend on which electrode is considered. heanodereceives the electrons, and the electronsmove upward from it. So the reactions T happening at the anode involve aloss of electronsfrom the reactant. They are therefore oxidationhalf-reactions. hecathodedonates the electrons, and the electronsmove downward from it. So the T reactions happening at the cathode involve againof electronsby the reactant. They are thereforereductionhalf-reactions. he key mnemonic for remembering this is to remember thatan T ode andoxidation both begin with vowels, andcathode andreduction bothbegin with consonants. You may combine this with theOIL RIGmnemonic:a-O ILc-RIG. This means: node →ox idation →loss of electrons a cathode →re duction →ga in of electrons 4) C onstruct a cell diagram for the voltaic cell powered by the following reactions, using 0.1 M solutions of each ion. Write the half-reactions involved at the cathode and the anode: a) Cu2+(aq) + Mg(s) → Mg2+(aq) + Cu(s) b) Fe3+(aq) + H2 (g) + OH-( aq) → Fe2+(aq) + 2 H2O (Usesolid platinum electrodes for both sides.) c) 2 Al(s) + 3 Zn2+(aq) + 8 OH-(aq) → 2 Al(OH)4-(aq) + 3 Zn(s) THE ACTIVITY SERIES AND THE ELECTROMOTIVE FORCE efore the invention of the Periodic Table, chemists used to classify metals based on their B “activity”, meaning how reactive they were toward water. They made some observations: Some metals, like potassium, react violently with cold liquid water. Other metals, like calcium, react vigorously, but not violently, with cold liquid water. There are metals, such as magnesium, which react quietly with cold liquid water, but more vigorously with hot liquid water. There are metals, such as aluminum, which do not react with liquid water at all, but react with steam (water vapor) Some metals, like iron, do not react with water, but react with acids like hydrochloric acid. Some metals, like copper, do not react even with hydrochloric acid, but react with nitric acid or sulfuric acid. Some metals, like silver, require concentrated nitric or sulfuric acid to dissolve. Finally, gold and platinum only dissolve in a mixture of concentrated nitric acid and concentrated hydrochloric acid. These observations led chemists to make the following observations as well: If copper metal is added to a solution of aluminum sulfate, there is no reaction. But if aluminum metal is added to a solution of copper sulfate, a reaction occurs and copper metal is formed. If sodium metal is added to a solution of zinc sulfate, a reaction occurs and zinc metal is formed. But if zinc metal is added to a solution of sodium sulfate, no reaction occurs. If silver is added to a solution of ferrous sulfate (or iron(II) sulfate), no reaction occurs. But if iron is added to a solution of silver nitrate, a reaction occurs and silver metal is formed. Chemists formulated the activity series, based on these observations: In general,metals with high reactivityreact onlywithmetal ions of the metals with lower reactivity. If a metal with low reactivity is mixedwith a metal ion of a metal with high reactivity, no reaction occurs. 5) B ased on the activity series, which pairs of reactants are expected to have a reaction? a) lithium metal and nickel(II) ion b) tin metal and copper(II) ion c) aluminum metal and iron(II) ion d) calcium metal and sodium ion e) cobalt metal and magnesium ion oday, we know that this activity series is based on the redox reactions occurring between T the metal and the metal ion, specifically if themetalis able tobe oxidizedby themetal ion, which will itself bereduced. or example, the reaction of lithium metal and nickel(II) ion can be written as two F half-reactions: i(s) → Li+(aq) + e- L (oxidation) Ni2+(aq) + 2 e- → Ni(s) (reduction) ote that the reduction reaction written is just the inverse of the oxidation reaction in the N above figure. The full reaction is, therefore, 2 Li(s) + Ni2+(aq) → 2 Li+( aq) + Ni(s) e also know that the “activity” of the more reactive metal is due to its giving electrons. W Faraday described this phenomenon by saying that the more reactive metal provides the electromotive force, the “push”, for the electronsto move. he electromotive force, oremf, is a quantityofenergy per charge,measured in volts. It T corresponds to the electric potential (or voltage) generated by a cell running on a particular redox reaction, so the emf of a cell is also called thecell potential. It is represented by E (for emf).
