UHS Tutoring. (4) Redox Reactions (02)

UHS Tutoring (4) Redox Reactions (02) 8739 1844 www.uhsinternational.com

UHS Tutoring 4. Oxidationreduction reactions are increasingly important as a source of energy Students learn to: A. Explain the displacement of metals from solution in terms of transfer of electrons B. Identify the relationship between displacement of metal ions in solution by other metals to the relative activity of metals C. Account for changes in the oxidation state of species in terms of their loss or gain of electrons D. Describe and explain galvanic cells in terms of oxidation/reduction reactions E. Outline the construction of galvanic cells and trace the direction of electron flow F. Define the terms anode, cathode, electrode and electrolyte to describe galvanic cells Students: Perform a first-hand investigation to identify the conditions under which a galvanic cell is produced Perform a first-hand investigation and gather firsthand information to measure the difference in potential of different combinations of metals in an electrolyte solution Gather and present information on the structure and chemistry of a dry cell or lead-acid cell and evaluate it in comparison to one of the following: Button cell Fuel cell Vanadium redox cell Lithium cell Liquid junction photovoltaic device (eg the Gratzel cell) In terms of: Chemistry Cost and practicality Impact on society Environmental impact Solve problems and analyse information to calculate the potential requirement of named electrochemical processes using tables of standard potentials and halfequations

What are redox reactions? Redox reactions are a type of displacement reactions (as you will see below). They are an important class of reactions in chemistry and give some students hard time. You need to make sure that you have understood a couple of essential concepts before you get into Redox reactions, some of these concepts include: a. Chemical reactions and electron transfer b. Assigning oxidation numbers (Important for knowing which species loose and which gain electrons when redox reactions happen. You will be asked to identify that in the HSC) c. Writing net ionic equations (Because redox reactions are essentially ionic equations). First let us ask: What is a net ionic equation? A net ionic equation is the equation that takes into consideration the ions that participate in the reaction (and nothing else). Can you remember what spectator ions are? - Steps involved in Redox reactions 1 Balance the equation 2 Assign oxidation numbers 3 Write net ionic equation 4 Notice changes in oxidation numbers 5 Write Redox half reactions 6 Balancing Redox reactions 7 Calculate cell potential

1 Balance the equation As learnt in year 11. 2 Assign Oxidation Numbers There are 6 main rules you use when assigning oxidation numbers. The rules are outlined in the table below: Rule 1 A Neutral element or compound H2 MgBr2 NaNO3 ZERO 1B Ions PO4 3- = -3 NH4 1+ = +1 Sum = charge on that ion 2A In compounds, Group 1A Elements in Group 1 +1 2B In compounds, Group 2A Elements in Group 2 +2 3 In compounds, Fluorine ALWAYS +1 4 In compounds, Hydrogen (Eg.). HCl +1 When hydrogen is bound to a metal, it is a -1 5 In most compounds, Oxygen -2 Exception in peroxides like H 2O 2 it is -1. 6 In 2-element compounds with metals Group 7A = -1 Group 6A = -2 Group 5A = -3 When assigning oxidation numbers, we follow these rules in PRIORITY ORDER. Which means that if the first rule for example is sufficient to assign the oxidation number, we don t have to look at the rule after it.

+ WORKED EXAMPLE What would be the oxidation number of SO 4 in Na 2SO 4? Step 1: See if rule 1A applies. Yes. Rule number 1 applies. Therefore the total charge of the molecule must ZERO. Is this enough to work out the oxidation number of SO 4? No. So keep going. Step 2: See if rule 1B applies. No. We do not have any charges displayed on our compound. Is this enough to work out the oxidation number of SO 4? No. So keep going. Step 3: See if rule 2A applies Yes. Rule 2A applies to Na because it is a metal from group 2 on the periodic table and it must have the charge. Is this enough to work out the oxidation number of SO 4? Yes. So stop here and do the maths. We know now that Na has a charge of +1 so Na 2 has a charge of +2. We know that the total charge is ZERO. (+2) + (SO 4) = 0 (SO 4) = 0 - (+2) (SO 4) = -2

TRY IT YOURSELF 1) Assign an oxidation number for Chlorine in: KClO 4 2) Assign an oxidation number for Sulphur in: K 2SO4 The oxidation number can be a fraction. It does not have to be an integer

3 Write net ionic equations WORKED EXAMPLE Write the complete ionic equation and the net ionic equation for the following reaction: Ca(s) + 2HBr(aq) CaBr2(aq) + H2(g) Solution Step 1: You break down all the aqueous solution compounds into individual ions (Remember: you do not breakdown solids or gases). Ca(s) + 2H + + 2Br - Ca 2+ + 2Br - + H2(g) Complete ionic equation Step 2: Just get rid of the spectator ions Ca(s) + 2H + + 2Br - Ca 2+ + 2Br - + H2(g) Ca(s) + 2H + Ca 2+ + H2(g).. Net ionic equation

TRY IT YOURSELF Write the net ionic equation for the following reaction: 1. BaBr 2 (aq) + Na 2SO 4 (aq) BaSO 4 (s) + 2 NaBr (aq) 2. CuSO4(aq) + Zn(s) ZnSO4(aq) + Cu(s) 3. Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate.

Let us now look at each dot point in details and make sure we understand the exam requirements. A. explain the displacement of metals from solution in terms of transfer of electrons When we say displacement of metals from solution, we mean that one metal is kicking out another metal from the solution and taking its place. You notice that there are two metals, each immersed in a liquid. The first metal (to the left) is losing electrons (e - ) (being converted from Zn(s) 0 into Zn 2+ ions) and the second metal (to the right) is gaining those electrons (being converted from Cu 2+ ions into Cu(s) 0 ). When the Solid Zinc (to the left) starts to convert into Zn 2+ ions (dissolved in water Aqueous), the weight of Solid Zinc will decrease. This means that it is taking the place of Cu 2+ in the solution and kicking it out (displacing it). So the Cu 2+ will have no choice but to return to the solid state Cu(s) 0. We say that Zinc displaced Copper (kicked it out) from its salt solution (We say salt solution because Cu 2+ is not just floating around the solution as an ion, it is actually bound, in the form of CuSO4). And we say that Zinc was able to kick out Cu 2+ because it is a more active metal than Copper (Recall metals activity series from Year 11), so it wins the battle of being in the salt solution form (ZnSO4). Zn(s) + CuSO4(aq) ZnS04(aq) + Cu(s)

What is a displacement reaction? For the following chemical reaction: iron + copper(ii) sulfate iron sulfate + copper a. Write the chemical equation b. Is this a displacement reaction? If yes, which metal is being displaced? Why does the following reaction not happen? iron + magnesium sulfate no reaction Could you keep a solution of copper sulfate in an iron tank? And Why?

B. identify the relationship between displacement of metal ions in solution by other metals to the relative activity of metals Active metals displace less active metals from its salt solution. The greater the difference in activity between the two metals, the more vigorous the displacement reaction As mentioned in the example of Zinc and Copper earlier, zinc is said to be more active that copper, therefore it is more likely to react in the solution and form an aqueous solution. Below is an illustration of the activity series of metals: http://archives.jesuitnola.org/upload/clark/refs/act_series.gif So why is this Activity series useful when it comes to displacement (redox) reactions?

4 Notice changes in oxidation numbers C. Account for changes in the oxidation state of species in terms of their loss or gain of electrons The oxidation state is also called the oxidation number. What is an oxidation state? Notice the difference between X 2+ X +2 When the charge is written before the number (-2) it means we are talking about oxidation number, and when it is written after the number (2-) it means we are talking about the charge. In most cases the oxidation number is the same as the charge (So we can use them interchangeably). A change in the oxidation number of compounds indicate the transfer of electrons (Redox reactions). Oxidation = Loss of electrons (e - ) Reduction = Gain of electrons (e - ) FOR EXAMPLE: 1. Mg 0 (S) + O2 0 (g) 2 Mg +2 O -2 Magnesium is changing its oxidation number from from 0 to +2 and Oxygen from 0 to -2. Does this mean that this reaction is a redox reaction?.. The best method of balancing redox equations is the half-equation method. What is an oxidising agent? What is a reducing agent?

More on Oxidation numbers Determine the oxidation number of the following: Chemical compound Chemical formula Oxidation number Phosphorus in phosphorus trichloride Phosphorus in phosphorus pentoxide Sulfur in sulfur dioxide Carbon in carbon monoxide

4 Write Redox half reactions D. describe and explain galvanic cells in terms of oxidation/reduction reactions Galvanic cells are also know sometimes as voltaic cells. They are a type of redox reactions that happen on their own without the need for a source of energy. Mark S. Cracolice, Edward I. Peters (2011) Introductory Chemistry: An active learning approach, 4 th Edition, Brooks/Cole, Cengage Learning, Page 587. Why do we connect the two half cells by a salt bridge? Examples of Voltaic (Galvanic) cells include: Figure 19.7 Nickel-cadmium (ni-cad) batteries. The fundamental components of these batteries are nickel oxide, nickel hydroxide, and nickel metal, cadmium hydroxide and cadmium metal, and potassium hydroxide. Figure 19.8 Lithium batteries. These batteries are typically made from lithium metal and manganese(iv) oxide. Mark S. Cracolice, Edward I. Peters (2011) Introductory Chemistry: An active learning approach, 4 th Edition, Brooks/Cole, Cengage Learning, Page 587.

E. outline the construction of galvanic cells and trace the direction of electron flow Oxidation happen at the ANODE (+ve) AN OX Reduction happens at the CATHODE (-ve) RED CAT The electrons flow from the Anode (Reducing agent) to the Cathode (Oxidising agent) Each side of the cell is called half-cell. One half represents the oxidation and one half represents the reduction reactions (See figure above). Sketch a Galvanic cell.

F. Define the terms anode, cathode, electrode and electrolyte to describe galvanic cells Galvanic cell: chemical energy electrical energy Electrode: part where electrons flow into/out of half-cells Electrolyte: any solution containing ions What is a cathode? What is an anode? Gather and present information on the structure and chemistry of a dry cell or lead-acid cell and evaluate it in comparison to one of the following: Button cell Fuel cell Vanadium redox cell Lithium cell Liquid junction photovoltaic device (eg the Gratzel cell) In terms of: Chemistry Cost and practicality Impact on society Environmental impact Suggestion: Make a table as follows Cell 1 (eg. Lithium) Cell 2 (Lead-acid cell) Cell 3 Chemistry Cost and practicality Impact on Society Environmental impact

Writing half-equations for redox reactions Remember that that all single displacement reactions are redox reactions. Recall polyatomic ions. What do half reactions represent? WORKED EXAMPLE Let us take the following single displacement (redox) reaction: Cu(s) + 2AgNO3 Cu(NO3)2 + 2Ag(s) Step 1: Assign oxidation numbers We do that to find out what is being oxidised and what is being reduced in order to assign the oxidation and reduction half reactions. According to the rules above (refer to assigning oxidation numbers above), we find that the oxidation numbers are as follows: Cu(s) 0 + 2Ag +1 NO3-1 Cu +2 (NO3-1 )2 + 2Ag(s) 0 Step2: Notice the changes in oxidation numbers Cu(s) 0 Cu +2... More positive, therefore loosing electrons (Oxidation). Ag +1 Ag(s) 0... More negative, therefore loosing electrons (Reduction) Step 3: Write the oxidation and reduction reactions taking into account the electrons being transferred. Cu(s) 0 Cu +2 + 2e -... Oxidation Ag +1 + 1e - Ag(s) 0... Reduction Notice that this reaction is not balanced (The oxidation reaction is losing 2 electrons but the reduction is gaining only 1).

TRY IT YOURSELF Write half reactions for the following redox reactiona: NH3 + O2 NO2 + H2O 2Na(s) + Cl2(g) 2NaCl

6 Balance Redox reactions Remember: Redox reactions happen in solutions (Ions floating around). In addition to balancing the atoms, we must balance the charges. To balance redox reactions, we use a methods called Half Equation Method. We separate the equation into two half-equations (Oxidation and Reduction). 1. Balancing redox reactions in neutral solutions You will notice in these equations that all protons (H + ) will cancel out. Worked example Balance the following reaction Cu (aq) + Fe(s) Fe 3 (aq) + Cu(s) Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions: Cu (aq) + e Cu(s) Fe 3 (aq) + 3e Fe(s) The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields: Cu (aq) + e Cu(s) Fe(s) Fe 3 (aq) + 3e Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire Cu+(aq)+e Cu(s) half-reaction by 3 and leaving the other half reaction as it is. This gives: 3 Cu (aq) + 3e 3Cu(s) Fe(s) Fe 3 (aq) + 3e Step 3: Adding the equations give: 3 Cu (aq) + 3e + Fe(s) 3Cu(s) + Fe 3 (aq) + 3e The electrons cancel out and the balanced equation is left. 3 Cu + (aq) + Fe(s) 3Cu(s) + Fe 3+ (aq)

2. Balancing redox reactions in acidic solutions In acidic we get hydrogen and oxygen atoms and in addition to balancing the charges (like we do in neutral solutions) we balance the Hydrogen and Oxygen. Problem-Solving Strategy The Half-Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Acidic Solution 1. Write separate equations for the oxidation and reduction half-reactions. 2. For each half-reaction: A. Balance all the elements except hydrogen and oxygen. B. Balance oxygen using H2O. C. Balance hydrogen using H. D. Balance the charge using electrons. 3. If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions. 4. Add the half-reactions, and cancel identical species. 5. Check that the elements and charges are balanced.

Worked example Balancing the equation for the reaction between Permanganate and iron(ii) ions in acidic solution: 1. Assign oxidation numbers and write half reactions noticing the oxidation (loss of electrons) and reduction (gain of electrons) Reduction reaction Oxidation reaction (Steven S. Zumdahl & Susan A. Zumdahl, 2014)

(Steven S. Zumdahl & Susan A. Zumdahl, 2014)

(Steven S. Zumdahl & Susan A. Zumdahl, 2014) Try it yourself Potassium dichromate (K2Cr2O7) is a bright orange compound that can be reduced to a blueviolet solution of Cr31 ions. Under certain conditions, K2Cr2O7 reacts with ethanol (C2H5OH) as follows: Balance this equation using the half-reaction method.

2. Balancing redox reactions in basic solutions Problem-Solving Strategy The Half-Reaction Method for Balancing Equations for Oxidation Reduction Reactions Occurring in Basic Solution 1. Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H ions were present. 2. To both sides of the equation obtained above, add a number of OH ions that is equal to the number of H ions. (We want to eliminate H by forming H2O.) 3. Form H2O on the side containing both H1 and OH ions, and eliminate the number of H2O molecules that appear on both sides of the equation. 4. Check that elements and charges are balanced. (Steven S. Zumdahl & Susan A. Zumdahl, 2014) Worked example Balance the following redox reaction in basic solution:

(Steven S. Zumdahl & Susan A. Zumdahl, 2014)

7 Calculate cell potential Solve problems and analyse information to calculate the potential requirement of named electrochemical processes using tables of standard potentials and half-equations. Cell potential is abbreviated with the symbole: E. The superscript indicates standardstate conditions. All standard Calculating E Potential for electrochemical cells relies heavily on understanding how to write half reactions for redox reactions, especially when you are not given the half reactions in the question. Therefore, we are going to learn how to write half equations for redox reactions. What is cell potential? What is standard cell potential? If a standard cell potential is E cell = +0.85 V at 25 C, is the redox reaction of the cell spontaneous? What are the factors that affect cell potential?

PRACTICE QUESTIONS 1. A galvanic cell under standard conditions is represented below. (5 marks) (a) On the diagram, clearly label the anode, the cathode and the direction of electron flow. (b) Write a balanced net ionic equation for the overall cell reaction. (c) Calculate the standard cell potential (E ).

(d) Explain any colour changes observed in this cell as the reaction proceeds. 2. An electrochemical cell is constructed using two half cells. One half cell consists of an inert platinum electrode and a solution of Fe 2+ and Fe 3+. The other half cell consists of a lead electrode and a solution of Pb 2+. (7 Marks) Current will flow from one electrode to the other electrode when the cell is completed using a voltmeter and a salt bridge. (a) Write relevant half equations and a balanced net ionic equation for the overall cell reaction. (b) Calculate the standard cell potential (E ).

(c) Identify the anode, cathode, metals and ions by labelling the following diagram (d) Identify an appropriate electrolyte to use in the salt bridge.