We think you have liked this presentation. If you wish to download it, please recommend it to your friends in any social system. Share buttons are a little bit lower. Thank you!
Presentation is loading. Please wait.
Published byAlexia Lovely
Modified over 2 years ago
1 of 35© Boardworks Ltd 2010
2 of 35© Boardworks Ltd 2010
3 of 35© Boardworks Ltd 2010 Recognising redox reactions
4 of 35© Boardworks Ltd 2010 Oxidation and reduction
5 of 35© Boardworks Ltd 2010 Oxidation states Oxidation states, sometimes called oxidation numbers, are a simple way of keeping track of redox reactions, so that it is easy to see which species has been oxidized and which reduced. sodium atomsodium ion + oxidation state = 0 oxidation state = +1 The oxidation state of an element in a compound is equivalent to the number of electrons it has lost or gained by forming bonds. They also help in balancing equations for redox reactions.
6 of 35© Boardworks Ltd 2010 Assigning oxidation states
7 of 35© Boardworks Ltd 2010 Assigning oxidation states Some elements have the same oxidation state in all their compounds, or in most with some specific exceptions. It is useful to memorise the following examples: +1 except in hydrides of group I/II metals, e.g. NaH, where it is -1 -2 except in peroxides (-1) and in compounds with fluorine (+2) -1 except in compounds with F and O where it has positive values hydrogen oxygen fluorine chlorine ElementOxidation state in compounds
8 of 35© Boardworks Ltd 2010 What is the oxidation state?
9 of 35© Boardworks Ltd 2010 Redox reactions a summary
10 of 35© Boardworks Ltd 2010
11 of 35© Boardworks Ltd 2010 Half equations Chloride ions can be oxidised to produce chlorine. The half equation for this reaction is: All redox reactions can be illustrated using half equations. Half equations can be combined to give the equation for the overall redox process. Half equations are used to show the loss or gain of electrons when a species undergoes oxidation or reduction. 2Cl – Cl 2 + 2e - 0 One element in a half equation changes oxidation state. Here chlorine has changed its oxidation state from -1 to 0.
12 of 35© Boardworks Ltd 2010 Combining half equations To combine half equations: Step 1: Write the half equations. (You may need to work these out if complex ions and other species such as H + are involved.) Step 2: Make sure that the number of electrons in each half equation is the same, so that the electrons cancel out. Do this by multiplying one or both equations to make the number of electrons the same in each case. Step 3: Add the half equations and cancel the electrons. It may be possible to cancel other species that appear on both sides – often H + or H 2 O.
13 of 35© Boardworks Ltd 2010 Combining half equations – example Chlorine oxidizes iron(II) to iron(III) and is itself reduced to chloride ions. Write a balanced equation for this reaction. Step 3: Add the half equations and cancel the electrons. Cl 2(aq) + 2Fe 2+ (aq) 2Cl - (aq) + 2Fe 3+ (aq) 2Fe 2+ (aq) 2Fe 3+ (aq) + 2e - Step 2: Eqn. A involves 2 electrons and Eqn. B involves 1 electron, so multiply both sides of Eqn. B by two. chlorine has been reduced Step 1: Write the half equations. Eqn. A Eqn. B Cl 2(aq) + 2e - 2Cl - (aq) Fe 2+ (aq) Fe 3+ (aq) + e - iron(II) has been oxidized
14 of 35© Boardworks Ltd 2010 Combining half equations
15 of 35© Boardworks Ltd 2010
16 of 35© Boardworks Ltd 2010 What is a half cell? If a rod of metal is dipped into a solution of its own ions, an equilibrium is set up. For example: Zn (s) Zn 2+ (aq) + 2e - This is a half cell and the strip of metal is an electrode. The position of the equilibrium determines the potential difference between the metal strip and the solution of metal. zinc sulfate solution (1 mol dm -3 ) zinc metal strip
17 of 35© Boardworks Ltd 2010 Cells and electrode potentials
18 of 35© Boardworks Ltd 2010 Combining half cells 1
19 of 35© Boardworks Ltd 2010 The standard hydrogen electrode
20 of 35© Boardworks Ltd 2010 Combining half cells 2
21 of 35© Boardworks Ltd 2010 Representing half cells: cell diagrams An electrochemical cell can be represented in a shorthand way by a cell diagram. The double vertical lines represents a salt bridge. The single lines represent a phase change between the solid metal and the aqueous metal ions. Zn (s) | Zn 2+ (aq) || Cu (aq) | Cu (s) The half cell with the greatest negative potential is on the left of the salt bridge, so E cell = E right cell – E left cell. In this case, E cell = +0.34 – -0.76 = +1.10 V. E ө = -0.76 VE ө = +0.34 V The left cell is being oxidized while the right is being reduced.
22 of 35© Boardworks Ltd 2010 The electrochemical series The electrochemical series is a list of standard electrode potentials (E ө ). The equilibria are written with the electrons on the left of the arrow, i.e. as a reduction. +0.80Ag + (aq) / Ag (s) +0.34Cu 2+ (aq) / Cu (s) 02H + (aq) / H 2(g) -0.76Zn 2+ (aq) / Zn (s) -2.36Mg 2+ (aq) / Mg (s) E ө / VHalf equation Half cell Mg 2+ (aq) + 2e - Mg (s) Zn 2+ (aq) + 2e - Zn (s) 2H + (aq) + 2e - H 2(g) Cu 2+ (aq) + 2e - Cu (s) Ag + (aq) + e - Ag (s) Electrodes with negative values of E ө are better at releasing electrons (i.e. better reducing agents) than hydrogen.
23 of 35© Boardworks Ltd 2010 Calculating E cell The e.m.f of an electrochemical cell, E cell, is the difference between the standard electrode potentials of the two half cells. E cell = E ө (positive electrode) – E ө (negative electrode) This can be worked out from the electrode potentials values in the electrochemical series. The positive electrode is taken to be the least negative half cell, and the negative electrode is the most negative half cell.
24 of 35© Boardworks Ltd 2010 Calculating E cell : worked example An electrochemical cell is set up using the two half reactions below. What potential difference E cell would this cell generate? E cell = E ө (positive electrode) – E ө (negative electrode) The zinc half cell has the more negative potential and so forms the negative electrode. Therefore: E cell = (+0.34) – (-0.76) Zn 2+ (aq) + 2e - Zn (s) E ө = -0.76 VCu 2+ (aq) + 2e - Cu (s) E ө = +0.34 V = +1.10 V
25 of 35© Boardworks Ltd 2010 Calculating E cell : combining half equations To find the overall reaction occurring in the cell as a whole, the two half equations are added together: Because the zinc half cell forms the negative electrode of the cell, oxidation occurs at this electrode and the half equation must be reversed: The two half equations are added to give the overall cell reaction: Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) Zn (s) Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e - Cu (s)
26 of 35© Boardworks Ltd 2010 Calculating E cell
27 of 35© Boardworks Ltd 2010
28 of 35© Boardworks Ltd 2010 Predicting the direction of redox reactions (1) It is possible to use standard electrode potentials to decide on the feasibility of a reaction. Electrodes with more negative electrode potentials have a lower tendency to accept electrons. Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) Zn 2+ (aq) + 2e - Zn (s) E ө = -0.76 VCu 2+ (aq) + 2e - Cu (s) E ө = +0.34 V When a pair of electrodes are connected, electrons flow from the more negative to the more positive. The signs of the electrodes can be used to predict the direction of the reaction. Zn 2+ (aq) + Cu (s) → Zn (s) + Cu 2+ (aq) feasible not feasible
29 of 35© Boardworks Ltd 2010 Predicting the direction of redox reactions (2) Worked example Fe 3+ (aq) + e - → Fe 2+ (aq) Electrons flow to the more positive terminal of a cell, which is Fe 3+ (aq) /Fe 2+ (aq). This half cell will accept electrons, a reduction reaction occurs, and the half equation is: Step 1: write the equations for the two half reactions: What reaction would occur if Fe 3+ (aq) /Fe 2+ (aq) and Cu 2+ (aq) /Cu (s) half cells were connected? Fe 3+ (aq) + e - Fe 2+ (aq) E θ = +0.77 VCu 2+ (aq) + 2e - Cu (s) E θ = +0.34 V
30 of 35© Boardworks Ltd 2010 Predicting the direction of redox reactions (3) An oxidation reaction occurs in the other half cell. Electrons are produced and the half equation is: 2Fe 3+ (aq) + Cu (s) → Cu 2+ (aq) + 2Fe 2+ (aq) Step 2: combine the two half equations to give a full equation for the reaction. Cu (s) → Cu 2+ (aq) + 2e - The number of electrons supplied and donated must be equal, so the reaction in the Fe 3+ (aq) /Fe 2+ (aq) half cell must occur twice for each Cu 2+ (aq) /Cu (s) half cell reaction: The reverse reaction is not feasible.
31 of 35© Boardworks Ltd 2010 Predicting the direction of redox reactions
32 of 35© Boardworks Ltd 2010
33 of 35© Boardworks Ltd 2010 Glossary
34 of 35© Boardworks Ltd 2010 What’s the keyword?
35 of 35© Boardworks Ltd 2010 Multiple-choice quiz
14 Redox Equilibria 14.1 Redox Equations (Review) 14.2 Electrode Potentials and the Electrochemical Series 14.3 Predicting the Direction of Redox Reactions.
Title: Lesson 7 Standard Electrode Potential Learning Objectives: – Describe the standard hydrogen electrode – Define the term standard electrode potential.
Redox reactions. Definitions of oxidation and reduction Oxidation.
Title: Lesson 4 Voltaic Cells
Topic 9 Oxidation and Reduction Introduction Oxidation numbers Redox equations Reactivity Voltaic cells Electrolytic cells.
Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Electrochemistry TEXT REFERENCE Masterton and Hurley Chapter 18.
Cell Potential L.O.: Construct redox equations using half- equations or oxidation numbers. Describe how to make an electrochemical cell.
3.1.7 Redox. A redox reaction is one in which both reduction and oxidation take place at the same time. The original definition of oxidation was the formation.
Zn Zn2+ + 2e- (oxidation) Cu e- Cu (reduction)
Redox Reactions Year 11 Chemistry ~ Unit 2.
GALVANIC AND ELECTROLYTIC CELLS
Redox Equilibria. Redox equilibria When a metal electrode is placed into a solution of one of its salts two things can happen; 1) Metal ions go into solution;
Oxidation Reduction Reactions
Objectives Define oxidation and reduction in terms of electron loss and gain. Deduce the oxidation number of an element in a compound. State the names.
Chapter 18 Electrochemistry Lesson 1. Electrochemistry 18.1Balancing Oxidation–Reduction Reactions 18.2 Galvanic Cells 18.3 Standard Reduction Potentials.
Objective: Determine the equivalence point. Equivalence point n OH - = n H + If 25.00mL of M NaOH is needed to react with mL of HCl. What is.
Oxidation and Reduction
Title: Lesson 2 Redox Equations
Oxidation Numbers & Redox Reactions How to Make Balancing Redox Reactions a Relatively Painless Process.
Copyright Sautter REVIEW OF ELECTROCHEMISTRY All electrochemical reactions involve oxidation and reduction. Oxidation means the loss of electrons.
© 2017 SlidePlayer.com Inc. All rights reserved.