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1 of 35© Boardworks Ltd 2010. 2 of 35© Boardworks Ltd 2010.

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Presentation on theme: "1 of 35© Boardworks Ltd 2010. 2 of 35© Boardworks Ltd 2010."— Presentation transcript:

1 1 of 35© Boardworks Ltd 2010

2 2 of 35© Boardworks Ltd 2010

3 3 of 35© Boardworks Ltd 2010 Recognising redox reactions

4 4 of 35© Boardworks Ltd 2010 Oxidation and reduction

5 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 6 of 35© Boardworks Ltd 2010 Assigning oxidation states

7 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 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 8 of 35© Boardworks Ltd 2010 What is the oxidation state?

9 9 of 35© Boardworks Ltd 2010 Redox reactions a summary

10 10 of 35© Boardworks Ltd 2010

11 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 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 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 14 of 35© Boardworks Ltd 2010 Combining half equations

15 15 of 35© Boardworks Ltd 2010

16 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 17 of 35© Boardworks Ltd 2010 Cells and electrode potentials

18 18 of 35© Boardworks Ltd 2010 Combining half cells 1

19 19 of 35© Boardworks Ltd 2010 The standard hydrogen electrode

20 20 of 35© Boardworks Ltd 2010 Combining half cells 2

21 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 = – = V. E ө = VE ө = V The left cell is being oxidized while the right is being reduced.

22 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 Ag + (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 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 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 = VCu 2+ (aq) + 2e - Cu (s) E ө = V = V

25 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 26 of 35© Boardworks Ltd 2010 Calculating E cell

27 27 of 35© Boardworks Ltd 2010

28 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 = VCu 2+ (aq) + 2e - Cu (s) E ө = 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 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 = VCu 2+ (aq) + 2e - Cu (s) E θ = V

30 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 31 of 35© Boardworks Ltd 2010 Predicting the direction of redox reactions

32 32 of 35© Boardworks Ltd 2010

33 33 of 35© Boardworks Ltd 2010 Glossary

34 34 of 35© Boardworks Ltd 2010 What’s the keyword?

35 35 of 35© Boardworks Ltd 2010 Multiple-choice quiz


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