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Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Electrochemistry TEXT REFERENCE Masterton and Hurley Chapter 18.

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Presentation on theme: "Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Electrochemistry TEXT REFERENCE Masterton and Hurley Chapter 18."— Presentation transcript:

1 Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Electrochemistry TEXT REFERENCE Masterton and Hurley Chapter 18

2 Chemistry 1011 Slot 52 18.1 Voltaic Cells YOU ARE EXPECTED TO BE ABLE TO: Identify oxidation-reduction reactions (Review Chapter 4.3) Assign oxidation numbers to elements in molecules and ions (Review Chapter 4.3) Balance equations for oxidation-reduction (Redox) reactions (Review Chapter 4.3) Identify the anode and cathode in an electrochemical cell Construct a labelled diagram to show the structure of an electrochemical voltaic cell and describe its operation Represent a voltaic cell using the symbolism M 1 | M 1 2+ | | M 2 | M 2 2+

3 Chemistry 1011 Slot 53 Oxidation-Reduction Reactions In simple terms, oxidation is the addition of oxygen to an element Mg (s) + O 2(g)  2MgO (s) Here the Mg is oxidized Mg  Mg 2+ + 2e  In this oxidation process, the magnesium loses electrons In general, in an oxidation-reduction (Redox) reaction, one species loses electrons and the other gains electrons The species gaining electrons is reduced The species losing electrons is oxidized

4 Chemistry 1011 Slot 54 Oxidation-Reduction Reactions When zinc reacts with hydrochloric acid: Zn (s) + 2HCl (aq)  ZnCl 2(aq) + H 2(g) The zinc atoms lose electrons – they are oxidized The H + ions gain electrons – they are reduced OXIDATION:Zn  Zn 2+ + 2e  REDUCTION:2H + + 2e    H 2 Oxidation and reduction take place together There is no net change in the total number of electrons

5 Chemistry 1011 Slot 55 Oxidation Number The oxidation number is an arbitrary number assigned to an element in a compound to indicate the charge that would be present if it were an ion The oxidation number of a pure element is 0 –Oxidation number of chlorine in Cl 2 is 0 Oxidation number of an element in a monatomic ion is equal to the charge on the ion –Oxidation number of chlorine in NaCl is –1 In compounds, the oxidation number of oxygen is usually –2 In compounds, the oxidation number of hydrogen is usually +1

6 Chemistry 1011 Slot 56 Oxidation Number What is the oxidation number of Mn in MnO 4  ? –Charge due to 4O will be –8 –Ion has overall charge of –1 –Mn is assigned oxidation number of +7 What is the oxidation number of N in NH 3, N 2, N 2 O, NO, NO 2 ? -3, 0, +1 +2 +4

7 Chemistry 1011 Slot 57 Balancing Simple Half Reactions For simple ions the balanced half reaction must be balanced for mass and charge –The numbers of atoms must be the same on both sides of the equation –The net charge must be the same on both sides of the equation Zn  Zn 2+ + 2e  Cl 2 + 2e   2Cl 

8 Chemistry 1011 Slot 58 Balancing More Complex Half Reactions Identify the element being oxidized or reduced Assign oxidation numbers to the element on both sides of the equation Identify change in oxidation number Add electrons to balance change in oxidation number Balance charge by adding H + or OH - Balance hydrogen by adding H 2 O

9 Chemistry 1011 Slot 59 Balancing More Complex Half Reactions Balance: MnO 4  (aq)  Mn 2+ (aq) (acid solution) –Oxidation number of Mn in MnO 4  (aq) is 7+ –Oxidation number of Mn in Mn 2+ (aq) is 2+ 5 electrons must be added to the Left Hand Side of the equation to change oxidation number by 5 MnO 4  (aq) + 5e   Mn 2+ (aq) 8 H + must be added to the LHS to balance for charge 8 H + (aq) + MnO 4  (aq) + 5e   Mn 2+ (aq) 4 H 2 O must be added to the RHS to balance completely 8 H + (aq) + MnO 4  (aq) + 5e   Mn 2+ (aq) + 4 H 2 O (l)

10 Chemistry 1011 Slot 510 Balancing More Complex Half Reactions Balance: SO 4 2  (aq)  SO 2(g) (acid solution) –Oxidation number of S in SO 4 2  (aq) is ? –Oxidation number of S in SO 2(g) is ? ?? electrons must be added to the LHS? or RHS? of the equation to change oxidation number by ?? SO 4 2  (aq) + ??  SO 2(g) ?? H + must be added to the LHS? or RHS? to balance for charge ?? H 2 O must be added to the LHS? or RHS? to balance completely

11 Chemistry 1011 Slot 511 Voltaic Cells In a voltaic cell, chemical energy is converted to electrical energy In a voltaic cell, an oxidation-reduction (redox) reaction takes place The voltaic cell is constructed so that the electrons produced by the oxidation half reaction must travel round an external circuit to become available for the reduction half reaction

12 Chemistry 1011 Slot 512 Voltaic Cells When a strip of zinc is placed in a solution of copper ions, a redox reaction takes place: Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s) Half reactions: Zn  Zn 2+ + 2e  oxidation Cu 2+ (aq) + 2e   Cu (s) reduction This reaction is spontaneous

13 Chemistry 1011 Slot 513 Designing the Voltaic Cell Dip zinc metal into zinc ions (zinc sulfate solution) in one beaker Dip copper metal into copper ions (copper sulfate solution) in a second beaker Link the two solutions with a “salt bridge” Link the two metal strips with a wire Electrons flow from the zinc to the copper through the wire Ions flow through the “bridge”

14 Chemistry 1011 Slot 514 The Zinc – Copper Voltaic Cell Cell representation: Zn | Zn 2+ || Cu 2+ | Cu –anode (oxidation) on left; –cathode (reduction) on right; –|| indicates salt bridge –| indicates phase boundary

15 Chemistry 1011 Slot 515 Voltaic Cell Operation The zinc strip is the ANODE –(oxidation occurs - Zn  Zn 2+ + 2e  ) The copper strip is the CATHODE –(reduction occurs - Cu 2+ (aq) + 2e   Cu (s) ) The electrons flow around the external circuit from the anode to the cathode Zinc ions are produced in solution around the zinc strip Copper ions are removed from solution around the copper strip Ions migrate across the salt bridge to maintain electrical neutrality

16 Chemistry 1011 Slot 516 Other Metal | Metal Ion Voltaic Cells Any two different metals can be used to produce an electric current Cu | Cu 2+ || Ag + | Ag Zn | Zn 2+ || Fe 2+ | Fe Al | Al 3+ || Cu 2+ | Cu LEMON Zn Cu

17 Chemistry 1011 Slot 517 The Zinc – Hydrogen Voltaic Cell Zn (s) + 2HCl (aq)  ZnCl 2(aq) + H 2(g) Zn (s) + 2H + (aq)  Zn 2+ (aq) + H 2(g) Half reactions: Zn  Zn 2+ + 2e  oxidation 2H + (aq) + 2e   H 2(g) reduction Zn | Zn 2+ || H + | H 2 | Pt The hydrogen half cell consists of hydrogen gas bubbled over a platinum electrode, submerged in a solution of acid

18 Chemistry 1011 Slot 518 The Zinc – Hydrogen Voltaic Cell

19 Chemistry 1011 Slot 519 Structure and Operation of Voltaic Cells A Summary A voltaic cell consists of two half cells joined by – an external circuit through which electrons flow, and by – a salt bridge through which ions move Each half cell consists of an electrode dipping into an aqueous solution –If a metal participates in the cell reaction it will usually be the electrode; otherwise an inert platinum electrode is used In one half cell oxidation occurs at the anode and reduction occurs at the cathode –The overall cell reaction is the sumof the half reactions taking place at anode and cathode


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