1 / 19

Chemistry 1011

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

arvid
Télécharger la présentation

Chemistry 1011

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chemistry 1011 TOPIC Electrochemistry TEXT REFERENCE Masterton and Hurley Chapter 18 Chemistry 1011 Slot 5

  2. 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 M1 | M12+ | | M2| M22+ Chemistry 1011 Slot 5

  3. Oxidation-Reduction Reactions • In simple terms, oxidation is the addition of oxygen to an element Mg(s) + O2(g) 2MgO(s) • Here the Mg is oxidized Mg  Mg2+ + 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 Chemistry 1011 Slot 5

  4. Oxidation-Reduction Reactions • When zinc reacts with hydrochloric acid: Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) • The zinc atoms lose electrons – they are oxidized • The H+ ions gain electrons – they are reduced OXIDATION: Zn  Zn2+ + 2e- REDUCTION: 2H+ + 2e-  H2 • Oxidation and reduction take place together • There is no net change in the total number of electrons Chemistry 1011 Slot 5

  5. 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 Cl2 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 Chemistry 1011 Slot 5

  6. Oxidation Number • What is the oxidation number of Mn in MnO4- ? • 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 NH3, N2, N2O, NO, NO2? • -3, 0, +1 +2 +4 Chemistry 1011 Slot 5

  7. 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  Zn2+ + 2e- Cl2 + 2e-  2Cl- Chemistry 1011 Slot 5

  8. 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 H2O Chemistry 1011 Slot 5

  9. Balancing More Complex Half Reactions • Balance: MnO4-(aq) Mn2+(aq) (acid solution) • Oxidation number of Mn in MnO4-(aq) is 7+ • Oxidation number of Mn in Mn2+(aq) is 2+ • 5 electrons must be added to the Left Hand Side of the equation to change oxidation number by 5 MnO4-(aq) + 5e- Mn2+(aq) • 8 H+ must be added to the LHS to balance for charge 8H+(aq) + MnO4-(aq) + 5e- Mn2+(aq) • 4 H2O must be added to the RHS to balance completely 8H+(aq) + MnO4-(aq) + 5e- Mn2+(aq) + 4H2O(l) Chemistry 1011 Slot 5

  10. Balancing More Complex Half Reactions • Balance: SO42-(aq) SO2(g) (acid solution) • Oxidation number of S in SO42-(aq) is ? • Oxidation number of S in SO2(g) is ? • ?? electrons must be added to the LHS? or RHS? of the equation to change oxidation number by ?? SO42-(aq) + ??  SO2(g) • ?? H+ must be added to the LHS? or RHS? to balance for charge • ?? H2O must be added to the LHS? or RHS? to balance completely Chemistry 1011 Slot 5

  11. 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 Chemistry 1011 Slot 5

  12. Voltaic Cells • When a strip of zinc is placed in a solution of copper ions, a redox reaction takes place: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Half reactions: Zn  Zn2+ + 2e- oxidation Cu2+(aq) + 2e- Cu(s) reduction • This reaction is spontaneous Chemistry 1011 Slot 5

  13. 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” Chemistry 1011 Slot 5

  14. The Zinc – Copper Voltaic Cell • Cell representation: Zn | Zn2+|| Cu2+| Cu • anode (oxidation) on left; • cathode (reduction) on right; • || indicates salt bridge • | indicates phase boundary Chemistry 1011 Slot 5

  15. Voltaic Cell Operation • The zinc strip is the ANODE • (oxidation occurs - Zn  Zn2+ + 2e- ) • The copper strip is the CATHODE • (reduction occurs - Cu2+(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 Chemistry 1011 Slot 5

  16. Other Metal | Metal Ion Voltaic Cells • Any two different metals can be used to produce an electric current Cu | Cu 2+|| Ag+| Ag Zn | Zn2+|| Fe2+| Fe Al | Al3+|| Cu2+| Cu Zn Cu LEMON Chemistry 1011 Slot 5

  17. The Zinc – Hydrogen Voltaic Cell • Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Half reactions: Zn  Zn2+ + 2e- oxidation 2H+(aq) + 2e- H2(g) reduction Zn | Zn2+|| H+| H2 | Pt • The hydrogen half cell consists of hydrogen gas bubbled over a platinum electrode, submerged in a solution of acid Chemistry 1011 Slot 5

  18. The Zinc – Hydrogen Voltaic Cell Chemistry 1011 Slot 5

  19. 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 Chemistry 1011 Slot 5

More Related