Chapter 20: Electrochemistry

Chia sẻ: Lê Minh Huy | Ngày: | Loại File: PPT | Số trang:59

Thêm vào BST

Báo xấu

39
lượt xem 3
download

Download Vui lòng tải xuống để xem tài liệu đầy đủ

Balancing oxidation-reduction equations, oxidation-reduction reactions, voltaic cells, cell EMF, effect of concentration on cell EMF is the main content of the lecture "Chapter 20: Electrochemistry". Invite you to consult the detailed content lectures to capture details.

Chủ đề:

Bình luận(0) Đăng nhập để gửi bình luận!

Lưu

Nội dung Text: Chapter 20: Electrochemistry

Electrochemistry David P. White University of North Carolina, Wilmington Chapter 20 Copyright 1999, PRENTICE HALL Chapter 20 1
OxidationReduction Reactions • Zn added to HCl yields the spontaneous reaction Zn(s) + 2H+(aq) Zn2+(aq) + H2(g). • The oxidation number of Zn has increased from 0 to 2+. • The oxidation number of H has reduced from 1+ to 0. • Therefore, Zn is oxidized to Zn2+ while H+ is reduced to H2. • H+ causes Zn to be oxidized and is the oxidizing agent. • Zn causes H+ to be reduced and is the reducing agent. • Note that the reducing agent is oxidized and the oxidizing agent is reduced. Copyright 1999, PRENTICE HALL Chapter 20 2
Balancing OxidationReduction Equations • Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end. • Conservation of charge: electrons are not lost in a chemical reaction. • In complicated redox reactions, we need to look at the transfer of electrons carefully. HalfReactions • Halfreactions are a convenient way of separating oxidation and reduction reactions. Copyright 1999, PRENTICE HALL Chapter 20 3
Balancing OxidationReduction Equations HalfReactions • The halfreactions for Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe3+(aq) are Sn2+(aq) Sn4+(aq) +2e 2Fe3+(aq) + 2e 2Fe2+(aq) • Oxidation: electrons are products. • Reduction: electrons are reagents. Copyright 1999, PRENTICE HALL Chapter 20 4
Balancing OxidationReduction Equations Balancing Equations by the Method of Half Reactions • Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple). • MnO4 is reduced to Mn2+ (pale pink) while the C2O42 is oxidized to CO2. • The equivalence point is given by the presence of a pale pink color. • If more KMnO4 is added, the solution turns purple due to the excess KMnOChapter 20 Copyright 1999, PRENTICE HALL 4. 5
Balancing OxidationReduction Equations Balancing Equations by the Method of Half Reactions Copyright 1999, PRENTICE HALL Chapter 20 6
Balancing OxidationReduction Equations Balancing Equations by the Method of Half Reactions • What is the balanced chemical equation? 1. Write down the two half reactions. 2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H+. d. Finish by balancing charge by adding electrons. Copyright 1999, PRENTICE HALL Chapter 20 7
Balancing OxidationReduction Equations Balancing Equations by the Method of Half Reactions 3. Multiply each half reaction to make the number of electrons equal. 4. Add the reactions and simplify. 5. Check! For KMnO4 + Na2C2O4: Copyright 1999, PRENTICE HALL Chapter 20 8
Balancing OxidationReduction Equations Balancing Equations by the Method of Half Reactions 1. The two incomplete half reactions are MnO4(aq) Mn2+(aq) C2O42(aq) 2CO2(g) 2. Adding water and H+ yields 8H+ + MnO4(aq) Mn2+(aq) + 4H2O • There is a charge of 7+ on the left and 2+ on the right. Therefore, 5 electrons need to be added to the left: 5e + 8H+ + MnO4(aq) Mn2+(aq) + 4H2O Copyright 1999, PRENTICE HALL Chapter 20 9
Balancing OxidationReduction Equations Balancing Equations by the Method of Half Reactions • In the oxalate reaction, there is a 2 charge on the left and a 0 charge on the right, so we need to add two electrons: C2O42(aq) 2CO2(g) + 2e 3. To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying gives: Copyright 1999, PRENTICE HALL Chapter 20 10
Balancing OxidationReduction Equations Balancing Equations by the Method of Half Reactions 10e + 16H+ + 2MnO4(aq) 2Mn2+(aq) + 8H2O 5C2O42(aq) 10CO2(g) + 10e 4. Adding gives: 16H+(aq) + 2MnO4(aq) + 5C2O42(aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g) 5. Which is balanced! Copyright 1999, PRENTICE HALL Chapter 20 11
Balancing OxidationReduction Equations Balancing Equations for Reactions Occurring in Basic Solution • We use OH and H2O rather than H+ and H2O. • The same method as above is used, but OH is added to “neutralize” the H+ used. Copyright 1999, PRENTICE HALL Chapter 20 12
Voltaic Cells • The energy released in a spontaneous redox reaction is used to perform electrical work. • Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit. • Voltaic cells are spontaneous. • If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+. • Zn is spontaneously oxidized to Zn2+ by Cu2+. • The Cu2+ is spontaneously reduced to Cu0 by Zn. • The entire process is spontaneous. Copyright 1999, PRENTICE HALL Chapter 20 13
Voltaic Cells Copyright 1999, PRENTICE HALL Chapter 20 14
Voltaic Cells • Voltaic cells consist of – Anode: Zn(s) Zn2+(aq) + 2e2 – Cathode: Cu2+(aq) + 2e Cu(s) – Salt bridge (used to complete the electrical circuit): cations move from anode to cathode, anions move from cathode to anode. • The two solid metals are the electrodes (cathode and anode). • As oxidation occurs, Zn is converted to Zn2+ and 2e. The electrons flow towards the anode where they are used in the reduction reaction. Copyright 1999, PRENTICE HALL Chapter 20 15
Voltaic Cells • We expect the Zn electrode to lose mass and the Cu electrode to gain mass. • “Rules” of voltaic cells: 1. At the anode electrons are products. (Oxidation) 2. At the cathode electrons are reagents. (Reduction) 3. Electrons cannot swim. • Electrons flow from the anode to the cathode. • Therefore, the anode is negative and the cathode is positive. • Electrons cannot flow through the solution, they have to be transported through an external wire. (Rule 3.) Copyright 1999, PRENTICE HALL Chapter 20 16
Voltaic Cells Copyright 1999, PRENTICE HALL Chapter 20 17
Voltaic Cells • Anions and cations move through a porous barrier or salt bridge. • Cations move into the cathodic compartment to neutralize the excess negatively charged ions (Cathode: Cu2+ + 2e Cu, so the counterion of Cu is in excess). • Anions move into the anodic compartment to neutralize the excess Zn2+ ions formed by oxidation. Copyright 1999, PRENTICE HALL Chapter 20 18
Voltaic Cells A Molecular View of Electrode Processes • Consider the spontaneous redox reaction between Zn(s) and Cu2+(aq). • During the reaction, Zn(s) is oxidized to Zn2+(aq) and Cu2+(aq) is reduced to Cu(s). • On the atomic level, a Cu2+(aq) ion comes into contact with a Zn(s) atom on the surface of the electrode. • Two electrons are directly transferred from the Zn(s) (forming Zn2+(aq)) to the Cu2+(aq) (forming Cu(s)). Copyright 1999, PRENTICE HALL Chapter 20 19
Cell EMF • The flow of electrons from anode to cathode is spontaneous. • Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode. • Potential difference: difference in electrical potential. Measured in volts. • One volt is the potential difference required to impart one joule of energy to a charge of one coulomb: 1 J 1 V 1 C Copyright 1999, PRENTICE HALL Chapter 20 20