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Galvanic Cell. Galvanic cell A galvanic cell is a device in which chemical energy is changed to electrical energy. A galvanic cell uses a spontaneous.

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Presentation on theme: "Galvanic Cell. Galvanic cell A galvanic cell is a device in which chemical energy is changed to electrical energy. A galvanic cell uses a spontaneous."— Presentation transcript:

1 Galvanic Cell

2 Galvanic cell A galvanic cell is a device in which chemical energy is changed to electrical energy. A galvanic cell uses a spontaneous redox reaction to produce an electric current that can be used to do work. The system does work on the surroundings. A redox reaction involves the transfer of electrons from the reducing agent to the oxidizing agent.

3 Oxidation v reduction Oxidation. * Involves a loss of electrons. * Increase in oxidation number. * “To get more positive.” * Occurs at the anode of a galvanic cell. Reduction. * Involves a gain of electrons. * Decrease in oxidation number. * “To get more negative.” * Occurs at the cathode of a galvanic cell.

4 Production of Current Oxidation Reactions involve a transfer of electrons. Electric current is a movement of electrons. In order to produce a usable current, the electrons must be forced across a set path (circuit). In order to accomplish this, an oxidizing agent and something to oxidize must be separated from a reducing agent with something to reduce.

5 Pictures Oxidation reduction reaction in the same container will have electrons transferring, but we can’t harness them. Separating the oxidation reaction from the reduction reaction, but connecting them by a wire would allow only electrons to flow. Oxidation Reduction oxidation reduction

6 Closer look We now have excess electrons being formed in the oxidizing solution and a need for electrons in the reducing solution with a path for them to flow through. However, if electrons did flow through the wire it would cause a negative and positive solution to form. OxidationReduction X → X + + e-X + + e- → X

7 That’s not possible Or at least it would require a lot of energy. A negative solution would theoretically be formed by adding electrons, and a positive one by removing electrons. The negative solution would then repel the electrons and stop them from flowing in, and a positive solution would attract the electrons pulling them back where they came from. Making it so the charged solutions wouldn’t form. In order for this to work, I would need a way for ions to flow back and forth but keeping the solutions mostly separated.

8 The Salt Bridge or the Porous Disk. These devices allow ion flow to occur (circuit completion) without mixing the solutions. They are typically made of sodium sulfate or potassium nitrate

9 Closer look Now electrons can flow across the wire from the oxidation reaction to the reduction reaction. As the oxidation reaction becomes positive, it removes negative ions and adds positive ions to the salt bridge. The reduction reaction does the reverse. Oxidation Reduction X → X + + e-X + + e- → X Salt Bridge e-e- e-e- e-e-

10 Closer look Zooming in on the oxidizing side This would make the salt bridge positive… Oxidation side e- Salt Bridge e-e- e-e- + ion - ion

11 Closer look (Zooming in on the reducing side) if the reverse wasn’t happening on this side. Reducing side e- Salt Bridge e-e- e-e- + ion - ion

12 Close up of salt bridge The ions keep flowing in the salt bridge to keep everything neutral. Electrons do also travel across the salt bridge. This decreases the cell’s effectiveness. + ion - ion + ion - ion

13 Electrochemical cell This is the basic unit of a battery. It is also called a galvanic cell, most commercial batteries have several galvanic cells linked together. Batteries always have two terminals. The terminal where oxidation occurs is called the anode. The terminal where reduction occurs is called the cathode.

14 Galvanic Cell

15 Copyright © Cengage Learning. All rights reserved 15 Voltaic Cell: Cathode Reaction

16 Copyright © Cengage Learning. All rights reserved 16 Voltaic Cell: Anode Reaction

17 Copyright © Cengage Learning. All rights reserved 17 Electrochemical Half-Reactions in a Galvanic Cell

18 Cell Potential (E cell ) Cell potential (electromotive force, emf) is the driving force in a galvanic cell that pulls electrons from the reducing agent in one compartment to the oxidizing agent in the other. The volt (V) is the unit of electrical potential. Electrical charge is measured in coulombs (C). A volt is 1 joule of work per coulomb of charge transferred: 1 V = 1 J/C. A voltmeter is a device which measures cell potential.

19 Standard Reduction Potentials The measured potential of a voltaic cell is affected by changed in concentration of the reactants as the reaction proceeds and by energy losses due to heating of the cell and external circuit. In order to compare the output of different cells, the standard cell potential (E o cell ) is obtained at 298 K, 1 atm for gases, 1 M for solutions, and the pure solid for electrodes.

20 The Standard Hydrogen Electrode is considered the reference half-cell electrode, with a potential equal to 0.00 V. It is obtained when platinum is immersed in 1 M H + (aq), through which H 2 (g) is bubbled.

21 The Standard Electrode (Half-Cell) Potential (E half-cell ) A standard electrode potential always refers to the half-reaction written as a reduction. Oxidized form + n e -  reduced form E o half-cell If you need the oxidation, you will have to reverse the reaction Reversing a reaction changes the sign of the potential. E o cell = E o reduction + E o oxidation

22 Spontaneous reactions As the potential increases in value (more positive), the reaction is more likely to occur (spontaneity occurs). E o cell must be positive for the cell to produce electricity. A substance will have a spontaneous reaction another substance with a lower E o cell. Although some half-reactions must be manipulated with coefficients, NEVER MULTIPLY THE E o cell BY THE COEFFICIENT!!!

23 Galvanic Cell Problems Consider a galvanic cell based on the following reactions. For each give the balanced half cell reactions and calculate E o cell. Cu 2+ (aq) + Zn (s)  Zn 2+ (aq) + Cu (s) Fe 3+ (aq) + Cu (s)  Fe 2+ (aq) + Cu 2+ (aq) Al (s) + Cd 2+ (aq)  Cd(s) + Al 3+ (aq)

24 Calculating an Unknown E o half-cell from E o cell. A voltaic cell based on the reaction between aqueous Br 2 and vanadium (III) ions has E o cell = 1.39 V: Br 2 (aq) +2V 3+ (aq) + 2H 2 O(l)  2VO 2+ (aq) + 4H + (aq) +2Br - (aq) What is the standard electrode potential for the reduction of VO 2+ to V 3+ ?

25 Line Notations. The components of the anode compartment are written to the left of the cathode compartment. Double vertical lines separate the half-cells and represents the wire and salt bridge. Within each half-cell, a single vertical line represents a phase boundary. A comma separates half-cell components in the same phase. Half-cell components appear in the same order as in the half-reaction, while electrodes appear at the extreme left and right of the notation. Mg (s) Mg 2+ (aq) Al 3+ (aq) Al (s) Fe (s) Fe 2+ (aq) H +, MnO 4 - (aq), Mn 2+ (aq) Pt (s)

26 Describing a Galvanic Cell. A cell will always run spontaneously in the direction that produces a positive cell potential. A complete description of a galvanic cell always includes four items: 1. The cell potential (always positive for a galvanic cell) and the balanced cell reaction. 2. The direction of electron flow, obtained by inspecting the half-reactions and using the direction that gives a positive E o cell.

27 3. Designation of the anode and cathode. 4. The nature of each electrode and the ions present in each compartment. A chemically inert conductor is required if none of the substances participating in the half-reaction is a conducting solid.

28 Diagramming Voltaic Cells In one compartment of a voltaic cell, a graphite rod dips into an acidic solution of K 2 Cr 2 O 7 and Cr(NO 3 ) 3 ; in the other, a tin bar dips into a Sn(NO 3 ) 2 solution. A KNO 3 salt bridge joins the half-cells. The tin electrode is negative relative to the graphite. Diagram the cell, show balanced equations, and write the cell notation.

29 Description of a Galvanic Cell. Describe completely the galvanic cell based on the following half-reactions under standard conditions: Ag + + e -  AgE o cell = 0.80 V (1) Fe 3+ + e -  Fe 2+ E o cell = 0.77 V (2) In addition, draw the cell and write the line notation.


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