1. Voltaic Cells/Galvanic Cells and Batteries

2. Background Information Electricity is the movement of electrons, and batteries are an important source of electrical energy. Batteries work because they have chemicals inside that undergo a reaction where electrons can be transferred from one place to another. These reactions are always REDOX reactions where electrons are transferred. We are going to examine how these reactions take place, and also how to calculate how much energy can be obtained from them.

3. Voltaic Cells Voltaic cells are one type of system used in rechargeable batteries Invented by Alessandro Volta in 1800. A voltaic cell is an electrochemical cell that use chemical reactions to produce electrical energy. Voltaic Cells use the spontaneous movement of electrons to perform electrical work. The electrons will flow between two parts of the cell called the electrodes.

4. The Basics The light bulb is a example of a load where current (the flow of electrons) is used to do work. In this case, light a bulb. Two different metals and different solutions are used on each side. Electrons flow along a wire connecting the two metals because of the Redox reactions happening on each side.

5. The Important Parts The anode is an electrode where oxidation occurs. The cathode is an electrode where reduction occurs. A salt bridge is a chamber of electrolytes necessary to balance charges and complete the circuit. The oxidation and reduction reactions are separated into compartments called half-cells.

6. The Big Picture!

7. Steps of a Voltaic Cell Part 1: Electrons are lost by the metal as they go into solution. If electrons are lost, then the Oxidation reaction is occuring. This always happens at the ANODE. –Shown in oxidation half-reaction The mass of the Zinc metal decreases because the atoms are dissolving into the solution. Electrons enter the wire connecting the two half cells.

8. Steps of a Voltaic Cell Part 2: Electrons flow along the wire to the other half cell. The electrons flow down into the metal and are gained by the positive ions in the solution If electrons are gained, then the Reduction reaction is occurring. This always occurs at the CATHODE. –Shown in the reduction half reaction The ions in solution are attracted to the electrons and become part of the metal. This causes the Mass to increase!

9. The Salt Bridge The salt bridge connects the two half cells together and allows ions to flow. It can be filled with ions that are liquid or gel, and that allows ions to move. The salt ions do not react with the solutions. These ions BALANCE the charges that develop in the solutions in each half cell.

11. Notation for a voltaic cell Part of the anode (oxidation half-cell) Parts of the cathode (reduction half-cell) lower oxidation state (Metal) higher oxidation state (Solution) lower oxidation state (Metal) higher oxidation state (Solution) Oxidation half cell is listed first! (Alphabetical) The elements are separated by a single line. Double line separates the two half cell reactions and represents a salt bridge. Zn|Zn 2+ ||Cu 2+ |Cu

12. What do we need to know when solving problems? For voltaic cell problems, you need to know what metals are involved and what is happening to their electrons. One metal will be part of the oxidation reaction while the other metal will be part of the reduction reaction. To help determine which metal does what, we have a table of reduction potentials. AKA how motivated is a metal to undergo “Reduction”?

13. Reduction Potential Chart The table shows just the REDUCTION reactions for substances. The E o value represents the voltage when that reaction happens.

14. Reading the Tables A positive voltage value indicates the metal will favorably undergo the reduction reaction. What do you think a negative value indicates? Negative values indicate the metal does not want to undergo reduction, and is more inclined to undergo an oxidation reaction. This chart ONLY shows values for reduction reactions. If we want to determine the oxidation energy we will have to flip the reaction, and what happens when we flip reactions around?

15. How to use the Table: If we are using two metals such as Zn and Cu in a voltaic cell, we must decide which one will oxidize and which will reduce. Find Copper(II) and Zinc(II) on your charts: Cu +2 + 2e - => Cu +0.34v Zn +2 + 2e - => Zn -0.76v Copper is higher on the chart, so it has a greater tendency to be reduced. Zinc is lower, so it will be oxidized.

16. Cell Potential / Voltage Cell Potential is the “driving force” of energy through a circuit. This is also known as the Voltage. It is represented by E cell or E O. To determine the cell potential in a voltaic cell, add the reduction potential of the cathode to the oxidation potential of the anode.

17. Determine the Cell Potential: Find Copper(II) and Zinc(II) on your charts: Cu +2 + 2e - => Cu +0.34 v Zn +2 + 2e - => Zn -0.76 v Remember that the oxidation reaction needs to be flipped from the chart, and the sign must flip as well! Cu +2 + 2e - => Cu +0.34 v Zn => Zn +2 + 2e - +0.76 v Total Cell Potential : = 1.10 v

18. Practice Problems Problem 1: Draw the shorthand notation for a cell containing magnesium and copper(II). 1) Figure out the Half Reactions –Find Mg and Cu on the list and determine which is reduced and which is oxidized. The higher half reaction will be reduced. –Remember you need to flip the equation for Mg. It’s a reduction chart, so flip for oxidation.

19. 2) Oxidation is written first in shorthand notation. Separate reactions by one line: Mg|Mg 2+ 3) Reduction comes next. Separate Oxidation from Reduction with two lines. Mg|Mg 2+ ||Cu 2+ |Cu

20. Part 2 Determine the cell potential for the previous cell setup: Mg|Mg 2+ ||Cu 2+ |Cu 1)Remember the oxidation is written first. The oxidation potential value needs to be flipped from the chart: Mg(s)  2e - + Mg 2+ =2.37 volts 2) The reduction potential can come straight from the chart: Cu 2+ + 2e -  Cu(s)=0.34 volts 2.37v + 0.34v = 2.71 volts

21. Part 3 Draw an illustration of a voltaic cell based on this cell notation: Mg|Mg 2+ ||Cu 2+ |Cu Include in your illustration: I) Location of Cathode and Anode II) Location of atoms and ions III) Flow of ions and electrons

22. Anode is oxidation Mg Cu 2+ Mg 2+ Cathode is Reduction Cu Mg 2+ Cu 2+ 2.71 volts E-E- E-E- K+K+ Anion Salt Bridge NO 3 - Cation

23. Sample Problem 21.2Diagramming voltaic cells PROBLEM:Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO 3 ) 3 solution, another half-cell with an Ag bar in an AgNO 3 solution, and a KNO 3 salt bridge. PLAN: SOLUTION: Identify the oxidation and reduction reactions and write each half- reaction. Identify which metal is anode and cathode. voltmeter oxidation half-reaction: Cr( s ) Cr 3+ ( aq ) + 3e - reduction half-reaction: Ag + ( aq ) + e - Ag( s ) overall (cell) reaction: Cr( s ) + Ag + ( aq ) Cr 3+ ( aq ) + Ag( s ) Cr Cr 3+ Ag Ag + K+K+ NO 3 - salt bridge e-e- Cr( s ) | Cr 3+ ( aq ) || Ag + ( aq ) | Ag( s )