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Electrochemistry Electrochemical reactions are oxidation-reduction reactions. The two parts of the reaction are physically separated. –The oxidation reaction.

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Presentation on theme: "Electrochemistry Electrochemical reactions are oxidation-reduction reactions. The two parts of the reaction are physically separated. –The oxidation reaction."— Presentation transcript:

1 Electrochemistry Electrochemical reactions are oxidation-reduction reactions. The two parts of the reaction are physically separated. –The oxidation reaction occurs in one cell. –The reduction reaction occurs in the other cell. There are two kinds electrochemical cells. 1.Electrochemical cells containing nonspontaneous chemical reactions are called electrolytic cells. 2.Electrochemical cells containing spontaneous chemical reactions are called voltaic or galvanic cells.

2 Electrical Conduction Metals conduct electric currents well in a process called metallic conduction. In metallic conduction there is electron flow with no atomic motion. In ionic or electrolytic conduction ionic motion transports the electrons. –Positively charged ions, cations, move toward the negative electrode, cathode. –Negatively charged ions, anions, move toward the positive electrode, anode.

3 Electrodes The following convention for electrodes is correct for either electrolytic or voltaic cells: The cathode is the electrode at which reduction occurs. The cathode is negative in electrolytic cells and positive in voltaic cells. The anode is the electrode at which oxidation occurs. The anode is positive in electrolytic cells and negative in voltaic cells.

4 The Half-Reaction Method Half reaction method rules: 1.Write the unbalanced reaction. 2.Break the reaction into 2 half reactions: ―One oxidation half-reaction and ―One reduction half-reaction Each reaction must have complete formulas for molecules and ions. 3.Mass balance each half reaction by adding appropriate stoichiometric coefficients. To balance H and O we can add: ―H + or H 2 O in acidic solutions. ―OH - or H 2 O in basic solutions. 4.Charge balance the half reactions by adding appropriate numbers of electrons.  Electrons will be products in the oxidation half-reaction.  Electrons will be reactants in the reduction half-reaction. 5.Multiply each half reaction by a number to make the number of electrons in the oxidation half-reaction equal to the number of electrons reduction half- reaction. 6.Add the two half reactions. 7.Eliminate any common terms and reduce coefficients to smallest whole numbers.

5 The Half-Reaction Method Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation.

6 The Half-Reaction Method Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction.

7 The Half-Reaction Method In basic solution hydrogen peroxide oxidizes chromite ions, Cr(OH) 4 -, to chromate ions, CrO The hydrogen peroxide is reduced to hydroxide ions. Write and balance the net ionic equation for this reaction.

8 The Half-Reaction Method When chlorine is bubbled into basic solution, it forms hypochlorite ions and chloride ions. Write and balance the net ionic equation. This is a disproportionation redox reaction. The same species, in this case Cl 2, is both reduced and oxidized.

9 Stoichiometry of Redox Reactions Just as we have done stoichiometry with acid-base reactions, it can also be done with redox reactions. What volume of M KMnO 4 is required to oxidize 35.0 mL of M HCl? The balanced reaction is:

10 Stoichiometry of Redox Reactions A volume of 40.0 mL of iron (II) sulfate is oxidized to iron (III) by 20.0 mL of M potassium dichromate solution. What is the concentration of the iron (II) sulfate solution? The balanced equation is:

11 Voltaic or Galvanic Cells Electrochemical cells in which a spontaneous chemical reaction produces electrical energy. Cell halves are physically separated so that electrons (from redox reaction) are forced to travel through wires and creating a potential difference. Examples of voltaic cells include:

12 The Construction of Simple Voltaic Cells Voltaic cells consist of two half-cells which contain the oxidized and reduced forms of an element (or other chemical species) in contact with each other. A simple half-cell consists of: –A piece of metal immersed in a solution of its ions. –A wire to connect the two half-cells. –And a salt bridge to complete the circuit, maintain neutrality, and prevent solution mixing.

13 The Zinc-Copper Cell Cell components for the Zn-Cu cell are: 1.A metallic Cu strip immersed in 1.0 M copper (II) sulfate. 2.A metallic Zn strip immersed in 1.0 M zinc (II) sulfate. 3.A wire and a salt bridge to complete circuit The cell’s initial voltage is 1.10 volts

14 The Zinc-Copper Cell In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode).

15 The Zinc-Copper Cell There is a commonly used short hand notation for voltaic cells. –The Zn-Cu cell provides a good example.

16 The Copper - Silver Cell Cell components: 1.A Cu strip immersed in 1.0 M copper (II) sulfate. 2.A Ag strip immersed in 1.0 M silver (I) nitrate. 3.A wire and a salt bridge to complete the circuit. The initial cell voltage is 0.46 volts.

17 The Copper - Silver Cell Compare the Zn-Cu cell to the Cu-Ag cell –The Cu electrode is the cathode in the Zn-Cu cell. –The Cu electrode is the anode in the Cu-Ag cell. Whether a particular electrode behaves as an anode or as a cathode depends on what the other electrode of the cell is.

18 The Copper - Silver Cell These experimental facts demonstrate that Cu 2+ is a stronger oxidizing agent than Zn 2+. –In other words Cu 2+ oxidizes metallic Zn to Zn 2+. Similarly, Ag + is is a stronger oxidizing agent than Cu 2+. – Because Ag + oxidizes metallic Cu to Cu 2+. If we arrange these species in order of increasing strengths, we see that:

19 Easiest to reduce is Strongest Oxidizing Agent Ag + Cu 2+ Zn 2+ (find order using left hand side of table) Least likely to be reduced is Strongest Reducing Agent Zn Cu Ag (find order using right hand side of table)

20 Standard Electrode Potential To measure relative electrode potentials, we must establish an arbitrary standard. That standard is the Standard Hydrogen Electrode (SHE). –The SHE is assigned an arbitrary voltage of … V

21 The Zinc-SHE Cell For this cell the components are: 1.A Zn strip immersed in 1.0 M zinc (II) sulfate. 2.The other electrode is the Standard Hydrogen Electrode. 3.A wire and a salt bridge to complete the circuit. The initial cell voltage is volts.

22 The Zinc-SHE Cell The cathode is the Standard Hydrogen Electrode. –In other words Zn reduces H + to H 2. The anode is Zn metal. –Zn metal is oxidized to Zn 2+ ions.

23 The Copper-SHE Cell The cell components are: 1.A Cu strip immersed in 1.0 M copper (II) sulfate. 2.The other electrode is a Standard Hydrogen Electrode. 3.A wire and a salt bridge to complete the circuit. The initial cell voltage is volts.

24 The Copper-SHE Cell In this cell the SHE is the anode –The Cu 2+ ions oxidize H 2 to H +. The Cu is the cathode. –The Cu 2+ ions are reduced to Cu metal.

25 Primary Voltaic Cells As a voltaic cell discharges, its chemicals are consumed. Once the chemicals are consumed, further chemical action is impossible. The electrodes and electrolytes cannot be regenerated by reversing current flow through cell. –These cells are not rechargable.

26 One example of a dry cell is flashlight and radio batteries. The cell’s container is made of zinc which acts as an electrode. A graphite rod is in the center of the cell which acts as the other electrode. The space between the electrodes is filled with a mixture of: 1.ammonium chloride, NH 4 Cl 2.manganese (IV) oxide, MnO 2 3.zinc chloride, ZnCl 2 4.and a porous inactive solid. The Dry Cell

27 The anode reaction is: The graphite rod is the positive electrode (cathode). Ammonium ions from the NH 4 Cl are reduced at the cathode.

28 The Dry Cell The other components in the cell are included to remove the byproducts of the reaction. MnO 2 prevents H 2 from collecting on graphite rod. At the anode, NH 3 combines with Zn 2+ to form a soluble complex and removing the Zn 2+ ions from the reaction.

29 The Dry Cell Alkaline dry cells are similar to ordinary dry cells except that KOH, an alkaline substance, is added to the mixture. Half reactions for an alkaline cell are:

30 Secondary Voltaic Cells Secondary cells are reversible, rechargeable. The electrodes in a secondary cell can be regenerated by the addition of electricity. –These cells can be switched from voltaic to electrolytic cells. One example of a secondary voltaic cell is the lead storage or car battery. The Lead Storage Battery In the lead storage battery the electrodes are two sets of lead alloy grids (plates). Holes in one of the grids are filled with lead (IV) oxide, PbO 2. The other holes are filled with spongy lead. The electrolyte is dilute sulfuric acid.

31 The Lead Storage Battery Diagram of the lead storage battery.

32 The Lead Storage Battery As the battery discharges, spongy lead is oxidized to lead ions and the plate becomes negatively charged. The Pb 2+ ions that are formed combine with SO 4 2- from sulfuric acid to form solid lead sulfate on the Pb electrode.

33 The Lead Storage Battery The net reaction at the anode during discharge is: The cell reaction for a discharging lead storage battery is: As the cell discharges, the cathode reaction is:

34 The Lead Storage Battery What happens at each electrode during recharging? At the lead (IV) oxide, PbO 2, electrode, lead ions are oxidized to lead (IV) oxide. The concentration of the H 2 SO 4 decreases as the cell discharges. Recharging the cell regenerates the H 2 SO 4.

35 The Nickel-Cadmium (Nicad) Cell Nicad batteries are the rechargeable cells used in calculators, cameras, watches, etc. As the battery discharges, the half-reactions are:

36 The Hydrogen-Oxygen Fuel Cell Fuel cells are batteries that must have their reactants continuously supplied in the presence of appropriate catalysts. A hydrogen-oxygen fuel cell is used in the space shuttle –The fuel cell is what exploded in Apollo 13. Hydrogen is oxidized at the anode. Oxygen is reduced at the cathode.

37 The Hydrogen-Oxygen Fuel Cell Notice that the overall reaction is the combination of hydrogen and oxygen to form water. –The cell provides a drinking water supply for the astronauts as well as the electricity for the lights, computers, etc. on board. Fuel cells are very efficient. –Energy conversion rates of 60-70% are common!

38 Uses of Standard Electrode Potentials Electrodes that force the SHE to act as an anode are assigned positive standard reduction potentials. Electrodes that force the SHE to act as the cathode are assigned negative standard reduction potentials. Standard electrode (reduction) potentials tell us the tendencies of half-reactions to occur as written. For example, the half-reaction for the standard potassium electrode is:

39 Uses of Standard Electrode Potentials Compare the potassium half-reaction to fluorine’s half- reaction:

40 Uses of Standard Electrode Potentials Use standard electrode potentials to predict whether an electrochemical reaction at standard state conditions will occur spontaneously. Will silver ions, Ag +, oxidize metallic zinc to Zn 2+ ions, or will Zn 2+ ions oxidize metallic Ag to Ag + ions? Steps for obtaining the equation for the spontaneous reaction. 1.Choose the appropriate half-reactions from a table of standard reduction potentials. 2.Write the equation for the half-reaction with the more positive E 0 value first, along with its E 0 value. 3.Write the equation for the other half-reaction as an oxidation with its oxidation potential, i.e. reverse the tabulated reduction half-reaction and change the sign of the tabulated E 0. 4Balance the electron transfer. 5Add the reduction and oxidation half-reactions and their potentials. This produces the equation for the reaction for which E 0 cell is positive, which indicates that the forward reaction is spontaneous.

41 Uses of Standard Electrode Potentials

42 Electrode Potentials for Other Half-Reactions Will permanganate ions, MnO 4 -, oxidize iron (II) ions to iron (III) ions, or will iron (III) ions oxidize manganese(II) ions to permanganate ions in acidic solution? Follow the steps outlined in the previous slides. Note that E 0 values are not multiplied by any stoichiometric relationships in this procedure.

43 Electrode Potentials for Other Half-Reactions Will nitric acid, HNO 3, oxidize arsenous acid, H 3 AsO 3, in acidic solution? The reduction product of HNO 3 is NO in this reaction.

44 Effect of Concentrations (or Partial Pressures) on Electrode Potentials The Nernst Equation Standard electrode potentials, those compiled in appendices, are determined at thermodynamic standard conditions. Reminder of standard conditions.

45 The Nernst Equation The value of the cell potentials change if conditions are nonstandard. The Nernst equation describes the electrode potentials at nonstandard conditions. The Nernst equation is: Substitution of the values of the constants into the Nernst equation at 25 o C gives: For this half-reaction: The corresponding Nernst equation is:

46 The Nernst Equation Substituting E 0 into the above expression gives: If [Cu 2+ ] and [Cu + ] are both 1.0 M, i.e. at standard conditions, then E = E 0 because the concentration term equals zero.

47 The Nernst Equation Calculate the potential for the Cu 2+ / Cu + electrode at 25 0 C when the concentration of Cu + ions is three times that of Cu 2+ ions.

48 The Nernst Equation Calculate the potential for the Cu 2+ /Cu + electrode at 25 0 C when the Cu + ion concentration is 1/3 of the Cu 2+ ion concentration.

49 The Nernst Equation Calculate the electrode potential for a hydrogen electrode in which the [H + ] is 1.0 x M and the H 2 pressure is 0.50 atmosphere.

50 The Nernst Equation The Nernst equation can also be used to calculate the potential for a cell that consists of two nonstandard electrodes. Calculate the initial potential of a cell that consists of an Fe 3+ /Fe 2+ electrode in which [Fe 3+ ]=1.0 x M and [Fe 2+ ]=0.1 M connected to a Sn 4+ /Sn 2+ electrode in which [Sn 4+ ]=1.0 M and [Sn 2+ ]=0.10 M. A wire and salt bridge complete the circuit.

51 The Nernst Equation Calculate the E 0 cell by the usual procedure.

52 Counting Electrons: Coulometry and Faraday’s Law of Electrolysis A coulomb is the amount of charge that passes a given point when a current of one ampere (A) flows for one second. 1 amp = 1 coulomb/second

53 Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Faraday’s Law states that during electrolysis, one faraday of electricity (96,487 coulombs) reduces and oxidizes, respectively, one equivalent of the oxidizing agent and the reducing agent. –This corresponds to the passage of one mole of electrons through the electrolytic cell.

54 Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Calculate the mass of palladium produced by the reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.

55 Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Calculate the volume of oxygen (measured at STP) produced by the oxidation of water in the previous example.


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