Presentation is loading. Please wait.

Presentation is loading. Please wait.

Oxidation is a loss of an electron or electrons by an atom or group of atoms.Oxidation is a loss of an electron or electrons by an atom or group of atoms.

Similar presentations


Presentation on theme: "Oxidation is a loss of an electron or electrons by an atom or group of atoms.Oxidation is a loss of an electron or electrons by an atom or group of atoms."— Presentation transcript:

1 Oxidation is a loss of an electron or electrons by an atom or group of atoms.Oxidation is a loss of an electron or electrons by an atom or group of atoms. Zn (s) → Zn 2+ (aq) + 2e – Zn (s) is oxidized to Zn 2+ (aq) in this half-reaction.Zn (s) is oxidized to Zn 2+ (aq) in this half-reaction. Reduction is a gain of an electron or electrons by an atom or group of atoms.Reduction is a gain of an electron or electrons by an atom or group of atoms. Cu 2+ (aq) + 2e – → Cu (s) Cu 2+ (aq) is reduced to Cu (s) in this half-reaction.Cu 2+ (aq) is reduced to Cu (s) in this half-reaction. These 2 half-reactions make an oxidation-reduction reaction. That is, electrons are transferred from 1 reactant to another reactant.These 2 half-reactions make an oxidation-reduction reaction. That is, electrons are transferred from 1 reactant to another reactant. Oxidation: Zn (s) → Zn 2+ (aq) + 2e – Reduction: Cu 2+ (aq) + 2e – → Cu (s) ___________ Oxidation-reduction: Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) CH 104: Voltaic Cells

2 Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) MAKING A BATTERY ReactantsProducts

3 An electrochemical cell (or battery) is made of 2 half-cells with electrodes joined by a wire and solutions joined by a salt bridge.An electrochemical cell (or battery) is made of 2 half-cells with electrodes joined by a wire and solutions joined by a salt bridge. The salt bridge lets ions move from 1 half-cell to the other half-cell; however, it does not let the bulk solutions move.The salt bridge lets ions move from 1 half-cell to the other half-cell; however, it does not let the bulk solutions move. A potentiometer (or voltmeter) measures the difference in electric potential (or voltage) between these 2 electrodes. In this battery the difference is volts.A potentiometer (or voltmeter) measures the difference in electric potential (or voltage) between these 2 electrodes. In this battery the difference is volts. MAKING A BATTERY

4 Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) Oxidation occurs at the anode. Which electrode is the anode?Oxidation occurs at the anode. Which electrode is the anode? Hint: oxidation and anode both start with vowels.Hint: oxidation and anode both start with vowels. Reduction occurs at the cathode. Which electrode is the cathode?Reduction occurs at the cathode. Which electrode is the cathode? Hint: reduction and cathode both start with consonants.Hint: reduction and cathode both start with consonants. MAKING A BATTERY

5 Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) What is the oxidation half-reaction?What is the oxidation half-reaction? What is the reduction half-reaction?What is the reduction half-reaction? Therefore, electrons flow from theto the.Therefore, electrons flow from theto the. Electrons are produced at the anode and are consumed at the cathode.Electrons are produced at the anode and are consumed at the cathode. The mass of Zn (s) at the anode will increase, decrease, or stay the same?The mass of Zn (s) at the anode will increase, decrease, or stay the same? The mass of Cu (s) at the cathode will increase, decrease, or stay the same?The mass of Cu (s) at the cathode will increase, decrease, or stay the same? MAKING A BATTERY Zn (s) → Zn 2+ (aq) + 2e – Cu 2+ (aq) + 2e – → Cu (s) Decrease, Zn (s) is converted into Zn 2+ (aq). Increase, Cu 2+ (aq) is converted into Cu (s). anodecathode

6 Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) Therefore, anions (NO 3 – ) flow from the salt bridge to the.Therefore, anions (NO 3 – ) flow from the salt bridge to the. That is, anions from the salt bridge are attracted to the Zn 2+ (aq) produced at the anode.That is, anions from the salt bridge are attracted to the Zn 2+ (aq) produced at the anode. Therefore, cations (K + ) flow from the salt bridge to the.Therefore, cations (K + ) flow from the salt bridge to the. That is, cations from the salt bridge are attracted to the deficit of Cu 2+ (aq) at the cathode.That is, cations from the salt bridge are attracted to the deficit of Cu 2+ (aq) at the cathode. What is the voltage of this battery?What is the voltage of this battery? volts (V) volts (V). MAKING A BATTERY anode cathode

7 Now let’s examine a different oxidation-reduction reaction: Cu (s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag (s) ______ is being oxidized to _______. ______ is being reduced to _______. MAKING A SECONDBATTERY MAKING A SECOND BATTERYReactantsProducts Cu (s) Cu 2+ (aq) Ag (s) Ag + (aq)

8 Cu (s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag (s) Notice that replacing Zn (s) and 1.00 M Zn(NO 3 ) 2(aq) with Ag (s) and 1.00 M AgNO 3(aq) decreases the voltage from V to V.Notice that replacing Zn (s) and 1.00 M Zn(NO 3 ) 2(aq) with Ag (s) and 1.00 M AgNO 3(aq) decreases the voltage from V to V. Which electrode is the anode?Which electrode is the anode? Hint: oxidation and anode both start with vowels.Hint: oxidation and anode both start with vowels. Which electrode is the cathode?Which electrode is the cathode? Hint: reduction and cathode both start with consonants.Hint: reduction and cathode both start with consonants. Notice the first battery had a Cu (s) |Cu 2+ (aq) cathode. This second battery has a Cu (s) |Cu 2+ (aq) anode.Notice the first battery had a Cu (s) |Cu 2+ (aq) cathode. This second battery has a Cu (s) |Cu 2+ (aq) anode. MAKING A SECONDBATTERY MAKING A SECOND BATTERY

9 Cu (s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag (s) What is the oxidation half-reaction?What is the oxidation half-reaction? What is the reduction half-reaction?What is the reduction half-reaction? Therefore, electrons flow from theto the.Therefore, electrons flow from theto the. Electrons are produced at the anode and are consumed at the cathode.Electrons are produced at the anode and are consumed at the cathode. The mass of Cu (s) at the anode will increase, decrease, or stay the same?The mass of Cu (s) at the anode will increase, decrease, or stay the same? The mass of Ag (s) at the cathode will increase, decrease, or stay the same?The mass of Ag (s) at the cathode will increase, decrease, or stay the same? MAKING A SECONDBATTERY MAKING A SECOND BATTERY Cu (s) → Cu 2+ (aq) + 2e – Ag + (aq) + e – → Ag (s) Decrease, Cu (s) is converted into Cu 2+ (aq). Increase, Ag + (aq) is converted into Ag (s). anodecathode

10 Cu (s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag (s) Therefore, anions (NO 3 – ) flow from the salt bridge to the.Therefore, anions (NO 3 – ) flow from the salt bridge to the. That is, anions from the salt bridge are attracted to the Cu 2+ (aq) produced at the anode.That is, anions from the salt bridge are attracted to the Cu 2+ (aq) produced at the anode. Therefore, cations (K + ) flow from the salt bridge to the.Therefore, cations (K + ) flow from the salt bridge to the. That is, cations from the salt bridge are attracted to the deficit of Ag + (aq) at the cathode.That is, cations from the salt bridge are attracted to the deficit of Ag + (aq) at the cathode. MAKING A SECONDBATTERY MAKING A SECOND BATTERYanode cathode

11 Standard reduction potentials (E° red ) are measured with all ions at 1.0 M concentration, all gases at 1.0 atmosphere pressure, all solids in their most thermodynamically stable forms, and the temperature at 298 K.Standard reduction potentials (E° red ) are measured with all ions at 1.0 M concentration, all gases at 1.0 atmosphere pressure, all solids in their most thermodynamically stable forms, and the temperature at 298 K. Therefore, the standard hydrogen electrode has pure H 2(g) at 1.0 atm bubbling into a solution of 1.0 M H + (aq) at 298 K. This electrode is assigned a standard reduction potential of zero volts.Therefore, the standard hydrogen electrode has pure H 2(g) at 1.0 atm bubbling into a solution of 1.0 M H + (aq) at 298 K. This electrode is assigned a standard reduction potential of zero volts. 2 H + (aq) + 2e – → H 2(g), E° red = 0 volts All standard reduction potentials are measured relative to the standard hydrogen electrode.All standard reduction potentials are measured relative to the standard hydrogen electrode. STANDARD REDUCTION POTENTIALS, E° red

12 What is the voltage of the anode?What is the voltage of the anode? The anode is a standard hydrogen electrode. By definition it has 0 volts.The anode is a standard hydrogen electrode. By definition it has 0 volts. What is the voltage of the cathode?What is the voltage of the cathode? The cathode is a standard Cu 2+ (aq) |Cu (s) electrode. It has 0.34 volts, according to the meter. This voltage agrees with the following table of standard reduction potentials (E° red ).The cathode is a standard Cu 2+ (aq) |Cu (s) electrode. It has 0.34 volts, according to the meter. This voltage agrees with the following table of standard reduction potentials (E° red ). STANDARD REDUCTION POTENTIALS, E° red

13 This table of standard reduction potentials (E° red ) is used to calculate standard cell potentials (E° cell ).This table of standard reduction potentials (E° red ) is used to calculate standard cell potentials (E° cell ). E° cell = E° ox + E° red STANDARD REDUCTION POTENTIALS, E° red

14 For example, use this table of standard reduction potentials (E° red ) to calculate the standard cell potential (E° cell ) of the following battery.For example, use this table of standard reduction potentials (E° red ) to calculate the standard cell potential (E° cell ) of the following battery. Oxidation: Cu (s) → Cu 2+ (aq) + 2e –, E° ox = –0.34 voltsOxidation: Cu (s) → Cu 2+ (aq) + 2e –, E° ox = –0.34 volts Reduction: 2Ag + (aq) + 2e – → 2Ag (s), E° red = 0.80 voltsReduction: 2Ag + (aq) + 2e – → 2Ag (s), E° red = 0.80 volts Net reaction:Cu (s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag (s)Net reaction:Cu (s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag (s) Important: Oxidation is the opposite of reduction; therefore, the sign of E° ox for a half-reaction is always the opposite of E° red. For example, the volts for the standard reduction potential of this oxidation half- reaction was changed to –0.34 volts.Important: Oxidation is the opposite of reduction; therefore, the sign of E° ox for a half-reaction is always the opposite of E° red. For example, the volts for the standard reduction potential of this oxidation half- reaction was changed to –0.34 volts. Important: Standard reduction potential is an intensive property; therefore, the values of E° ox and E° red are not changed if the stoichiometric coefficients are multiplied by a constant. For example, the 0.80 volts for Ag + (aq) + e – → Ag (s) was not changed when these stoichiometric coefficients were multiplied by 2 to balance the net reaction.Important: Standard reduction potential is an intensive property; therefore, the values of E° ox and E° red are not changed if the stoichiometric coefficients are multiplied by a constant. For example, the 0.80 volts for Ag + (aq) + e – → Ag (s) was not changed when these stoichiometric coefficients were multiplied by 2 to balance the net reaction. Important: If E° cell is +, then the cell reaction is spontaneous as written.Important: If E° cell is +, then the cell reaction is spontaneous as written. Important: If E° cell is –, then the cell reaction is non-spontaneous as written.Important: If E° cell is –, then the cell reaction is non-spontaneous as written. E° cell = E° ox + E° red E° cell = –0.34 volts volts = 0.46 volts STANDARD CELL POTENTIALS, E° cell

15 Calculate the standard cell potential (E° cell ) of the following battery.Calculate the standard cell potential (E° cell ) of the following battery. Oxidation: 4Ag (s) → 4Ag + (aq) + 4e –, E°ox = –0.80 voltsOxidation: 4Ag (s) → 4Ag + (aq) + 4e –, E°ox = –0.80 volts Reduction: O 2(g) + 4H + (aq) + 4e – → 2H 2 O (l), E° red = 1.23 voltsReduction: O 2(g) + 4H + (aq) + 4e – → 2H 2 O (l), E° red = 1.23 volts Net reaction:4Ag (s) + O 2(g) + 4H + (aq) → 4Ag + (aq) + 2H 2 O (l)Net reaction:4Ag (s) + O 2(g) + 4H + (aq) → 4Ag + (aq) + 2H 2 O (l) E° cell = E° ox + E° red E° cell = –0.80 volts volts = 0.43 volts Is this reaction spontaneous?Is this reaction spontaneous? Yes, E° cell is +.Yes, E° cell is +. STANDARD CELL POTENTIALS, E° cell

16 The Nernst equation is used to calculate cell potentials (E cell ) at non-standard conditions.The Nernst equation is used to calculate cell potentials (E cell ) at non-standard conditions. For the cell reaction: aA + bB = cC + dDFor the cell reaction: aA + bB = cC + dD WhereWhere E cell = the cell potential at non-standard conditionsE cell = the cell potential at non-standard conditions E° cell = the standard cell potentialE° cell = the standard cell potential R = the gas constant = JK –1 mol –1R = the gas constant = JK –1 mol –1 T = the temperature in KelvinT = the temperature in Kelvin n = the total number of electrons transferred in the equation as writtenn = the total number of electrons transferred in the equation as written F = the Faraday constant = 96,500 Jvolt –1 mol –1F = the Faraday constant = 96,500 Jvolt –1 mol –1 THE NERNST EQUATION

17 Use the Nernst equation to calculate E cell for this battery at 25° C.Use the Nernst equation to calculate E cell for this battery at 25° C. Oxidation: Fe 2+ (aq) → Fe 3+ (aq) + e –, E° ox = –0.77 voltsOxidation: Fe 2+ (aq) → Fe 3+ (aq) + e –, E° ox = –0.77 volts Reduction: Ag + (aq) + e – → Ag (s), E° red = 0.80 voltsReduction: Ag + (aq) + e – → Ag (s), E° red = 0.80 volts Net reaction:Fe 2+ (aq) + Ag + (aq) → Fe 3+ (aq) + Ag (s)Net reaction:Fe 2+ (aq) + Ag + (aq) → Fe 3+ (aq) + Ag (s) E° cell = E° ox + E° red = THE NERNST EQUATION –0.77 volts volts = 0.03 volts

18 Oxidation-reduction causes the corrosion of iron and steel. Steel is refined Fe containing less than 1.7% C.Oxidation-reduction causes the corrosion of iron and steel. Steel is refined Fe containing less than 1.7% C. The oxidation of Fe (s) to Fe 2+ (aq) occurs at the strained regions of these iron nails: the heads, tips, and bend. This generation of Fe 2+ (aq) is marked by the formation of a blue precipitate in the presence of K 3 [Fe(CN) 6 ].The oxidation of Fe (s) to Fe 2+ (aq) occurs at the strained regions of these iron nails: the heads, tips, and bend. This generation of Fe 2+ (aq) is marked by the formation of a blue precipitate in the presence of K 3 [Fe(CN) 6 ]. The reduction of O 2(aq) to OH – (aq) occurs at the other regions of these nails. This generation of OH – (aq) is marked by the pink color of phenolphthalein.The reduction of O 2(aq) to OH – (aq) occurs at the other regions of these nails. This generation of OH – (aq) is marked by the pink color of phenolphthalein. Use these observations to write a balanced equation for the corrosion of iron or steel.Use these observations to write a balanced equation for the corrosion of iron or steel. 2Fe (s) + O 2(aq) + 2H 2 O (l) → 2Fe 2+ (aq) + 4OH – (aq) Notice that exposure to air and water causes the corrosion of iron and steel.Notice that exposure to air and water causes the corrosion of iron and steel. ELECTROCHEMISTRY AND CORROSION

19 Coatings, such as paints or Zn (s), are used to retard the corrosion of iron and steel by preventing this exposure to air and water.Coatings, such as paints or Zn (s), are used to retard the corrosion of iron and steel by preventing this exposure to air and water. Galvanized iron or steel is coated with Zn (s). This retards the corrosion of Fe (s).Galvanized iron or steel is coated with Zn (s). This retards the corrosion of Fe (s). Coating iron or steel with Cu (s) accelerates the corrosion of Fe (s).Coating iron or steel with Cu (s) accelerates the corrosion of Fe (s). ELECTROCHEMISTRY AND CORROSION

20 Why is stainless steel stainless?Why is stainless steel stainless? Stainless steel contains at least 10.5% Cr. This Cr reacts with the O 2(g) in air to form a complex chrome-oxide surface layer that prevents further corrosion. Unlike the rusting surface of iron, this chrome-oxide layer does not flake away and expose additional atoms to air and water.Higher levels of Cr and the addition of other alloying elements such as Ni and Mo further improve this surface layer and the corrosion resistance of stainless steels.Stainless steel contains at least 10.5% Cr. This Cr reacts with the O 2(g) in air to form a complex chrome-oxide surface layer that prevents further corrosion. Unlike the rusting surface of iron, this chrome-oxide layer does not flake away and expose additional atoms to air and water. Higher levels of Cr and the addition of other alloying elements such as Ni and Mo further improve this surface layer and the corrosion resistance of stainless steels. ELECTROCHEMISTRY AND CORROSION

21 Barnes, D.S., J.A. Chandler Chemistry Workbook and Laboratory Manual. Amherst, MA: University of Massachusetts.Barnes, D.S., J.A. Chandler Chemistry Workbook and Laboratory Manual. Amherst, MA: University of Massachusetts. McMurry, J., R.C. Fay Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall.McMurry, J., R.C. Fay Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall. Merriam-Webster, Inc Webster’s 9th New Collegiate Dictionary. Springfield, MA: Merriam-Webster, Inc.Merriam-Webster, Inc Webster’s 9th New Collegiate Dictionary. Springfield, MA: Merriam-Webster, Inc. Petrucci, R.H General Chemistry Principles and Modern Applications, 4th ed. New York, NY: Macmillan Publishing Company.Petrucci, R.H General Chemistry Principles and Modern Applications, 4th ed. New York, NY: Macmillan Publishing Company. Specialty Steel Industry of North America SSINA: Stainless Steel: About. Available: [accessed 12 October 2006].Specialty Steel Industry of North America SSINA: Stainless Steel: About. Available: [accessed 12 October 2006].http://www.ssina.com/index2.html SOURCES


Download ppt "Oxidation is a loss of an electron or electrons by an atom or group of atoms.Oxidation is a loss of an electron or electrons by an atom or group of atoms."

Similar presentations


Ads by Google