Redox Reactions oxidation reduction reactions

 Ch 22 sec 1  From the combustion of gasoline to the metabolism of food - oxidation is responsible  Oxidation - Loss of electrons  old definition - gain of oxygen  Reduction - Gain of electrons  old definition - loss of oxygen  ONE DOES NOT OCCUR WITHOUT THE OTHER  Ch 22 sec 1  From the combustion of gasoline to the metabolism of food - oxidation is responsible  Oxidation - Loss of electrons  old definition - gain of oxygen  Reduction - Gain of electrons  old definition - loss of oxygen  ONE DOES NOT OCCUR WITHOUT THE OTHER

LEO the lion goes GER  Losing Electrons is Oxidation  Gaining Electrons is Reduction  Example  Mg + S ---> Mg 2+ + S 2-  Magnesium is oxidized (aka reducing agent)  Sulfur is reduced (aka oxidizing agent)  Losing Electrons is Oxidation  Gaining Electrons is Reduction  Example  Mg + S ---> Mg 2+ + S 2-  Magnesium is oxidized (aka reducing agent)  Sulfur is reduced (aka oxidizing agent)

Redox reactions are usually presented as two components  Mg ----> Mg e -  S + 2e - ---> S 2-  Identifying transfers of electrons is easy for ionic reactions.  What about covalent where there is not a transfer of electrons but a sharing?  Mg ----> Mg e -  S + 2e - ---> S 2-  Identifying transfers of electrons is easy for ionic reactions.  What about covalent where there is not a transfer of electrons but a sharing?

Consider 2H 2 + O 2 ---> 2H 2 O  Which element is reduced and oxidized? Explain  Oxygen is the electron hog. The partial gain of electrons means it is reduced and hydrogen’s partial loss means it is oxidized.  Which element is reduced and oxidized? Explain  Oxygen is the electron hog. The partial gain of electrons means it is reduced and hydrogen’s partial loss means it is oxidized.

Your Turn 4Fe + 3O 2 ---> 2Fe 2 O 3  Write the 2 redox reactions. Which is oxidized and which is reduced.  answer  Fe --> Fe e - oxidized (reducing agent)  O 2 + 2e - ---> O 2- reduced (oxidizing agent)  Corrosion occurs more rapidly in the presence of salts and acids. Why?  These are conducting solutions that make electron transfer easier.  Write the 2 redox reactions. Which is oxidized and which is reduced.  answer  Fe --> Fe e - oxidized (reducing agent)  O 2 + 2e - ---> O 2- reduced (oxidizing agent)  Corrosion occurs more rapidly in the presence of salts and acids. Why?  These are conducting solutions that make electron transfer easier.

Sometimes Corrosion is good.  Aluminum oxidizes to form tightly packed aluminum oxide particles. This is a protective covering.  Iron needs to be coated to protect it because the oxide is not tightly packed.  Aluminum oxidizes to form tightly packed aluminum oxide particles. This is a protective covering.  Iron needs to be coated to protect it because the oxide is not tightly packed.

Ch 22 sec 2 Assigning Oxidation Numbers  Redox equations will be balanced using oxidation numbers  Rules  1. The oxidation number of a monatomic ion is equal to its ionic charge.  2. The oxidation number of hydrogen in a compound is +1 except in a metal hydride. Example - NaH H is -1  3. The oxidation number for oxygen is -2 unless in a peroxide H 2 O 2 where it is -1  Redox equations will be balanced using oxidation numbers  Rules  1. The oxidation number of a monatomic ion is equal to its ionic charge.  2. The oxidation number of hydrogen in a compound is +1 except in a metal hydride. Example - NaH H is -1  3. The oxidation number for oxygen is -2 unless in a peroxide H 2 O 2 where it is -1

more rules  The oxidation number of an uncombined atom (elemental form) is 0.  For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal 0.  For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion.  The oxidation number of an uncombined atom (elemental form) is 0.  For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal 0.  For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion.

What is the oxidation number of elements in the following compounds?  1. SO 2  S is +4 and O is -2  2. K 2 SO 4  K is +1 S is +6 O is -2  1. SO 2  S is +4 and O is -2  2. K 2 SO 4  K is +1 S is +6 O is -2

Oxidation numbers change in redox reaction. No change no redox.  An increase in the oxidation number indicates oxidation  A decrease in the oxidation number indicates reduction  Example - What is being oxidized and what is being reduced?  2AgNO 3 + Cu --> Cu(NO 3 ) 2 + 2Ag   An increase in the oxidation number indicates oxidation  A decrease in the oxidation number indicates reduction  Example - What is being oxidized and what is being reduced?  2AgNO 3 + Cu --> Cu(NO 3 ) 2 + 2Ag 

Ch 22 sec 3 Classifying Reactions  Either electrons are transferred or they are not.  Redox reactions include single-replacement, combination, decomposition and combustion.  Others - double replacement and acid/base reactions.  Color changes signify redox reactions  video example  Either electrons are transferred or they are not.  Redox reactions include single-replacement, combination, decomposition and combustion.  Others - double replacement and acid/base reactions.  Color changes signify redox reactions  video example

another redox clip

Balancing Redox Reactions Two Methods  Many reactions are too complex to be balanced by trial and error.  All reactions should be balanced in this manner. You were taught to balance atoms because for simple reactions it works. Balancing atoms may balance the equation and may not be the correct balanced equation. Atoms along with charges need to be balanced.  Many reactions are too complex to be balanced by trial and error.  All reactions should be balanced in this manner. You were taught to balance atoms because for simple reactions it works. Balancing atoms may balance the equation and may not be the correct balanced equation. Atoms along with charges need to be balanced.

oxidation-number-change method  balanced by comparing the increase and decrease of oxidation numbers  Example:  Step 1 - Assign Oxidation Numbers   K 2 Cr 2 O 7 + H 2 O + S --> KOH + Cr 2 O 3 + SO 2  balanced by comparing the increase and decrease of oxidation numbers  Example:  Step 1 - Assign Oxidation Numbers   K 2 Cr 2 O 7 + H 2 O + S --> KOH + Cr 2 O 3 + SO 2

  K 2 Cr 2 O 7 + H 2 O + S --> KOH + Cr 2 O 3 + SO 2  Step 2 - Identify which atoms are oxidized and which are reduced.  Cr is reduced and S is oxidized  Onto step 3  Use a bracket line to connect the atoms that undergo oxidation and another line to connect those that undergo reduction. Then write the oxidation-number change at the midpoint of each line.   K 2 Cr 2 O 7 + H 2 O + S --> KOH + Cr 2 O 3 + SO 2  Step 2 - Identify which atoms are oxidized and which are reduced.  Cr is reduced and S is oxidized  Onto step 3  Use a bracket line to connect the atoms that undergo oxidation and another line to connect those that undergo reduction. Then write the oxidation-number change at the midpoint of each line.

  K 2 Cr 2 O 7 + H 2 O + S --> KOH + Cr 2 O 3 + SO 2  Step 4 - Make the total increase in oxidation number equal to the total decrease in oxidation number by using appropriate coefficients.  K 2 Cr 2 O 7 + H 2 O + S --> KOH + Cr 2 O 3 + SO 2  The coefficient is the number of atoms needed.  2 K 2 Cr 2 O 7 + H 2 O + 3S --> KOH + 2Cr 2 O 3 + 3SO 2   K 2 Cr 2 O 7 + H 2 O + S --> KOH + Cr 2 O 3 + SO 2  Step 4 - Make the total increase in oxidation number equal to the total decrease in oxidation number by using appropriate coefficients.  K 2 Cr 2 O 7 + H 2 O + S --> KOH + Cr 2 O 3 + SO 2  The coefficient is the number of atoms needed.  2 K 2 Cr 2 O 7 + H 2 O + 3S --> KOH + 2Cr 2 O 3 + 3SO (4)(-3)=-12 (3)(+4)=+12

 Step 5 - Finish balancing by inspection  2K 2 Cr 2 O 7 + 2H 2 O + 3S --> 4KOH + 2Cr 2 O 3 + 3SO 2  Your Turn -  Balance using the oxidation-number-change method.  As 2 O 3 + Cl 2 + H 2 O --> H 3 AsO 4 + HCl  (2)(-1)=-2   As 2 O 3 + Cl 2 + H 2 O --> H 3 AsO 4 + HCl  (1)(+2)=+2  As 2 O 3 + 2Cl 2 +5H 2 O --> 2H 3 AsO 4 + 4HCl  Step 5 - Finish balancing by inspection  2K 2 Cr 2 O 7 + 2H 2 O + 3S --> 4KOH + 2Cr 2 O 3 + 3SO 2  Your Turn -  Balance using the oxidation-number-change method.  As 2 O 3 + Cl 2 + H 2 O --> H 3 AsO 4 + HCl  (2)(-1)=-2   As 2 O 3 + Cl 2 + H 2 O --> H 3 AsO 4 + HCl  (1)(+2)=+2  As 2 O 3 + 2Cl 2 +5H 2 O --> 2H 3 AsO 4 + 4HCl

Using Half-Reactions  Good for ionic reactions  Two equations used - one shows oxidation the other reduction.Then combined together in the last step.  Good for ionic reactions  Two equations used - one shows oxidation the other reduction.Then combined together in the last step.

Balance using 1/2 reactions Example:  KMnO 4 + HCl --> MnCl 2 + Cl 2 + H 2 O + KCl  Step 1 - Write the unbalanced equation in ionic form. Place oxidation numbers above.   K 1+ + MnO H 1+ + Cl 1- --> Mn Cl 1- + Cl 2 + H 2 O + K 1+ + Cl 1-  Step 2 - Write separate 1/2 reactions for the oxidation reduction process.  Cl 1- --> Cl 2  MnO > Mn 2+  KMnO 4 + HCl --> MnCl 2 + Cl 2 + H 2 O + KCl  Step 1 - Write the unbalanced equation in ionic form. Place oxidation numbers above.   K 1+ + MnO H 1+ + Cl 1- --> Mn Cl 1- + Cl 2 + H 2 O + K 1+ + Cl 1-  Step 2 - Write separate 1/2 reactions for the oxidation reduction process.  Cl 1- --> Cl 2  MnO > Mn 2+

 Step 3 - Balance the atoms in each 1/2 reaction. When reactions take place in an acid solution you will need to use water and H + to balance hydrogen and oxygen atoms.  2Cl 1- --> Cl 2  MnO H + --> Mn H 2 O  atoms are balanced but charges are not.  Step 4 - Add electrons to one side of each 1/2 reaction to balance the charges.  2Cl 1- --> Cl 2 + 2e -  MnO H + + 5e - --> Mn H 2 O  Step 3 - Balance the atoms in each 1/2 reaction. When reactions take place in an acid solution you will need to use water and H + to balance hydrogen and oxygen atoms.  2Cl 1- --> Cl 2  MnO H + --> Mn H 2 O  atoms are balanced but charges are not.  Step 4 - Add electrons to one side of each 1/2 reaction to balance the charges.  2Cl 1- --> Cl 2 + 2e -  MnO H + + 5e - --> Mn H 2 O

 Step 5 - numbers of electrons must be equal. Multiply each reaction by a number to make electrons equal. In this case make electrons equal 10.  10Cl 1- --> 5Cl e -  2MnO H e - --> 2Mn H 2 O  Step 6 - Add the 1/2 reactions to show an overall equation.  10Cl MnO H e - --> 5Cl e - + 2Mn H 2 O  Remove terms that are the same on both sides.  10Cl - + 2MnO H + --> 5Cl 2 + 2Mn H 2 O  Step 5 - numbers of electrons must be equal. Multiply each reaction by a number to make electrons equal. In this case make electrons equal 10.  10Cl 1- --> 5Cl e -  2MnO H e - --> 2Mn H 2 O  Step 6 - Add the 1/2 reactions to show an overall equation.  10Cl MnO H e - --> 5Cl e - + 2Mn H 2 O  Remove terms that are the same on both sides.  10Cl - + 2MnO H + --> 5Cl 2 + 2Mn H 2 O

 Step 7 - Add spectator ions and balance the equation.  From step 1  K 1+ + MnO H 1+ + Cl 1- --> Mn Cl 1- + Cl 2 + H 2 O + K 1+ + Cl 1-  10Cl - + 2K + + 2MnO H + + 6Cl - --> 5Cl 2 + 2Mn Cl - + 8H 2 O + 2K + + 2Cl -  Summarize spectator and nonspectator ions.  16Cl - + 2K + + 2MnO H + --> 5Cl 2 + 2Mn Cl - + 8H 2 O + 2K + The equation is now balanced for atoms and charge. Now you can rewrite it into: 2KMnO HCl --> 2MnCl 2 + 5Cl 2 + 8H 2 O + 2KCl  Step 7 - Add spectator ions and balance the equation.  From step 1  K 1+ + MnO H 1+ + Cl 1- --> Mn Cl 1- + Cl 2 + H 2 O + K 1+ + Cl 1-  10Cl - + 2K + + 2MnO H + + 6Cl - --> 5Cl 2 + 2Mn Cl - + 8H 2 O + 2K + + 2Cl -  Summarize spectator and nonspectator ions.  16Cl - + 2K + + 2MnO H + --> 5Cl 2 + 2Mn Cl - + 8H 2 O + 2K + The equation is now balanced for atoms and charge. Now you can rewrite it into: 2KMnO HCl --> 2MnCl 2 + 5Cl 2 + 8H 2 O + 2KCl

Balance using 1/2 reaction method Your Turn:  S + HNO 3 --> SO 2 + NO + H 2 O  Step 1 - Write the unbalanced equation in ionic form. Place oxidation numbers above.   S + H + + NO > SO 2 + NO + H 2 O  Identify what is oxidized and what is reduced.  S + HNO 3 --> SO 2 + NO + H 2 O  Step 1 - Write the unbalanced equation in ionic form. Place oxidation numbers above.   S + H + + NO > SO 2 + NO + H 2 O  Identify what is oxidized and what is reduced.

 Step 2 - Write separate half-reactions for the oxidation and reduction process.  S --> SO 2  NO > NO  Step 3 - Balance the atoms in each 1/2 reaction. When reactions take place in an acid solution you will need to use water and H + to balance hydrogen and oxygen atoms.  2H 2 O + S --> SO 2 + 4H +  4H + + NO > NO + 2H 2 O  note: charges are not balanced  Step 2 - Write separate half-reactions for the oxidation and reduction process.  S --> SO 2  NO > NO  Step 3 - Balance the atoms in each 1/2 reaction. When reactions take place in an acid solution you will need to use water and H + to balance hydrogen and oxygen atoms.  2H 2 O + S --> SO 2 + 4H +  4H + + NO > NO + 2H 2 O  note: charges are not balanced

 Step 4 - Add electrons to one side of each 1/2 reaction to balance the charges.  2H 2 O + S --> SO 2 + 4H + + 4e -  4H + + NO e - --> NO + 2H 2 O  Step 5 - numbers of electrons must be equal. Multiply each reaction by a number to make electrons equal. In this case make electrons equal 12.  6H 2 O + 3S --> 3SO H e -  16H + + 4NO e - --> 4NO + 8H 2 O  Step 4 - Add electrons to one side of each 1/2 reaction to balance the charges.  2H 2 O + S --> SO 2 + 4H + + 4e -  4H + + NO e - --> NO + 2H 2 O  Step 5 - numbers of electrons must be equal. Multiply each reaction by a number to make electrons equal. In this case make electrons equal 12.  6H 2 O + 3S --> 3SO H e -  16H + + 4NO e - --> 4NO + 8H 2 O

 Step 6 - Add the 1/2 reactions to show an overall equation.  6H 2 O + 3S + 16H + + 4NO e - --> 3SO H e - + 4NO + 8H 2 O  Remove terms that are the same on both sides.  3S + 4H + + 4NO > 3SO 2 + 4NO + 2H 2 O  Step 7 - Add spectator ions and balance the equation. This reaction does not have spectator ions. Therefore final answer is:  3S + 4HNO 3 --> 3SO 2 + 4NO + 2H 2 O  Step 6 - Add the 1/2 reactions to show an overall equation.  6H 2 O + 3S + 16H + + 4NO e - --> 3SO H e - + 4NO + 8H 2 O  Remove terms that are the same on both sides.  3S + 4H + + 4NO > 3SO 2 + 4NO + 2H 2 O  Step 7 - Add spectator ions and balance the equation. This reaction does not have spectator ions. Therefore final answer is:  3S + 4HNO 3 --> 3SO 2 + 4NO + 2H 2 O

Balancing Redox Reactions in a Basic Solution  MnO CN - --> MnO 2 + CNO -  Initially assume acid solution but eventually switch over to a base.  Two 1/2 reactions are:  MnO > MnO 2  CN - --> CNO -  Balance as if an acid:  4H + + MnO > MnO 2 + 2H 2 O  H 2 O + CN - --> CNO - + 2H +  MnO CN - --> MnO 2 + CNO -  Initially assume acid solution but eventually switch over to a base.  Two 1/2 reactions are:  MnO > MnO 2  CN - --> CNO -  Balance as if an acid:  4H + + MnO > MnO 2 + 2H 2 O  H 2 O + CN - --> CNO - + 2H +

 Now reality sets in. Switch over to a base by adding the same number of OH - ions as H +  4OH - + 4H + + MnO > MnO 2 + 2H 2 O + 4OH -  2OH - + H 2 O + CN - --> CNO - + 2H + + 2OH -  Next, combine H + and OH - to make water and cancel if possible  2 H 2 O + MnO > MnO 2 + 4OH -  2OH - + CN - --> CNO - + H 2 O  Mass balance has been achieved. Charge balance is next.  Now reality sets in. Switch over to a base by adding the same number of OH - ions as H +  4OH - + 4H + + MnO > MnO 2 + 2H 2 O + 4OH -  2OH - + H 2 O + CN - --> CNO - + 2H + + 2OH -  Next, combine H + and OH - to make water and cancel if possible  2 H 2 O + MnO > MnO 2 + 4OH -  2OH - + CN - --> CNO - + H 2 O  Mass balance has been achieved. Charge balance is next.

 3e H 2 O + MnO > MnO 2 + 4OH -  2OH - + CN - --> CNO - + H 2 O + 2e -  Electrons need to balance - multiply by LCM  6e H 2 O + 2 MnO > 2 MnO 2 + 8OH -  6OH - + 3CN - --> 3CNO - + 3H 2 O + 6e -  Sum the reactions canceling like terms to give the net ionic equation:  H 2 O + 2MnO CN - --> 2MnO 2 + 2OH - + 3CNO -  3e H 2 O + MnO > MnO 2 + 4OH -  2OH - + CN - --> CNO - + H 2 O + 2e -  Electrons need to balance - multiply by LCM  6e H 2 O + 2 MnO > 2 MnO 2 + 8OH -  6OH - + 3CN - --> 3CNO - + 3H 2 O + 6e -  Sum the reactions canceling like terms to give the net ionic equation:  H 2 O + 2MnO CN - --> 2MnO 2 + 2OH - + 3CNO -

Electrochem Intro

Ch 23 sec 1 Electrochemical Cells  When a strip of zinc metal is placed in CuSO 4. Electrons are transferred from Zn to Cu.  Zn (s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu (s)  The flow of electrons is an electric current.  Any conversion between electrical and chemical energy is an electrochemical process. The device that converts between the two is an electrochemical cell.  When a strip of zinc metal is placed in CuSO 4. Electrons are transferred from Zn to Cu.  Zn (s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu (s)  The flow of electrons is an electric current.  Any conversion between electrical and chemical energy is an electrochemical process. The device that converts between the two is an electrochemical cell.

Voltaic Cells  Convert chemical energy into electrical energy  Half-cell is part of the cell where oxidation or reduction takes place.  A half-cell consists of a metal rod or strip immersed in a solution of its ions.  Convert chemical energy into electrical energy  Half-cell is part of the cell where oxidation or reduction takes place.  A half-cell consists of a metal rod or strip immersed in a solution of its ions.

zinc-copper reaction  One 1/2 cell has a zinc rod immersed in zinc sulfate  One 1/2 cell has a copper rod immersed in copper (II) sulfate  Cells are separated by a salt bridge  a tube containing a strong electrolyte often K 2 SO 4  A wire carries the electrons in the external circuit from the zinc rod to the copper rod.  The driving force is the spontaneous redox reaction  One 1/2 cell has a zinc rod immersed in zinc sulfate  One 1/2 cell has a copper rod immersed in copper (II) sulfate  Cells are separated by a salt bridge  a tube containing a strong electrolyte often K 2 SO 4  A wire carries the electrons in the external circuit from the zinc rod to the copper rod.  The driving force is the spontaneous redox reaction

 Electrode - a conductor in a circuit that carries electrons to or from a substance.  Anode - electrode at which oxidation occurs. Electrons are produced at the anode so it it labeled the negative electrode.  Cathode - electrode at which reduction takes place. Electrons are consumed at the electrode. The cathode is labeled the positive electrode.  Neither electrode is really charged. The moving electrons balance any charge that might build up as oxidation and reduction occur.  Electrode - a conductor in a circuit that carries electrons to or from a substance.  Anode - electrode at which oxidation occurs. Electrons are produced at the anode so it it labeled the negative electrode.  Cathode - electrode at which reduction takes place. Electrons are consumed at the electrode. The cathode is labeled the positive electrode.  Neither electrode is really charged. The moving electrons balance any charge that might build up as oxidation and reduction occur.

The electrochemical process in a zinc-copper voltaic cell. These steps occur at the same time.  1. Electrons are produced at the zinc rod. Zn --> Zn e - (anode)  2. The electrons leave the the zinc anode and travel through the external circuit to the copper rod. If a bulb is in the circuit it will light.  3. Electrons enter the copper rod and interact with copper ions in solution.  Cu 2+ +2e - --> Cu  1. Electrons are produced at the zinc rod. Zn --> Zn e - (anode)  2. The electrons leave the the zinc anode and travel through the external circuit to the copper rod. If a bulb is in the circuit it will light.  3. Electrons enter the copper rod and interact with copper ions in solution.  Cu 2+ +2e - --> Cu

 To complete the circuit, both positive and negative ions move through the aqueous solutions via the salt bridge.  Summary:  Zn --> Zn e -  Cu e - --> Cu  Zn (s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu (s)  To complete the circuit, both positive and negative ions move through the aqueous solutions via the salt bridge.  Summary:  Zn --> Zn e -  Cu e - --> Cu  Zn (s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu (s)

Dry Cells (batteries)  A voltaic cell in which the electrolyte is a paste.  A zinc container is filled with a thick, moist electrolyte paste of manganese (IV) oxide, (MnO 2 ), ZnCl 2, NH 4 Cl and water.  A graphite rod is embedded in the paste.  The zinc container is the anode and the graphite rod is the cathode.  A voltaic cell in which the electrolyte is a paste.  A zinc container is filled with a thick, moist electrolyte paste of manganese (IV) oxide, (MnO 2 ), ZnCl 2, NH 4 Cl and water.  A graphite rod is embedded in the paste.  The zinc container is the anode and the graphite rod is the cathode.

 The thick paste and its surrounding paper liner prevent the contents of the cell from freely mixing.  Dry Cell Reactions:  Zn --> Zn e -  2MnO 2 + 2NH e - --> Mn 2 O 3 + 2NH 3 + H 2 O  The graphite serves as a conductor even though it is the cathode. The Mn is actually reduced.  Alkaline battery - same as dry cell but the paste is KOH to prevent buildup of ammonia gas.  The thick paste and its surrounding paper liner prevent the contents of the cell from freely mixing.  Dry Cell Reactions:  Zn --> Zn e -  2MnO 2 + 2NH e - --> Mn 2 O 3 + 2NH 3 + H 2 O  The graphite serves as a conductor even though it is the cathode. The Mn is actually reduced.  Alkaline battery - same as dry cell but the paste is KOH to prevent buildup of ammonia gas.

Lead Storage Batteries  A battery is actually a group of cells connected together.  A 12 volt car battery consists of 6 cells.  One set of grids packed with spongy lead (anode).  The other grid is (cathode) is packed with PbO 2.  The grids are immersed in concentrated sulfuric acid.  A battery is actually a group of cells connected together.  A 12 volt car battery consists of 6 cells.  One set of grids packed with spongy lead (anode).  The other grid is (cathode) is packed with PbO 2.  The grids are immersed in concentrated sulfuric acid.

 The reactions:  Pb + SO > PbSO 4 + 2e - (oxidation)  PbO 2 + 4H + + SO e - --> PbSO 4 + 2H 2 O (reduction)  Overall  Pb + PbO 2 + 2H 2 SO 4 --> 2PbSO 4 + 2H 2 O  The sulfate slowly builds up and concentration of acid decreases.  The car’s generator reverses the reaction.  The reactions:  Pb + SO > PbSO 4 + 2e - (oxidation)  PbO 2 + 4H + + SO e - --> PbSO 4 + 2H 2 O (reduction)  Overall  Pb + PbO 2 + 2H 2 SO 4 --> 2PbSO 4 + 2H 2 O  The sulfate slowly builds up and concentration of acid decreases.  The car’s generator reverses the reaction.

Fuel Cells  Cells with renewable electrodes.  A voltaic cell in which a fuel substance undergoes oxidation and from which electrical energy is continuously obtained.  Do not have to be recharged  emits no air pollutants, operates quieter, more cost effective than an electrical generator.  Cells with renewable electrodes.  A voltaic cell in which a fuel substance undergoes oxidation and from which electrical energy is continuously obtained.  Do not have to be recharged  emits no air pollutants, operates quieter, more cost effective than an electrical generator.

The hydrogen-oxygen fuel cell  3 compartments separated by 2 carbon electrodes.  Oxygen (the oxidizer) is fed into the cathode compartment.  Hydrogen (the fuel) is fed into the anode compartment.  The gases diffuse slowly.  3 compartments separated by 2 carbon electrodes.  Oxygen (the oxidizer) is fed into the cathode compartment.  Hydrogen (the fuel) is fed into the anode compartment.  The gases diffuse slowly.

 The electrolyte in the center is a hot, concentrated solution of KOH.  Electrons enter from the oxidation 1/2 reaction at the anode then pass through an external circuit to enter the reduction 1/2 reaction at the cathode.  Summary:  2H 2 + 4OH - --> 4H 2 O + 4e - (anode)  O 2 + 2H 2 O + 4e - -->4OH - (cathode)  The overall reaction is:  2H 2 + O 2 --> 2H 2 O  The electrolyte in the center is a hot, concentrated solution of KOH.  Electrons enter from the oxidation 1/2 reaction at the anode then pass through an external circuit to enter the reduction 1/2 reaction at the cathode.  Summary:  2H 2 + 4OH - --> 4H 2 O + 4e - (anode)  O 2 + 2H 2 O + 4e - -->4OH - (cathode)  The overall reaction is:  2H 2 + O 2 --> 2H 2 O

Ch 23 sec 2 Cell Potentials  Electrical Potential - cells ability to produce an electric current. Measured in Volts.  The electrical potential results from a competition for electrons.  E 0 = reduction potential  Cell Potential  E 0 = E 0 red - E 0 Oxid  Electrical Potential - cells ability to produce an electric current. Measured in Volts.  The electrical potential results from a competition for electrons.  E 0 = reduction potential  Cell Potential  E 0 = E 0 red - E 0 Oxid

 Standard Cell Potential  measured potential when the ion concentrations in the half-cells are 1M, gases are at 101 kPa and temperature is 25 o C  Half-cell potentials cannot be measured. So, the 1/2 cells are compared to assigning a hydrogen cell 0.00 V where reduction of Hydrogen occurs.  Standard Cell Potential  measured potential when the ion concentrations in the half-cells are 1M, gases are at 101 kPa and temperature is 25 o C  Half-cell potentials cannot be measured. So, the 1/2 cells are compared to assigning a hydrogen cell 0.00 V where reduction of Hydrogen occurs.

 Reduction takes place at the cathode  Oxidation takes place at the anode  Calculating 1/2 cell potential examples:  2 examples  Zinc and Hydrogen half cells.  A voltmeter reads V.  The zinc is oxidized (anode)  Hydrogen Ions are reduced (cathode)  oxidation Zn --> Zn e -  reduction 2H + + 2e - --> H 2  cell reaction Zn + 2H + --> Zn 2+ + H 2  Reduction takes place at the cathode  Oxidation takes place at the anode  Calculating 1/2 cell potential examples:  2 examples  Zinc and Hydrogen half cells.  A voltmeter reads V.  The zinc is oxidized (anode)  Hydrogen Ions are reduced (cathode)  oxidation Zn --> Zn e -  reduction 2H + + 2e - --> H 2  cell reaction Zn + 2H + --> Zn 2+ + H 2

 therefore,  E 0 = E 0 red - E 0 Oxid  E 0 = E 0 H+ - E 0 Zn2+  +0.76V = 0.00V - E 0 Zn2+  E 0 Zn2+ = -0.76V  The value is negative because the tendency for zinc ions to be reduced to zinc metal is less than the tendency of hydrogen ions to be reduced to hydrogen gas. Therefore it is negative because  Zinc is oxidized.  therefore,  E 0 = E 0 red - E 0 Oxid  E 0 = E 0 H+ - E 0 Zn2+  +0.76V = 0.00V - E 0 Zn2+  E 0 Zn2+ = -0.76V  The value is negative because the tendency for zinc ions to be reduced to zinc metal is less than the tendency of hydrogen ions to be reduced to hydrogen gas. Therefore it is negative because  Zinc is oxidized.

 Second example this time for copper.  Reduction Cu e - --> Cu  Oxidation H 2 --> 2H + + 2e -  Cell reaction Cu 2+ + H 2 --> Cu + 2H +  Cell potential measured at +0.34V.  So, E 0 = E 0 red - E 0 Oxid  +0.34V = E 0 Cu H+  E 0 Cu2+ = +0.34V  The potential for Cu ions to be reduced is higher than the potential for H ions to be reduced. Therefore it is positive because  Copper is reduced.  Table 23.2 in your book on page 688 has a list of all the 1/2 cell potentials.  Second example this time for copper.  Reduction Cu e - --> Cu  Oxidation H 2 --> 2H + + 2e -  Cell reaction Cu 2+ + H 2 --> Cu + 2H +  Cell potential measured at +0.34V.  So, E 0 = E 0 red - E 0 Oxid  +0.34V = E 0 Cu H+  E 0 Cu2+ = +0.34V  The potential for Cu ions to be reduced is higher than the potential for H ions to be reduced. Therefore it is positive because  Copper is reduced.  Table 23.2 in your book on page 688 has a list of all the 1/2 cell potentials.

Calculating Standard Cell Potentials  To function a cell must be constructed of 2 half-cells.  If the cell potential for a given redox reaction is positive then the reaction is spontaneous.  To function a cell must be constructed of 2 half-cells.  If the cell potential for a given redox reaction is positive then the reaction is spontaneous.

Calculating Standard Cell Potentials  Example  Determine the cell reaction, the standard cell potential, and the half-cell that acts as the cathode for a voltaic cell composed of the following half- cells.  Fe 3+ + e - --> Fe 2+ E 0 Fe3+ = +0.77V  Ni e - --> Ni E 0 Ni2+ = -0.25V  Answer:  Reduction takes place in the Fe 3+ half cell. So, this cell is the Cathode  Example  Determine the cell reaction, the standard cell potential, and the half-cell that acts as the cathode for a voltaic cell composed of the following half- cells.  Fe 3+ + e - --> Fe 2+ E 0 Fe3+ = +0.77V  Ni e - --> Ni E 0 Ni2+ = -0.25V  Answer:  Reduction takes place in the Fe 3+ half cell. So, this cell is the Cathode

The 1/2 reactions are: Ni --> Ni e - 2[Fe 3+ + e - --> Fe 2+ ] The cell reaction is: Ni + 2Fe 3+ --> Ni Fe 2+ Cell potential is: E 0 = E 0 red - E 0 Oxid +0.77V - (-0.25V) = +1.02V The reaction is spontaneous. The 1/2 reactions are: Ni --> Ni e - 2[Fe 3+ + e - --> Fe 2+ ] The cell reaction is: Ni + 2Fe 3+ --> Ni Fe 2+ Cell potential is: E 0 = E 0 red - E 0 Oxid +0.77V - (-0.25V) = +1.02V The reaction is spontaneous.

 Your Turn:  Calculate E 0 cell to determine whether the following redox reaction is spontaneous as written:  Ni + Fe 2+ --> Ni 2+ + Fe  From table 23.2 in your text  E 0 Ni2+ = -0.25V E 0 Fe2+ = -0.44V  answer  Nickel is oxidized Iron is reduced  E 0 = -0.44V - (-0.25V) =  Reaction will proceed in the reverse  Your Turn:  Calculate E 0 cell to determine whether the following redox reaction is spontaneous as written:  Ni + Fe 2+ --> Ni 2+ + Fe  From table 23.2 in your text  E 0 Ni2+ = -0.25V E 0 Fe2+ = -0.44V  answer  Nickel is oxidized Iron is reduced  E 0 = -0.44V - (-0.25V) =  Reaction will proceed in the reverse

Electrolytic Cells 23.3

 Electrolysis - making a nonspontaneous reaction go.  ex - silverplated dishes and utensils, gold- plated jewelry and chrome-plated auto parts  The electrolytic cell is an electrochemical cell used to cause a chemical change through the application of electrical energy (DC Current).  Similarities between voltaic and electric cells:  electrons flow from the anode to the cathode  reduction occurs at cathode  oxidation anode  Electrolysis - making a nonspontaneous reaction go.  ex - silverplated dishes and utensils, gold- plated jewelry and chrome-plated auto parts  The electrolytic cell is an electrochemical cell used to cause a chemical change through the application of electrical energy (DC Current).  Similarities between voltaic and electric cells:  electrons flow from the anode to the cathode  reduction occurs at cathode  oxidation anode

 Differences  Voltaic is spontaneous  Electrolytic is result of outside push of electrons  Cathode is negative electrode (connected to negative electrode of battery) and vice versa for anode.  Differences  Voltaic is spontaneous  Electrolytic is result of outside push of electrons  Cathode is negative electrode (connected to negative electrode of battery) and vice versa for anode.

 Electrolysis of water produces hydrogen and oxygen gas.  An electrolyte is often used to help conduct an electric current. The electrolyte is usually redoxed.  The electrolysis of Brine (salt water) produces Cl 2 + H 2 + NaOH  Electrolysis of molten NaCl produces:  liquid Na used in sodium vapor lamps and as the coolant in some nuclear reactors.  chlorine gas used to sterilize drinking water and the manufacturing of PVC and pesticides.  Which one comes off of which electrode?  sodium gains an electron - cathode  Electrolysis of water produces hydrogen and oxygen gas.  An electrolyte is often used to help conduct an electric current. The electrolyte is usually redoxed.  The electrolysis of Brine (salt water) produces Cl 2 + H 2 + NaOH  Electrolysis of molten NaCl produces:  liquid Na used in sodium vapor lamps and as the coolant in some nuclear reactors.  chlorine gas used to sterilize drinking water and the manufacturing of PVC and pesticides.  Which one comes off of which electrode?  sodium gains an electron - cathode

Electroplating  The depositing of a thin layer of a metal on an object  Common metal used to plate:  Ag, Au,Cu, Ni, Cr  An object to be silver plated is made the cathode. Why?  Anode is the silver to be deposited  Electrolyte is a silver salt.  Also used to purify metals and various other processes you will read about.  The depositing of a thin layer of a metal on an object  Common metal used to plate:  Ag, Au,Cu, Ni, Cr  An object to be silver plated is made the cathode. Why?  Anode is the silver to be deposited  Electrolyte is a silver salt.  Also used to purify metals and various other processes you will read about.