Electrochemistry Redox reactions.

Electrochemistry Redox reactions

Practical Applications of Redox Reactions
Have you ever chewed on a piece of aluminum foil (hopefully by accident…)? If you have amalgam fillings you may have experienced the transfer of electrons in your mouth. A redox reaction! Or if you have a grill…

Reduction – Oxidation Reactions “REDOX”
Is a chemical reaction in which electrons are transferred Must have both reduction and oxidation happening for the reaction to occur REDUCTION – a process in which electrons are gained by an entity OXIDATION – a process in which electrons are lost by an entity How can you remember this? “LEO the lion says GER” LEO = Losing Electrons = Oxidation GER = Gaining Electrons = Reduction Other memory devices: OIL RIG (Oxidation Is Losing electrons, Reduction Is Gaining electrons) ELMO (Electron Loss Means Oxidation)

Reduction – Oxidation Reactions “REDOX”
Examples of Redox Reactions: Formation, decomposition, combustion, single replacement, cellular respiration, photosynthesis, (NOT double replacement)

An Introduction to Redox #1
Imagine that a reaction is a combination of two parts called half-reactions. A half reaction represents what is happening to one reactant, it tells one part of the story. Another half-reaction is required to complete the description of the reaction. Example: When metal is placed into hydrochloric acid solution, gas bubbles form as the zinc slowly disappears. Zn(s) + Cl2(aq)  ZnCl2(aq) + H2(g) What happens to the zinc? To the HCl(aq)? Look at the half-reactions. Zn(s)  Zn 2+ (aq) e- 2 H+(aq) + 2 e-  H2 (g) Notice that both of these half-reactions are balanced by mass (same number of atoms/ions of each element on both sides) and by charge (same total charge on both sides) A half reaction is a balanced chemical equation that represents either a loss or gain of electrons by a substance You need to break down the two half reactions separately. DO NOT TRY TO DO EVERYTHING AT ONCE!!!! Go through this example on page 561. Notice that Cl is a spectator  doesn’t participate in the reaction goes from -1 on the reactant side and -1 on the product side

Let’s Try this one… In the following equation indicate the element that has been oxidized and the one that has been reduced. You should also label the oxidation state (charge) of each before and after the process as well as write the half reactions and net equation. 2Na + FeCl2  2NaCl + Fe Spectator: Cl remains the same on both sides (-1) therefore it is a spectator ion We are only concerned with Na and F Oxidation Numbers: Na goes from 0 (atoms do not have a charge) on the reactant side to +1(needs to balance out Cl) on the product side  becomes more positive therefore it is oxidized and is the reducing agent Fe goes from +2 (needs to balance Cl) on the reactant side to 0 (atom) on the product side  becomes more negative therefore it is reduced and is the oxidizing agent

Example Continued… We now need to look at our half reactions from the data booklet Fe2+(aq) + 2e-  Fe(s) Na(s)  Na+(aq) + e- (Na(s)  Na+(aq) + e-) x 2 Net Reaction: Fe2+(aq) + 2e-  Fe(s) 2Na(s)  2Na+(aq) + 2e- Fe2+(aq) + 2Na(s)  Fe(s) + 2Na+(aq) We need to multiply the Na half reaction by 2 in order to balance the electrons Half reactions  which you got from your data booklet. Reduced Oxidized

Example 3 Write the balanced half-reaction equation and a balanced net equation for the reaction of copper metal with aqueous silver nitrate

Example 4

Example 5

Identifying the OA and RA
Identify the oxidizing and reducing agents in the oxidation-reduction reaction 2 H2O(l) + Al(s) + MnO4–(aq) →Al(OH)4–(aq) + MnO2(s) Answer: Al(s) is the reducing agent; MnO4–(aq) is the oxidizing agent.

Your Task You need to read pages 558-571 and take your own notes.
Question # 2a-f page 559 Questions #7, 8ab,9a-d, 11 page 564 Questions #1, 2, 3,5-7 Thursday you will have a quiz with questions that will be taken directly from the Practice questions on page 559 and 571. I will not be changing the questions at all so if you want to be prepared you should maybe have a look at these questions. Happy Thought of the Day: When you feel down take comfort in the fact that you don’t look as pathetic as this  I will see you all tomorrow

Example: When a piece of copper is placed in a beaker of silver nitrate, the following changes occur. Cu(s) + AgNO3(aq)  Cu(NO3)2(aq) + Ag(s) Write the balanced half-reaction equations: To show that the number of electrons gained equals the number of electrons lost in two half-equations, it may be necessary to multiply one or both half-reaction equations by a coefficient to balance the electrons. Ex: Ag half reaction must be multiplied by 2 Cu(s)  Cu 2+ (aq) e- 2 [Ag+(aq) + e-  Ag (s)] Where is Oxidation occurring? Where is Reduction occurring? OXIDATION REDUCTION

Cu(s)  Cu 2+ (aq) e- 2 [Ag+(aq) + e-  Ag (s)] Now add the half-reactions and cancel the terms that appear on both sides of the equation to obtain the net-ionic equation 2 Ag+(aq) + 2 e- + Cu(s)  2 Ag(s) + Cu 2+ (aq) e- 2 Ag+(aq)+ Cu(s)  2 Ag(s) + Cu 2+ (aq) Silver ions are reduced to silver metal by reaction with copper metal. Simultaneously, copper metal is oxidized to copper(II) ions by reaction with silver ions. OXIDATION REDUCTION

Silver ions are reduced to silver metal by reaction with copper metal. Simultaneously, copper metal is oxidized to copper(II) ions by reaction with silver ions.

There are two methods for developing net ionic equations: 1) ½ reaction method we just learned OR 2) Using the net-ionic equation method from Chem 20 Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s) (dissociate high solubility and ionic compounds) Cu(s) Ag + (aq) + 2 NO3- (aq)  Cu2+(aq)+ 2NO3-(aq) + 2Ag(s) (cancel spectator ions) 2 Ag+(aq)+ Cu(s)  2 Ag(s) + Cu 2+ (aq) (Do we get the same net ionic reaction?? YES!) NON-IONIC TOTAL IONIC NET-IONIC

Summary “Electron Transfer Theory”
A redox reaction is a chemical reaction in which electrons are transferred between entities The total number of electrons gained in the reduction equals the total number of electrons lost in the oxidation Reduction is a process in which electrons are gained by an entity Oxidation is a process in which electrons are lost by an entity Both reduction and oxidation are represented by balanced half- reaction equations.

REDOX Reactions…. so far
Reduction Oxidation Historically, the formation of a metal from its “ore” (or oxide) Example: nickel(II) oxide is reduced by hydrogen gas to nickel metal NiO(s) + H2(g)  Ni(s) + H2O(l) Ni  Nio A gain of electrons occurs (so the entity becomes more negative) Electrons are shown as the reactant in the half-reaction Historically, reactions with oxygen Example: iron reacts with oxygen to produce iron(III) oxide 4 Fe(s) + O2(g)  Fe2O3(s) Fe  Fe+3 A loss of electrons occurs (so the entity becomes more positive) Electrons are shown as the product in the half-reaction

Redox Terms Review: “LEO the lion says GER”
Loss of electrons = entity being oxidized Gain of electrons = entity being reduced BUT…. Chemists don’t say “the reactant being oxidized” or “the reactant being reduced” Rather, they use the terms OXIDIZING AGENT (OA) and REDUCING AGENT (RA) OXIDIZING AGENT: causes oxidation by removing (gaining) electrons from another substance in a redox reaction REDUCING AGENT: causes reduction by donating (losing) electrons to another substance in a redox reaction

Redox Terms What does this mean? Let’s revisit our first example when zinc and hydrochloric acid reacted. Which reactant was reduced? Which was oxidized? So…. Which is the Oxidizing Agent (OA)? Which is the Reducing Agent (RA) LEO = Oxidized Zn(s)  Zn 2+ (aq) e- 2 H+(aq) + 2 e-  H2 (g) Reducing Agent GER = Reduced Oxidizing Agent

Redox Terms Let’s revisit Example #3….
Silver ions were reduced to silver metal by reaction with copper metal. Simultaneously, copper metal was oxidized to copper(II) ions by reaction with silver ions. If Ag+(aq) is reduced it is the: If Cu(s) is oxidized it is the: OXIDIZING AGENT (OA) REDUCING AGENT (RA) It is important to note that oxidation and reduction are processes, and oxidizing agents and reducing agents are substances.

REDOX Reactions … so far
Reduction Oxidation Historically, the formation of a metal from its “ore” (or oxide) I.e. nickel(II) oxide is reduced by hydrogen gas to nickel metal NiO(s) + H2(g)  Ni(s) + H2O(l) Ni  Nio A gain of electrons occurs (so the entity becomes more negative) Electrons are shown as the reactant in the half- reaction A species undergoing reduction will be responsible for the oxidation of another entity – and is therefore classified as an oxidizing agent (OA) Historically, reactions with oxygen I.e. iron reacts with oxygen to produce iron(III) oxide 4 Fe(s) + O2(g)  Fe2O3(s) Fe  Fe+3 A loss of electrons occurs (so the entity becomes more positive) Electrons are shown as the product in the half- reaction A species undergoing oxidation will be responsible for the reduction of another entity – and is therefore classified as an reducing agent (RA)

The substance has a very strong attraction for electrons.
Redox Terms Summary so far: The substance that is reduced (gains electrons) is also known as the oxidizing agent The substance that is oxidized (loses electrons) is also knows as the reducing agent Question: If a substance is a very strong oxidizing agent, what does this mean in terms of electrons? The substance has a very strong attraction for electrons. Question: If a substance is a very strong reducing agent, what does this mean in terms of electrons? The substance has a weak attraction for its electrons, which are easily removed

Redox Table A reaction is considered spontaneous if it occurs on its own A reduction ½ reaction table is useful in predicting the spontaneity of a reaction Reduction Tables show reduction ½ reactions in the forward direction, therefore all the reactants will be oxidizing agents If we list the OA’s from our intro lab in decreasing order of strength, we create a reduction ½ reaction table: Ag+(aq) e-  Ag(s) Cu2+(aq) e-  Cu(s) Zn2+(aq) e-  Zn(s) Mg2+(aq) e-  Mg(s) SOA SRA

Building Redox Tables #1
Consider the following experimental information and add half-reactions to the redox table you have created Au3+(aq) e-  Au(s) Hg2+(aq) e-  Hg(s) Ag+(aq) e-  Ag(s) Cu2+(aq) e-  Cu(s) Hg2+(aq) Cu2+(aq) Ag+(aq) Au3+(aq) Hg(s) ✗ ✓ Cu(s) Ag(s) Au(s) SOA SRA

Check page 7 of your data booklet. Does our ranking order match up with theirs? Au3+(aq) e-  Au(s) Hg2+(aq) e-  Hg(s) Ag+(aq) e-  Ag(s) Cu2+(aq) e-  Cu(s) Zn2+(aq) e-  Zn(s) Mg2+(aq) e-  Mg(s) YES! Because of the spontaneity rule! A reaction will be spontaneous if on a redox table: OA RA above = Spontaneous below = Non-spontaneous RA Reaction OA Reaction SOA SRA A spontaneous reaction proceeds w/o additional energy or stimulus

Building Redox Tables Au3+(aq) + 3 e-  Au(s) Hg2+(aq) + 2 e-  Hg(s)
Ag+(aq) e-  Ag(s) Cu2+(aq) e-  Cu(s) Zn2+(aq) e-  Zn(s) Mg2+(aq) e-  Mg(s) SOA SRA

Use the following information to create a table of reduction ½ reactions 3 Co 2+ (aq) + 2 In(s)  2 In 3+ (aq) + 3 Co(s) Cu 2+ (aq) + Co(s)  Co 2+ (aq) + Cu(s) Cu 2+ (aq) + Pd(s)  no reaction

Use the following information to create a table of reduction ½ reactions 2 A 3+ (aq) + 3 D(s)  3 D2+ (aq) + 2 A(s) G + (aq) + D(s)  no reaction 3 D 2+ (aq) + 2 E(s)  3 D(s) + 2 E3+(aq) G +(aq) E(s)  no reaction

Grand Exalted Ruler You
Need To Add a Chart

Building Redox Tables So far we have been using examples where the oxidizing agents are metal ions and the reducing agents are metal atoms. What else could gain or lose electrons? Non-metal atoms Example: Cl2(g) + 2e-  2 Cl-(aq) (Cl2(g) could act as a Reducing Agent) Non-metal ions Example: 2 Br- (aq)  Br2(l) + 2 e (2Br-(aq) could act as an Oxidizing Agent) Redox Table Trend OA’s tend to be metal ions and non-metal atoms RA’s tend to be metal atoms and non-metal ions Also, are there any entities that could act as both OA or RA? Multivalent metals

Try This… Use the following information to create a table of reduction ½ reactions Ag(s) Br2(l)  AgBr(s) Ag(s) I2(s)  no evidence of reaction Cu2+(aq) I-(aq)  no redox reaction Br2(l) Cl-(aq)  no evidence of reaction

Your Task Page 571 Questions 5, 6, 7,8 Page 573 Questions 11, 12ab, 14
Page 574 Questions 18(challenge question), 20a-f

Predicting Redox Reactions in Solution
If solid copper is placed in nitric acid, what is the net ionic equations?

Writing Complex Redox Reactions
A method of writing half reactions for polyatomic ions and molecular compounds requires that water molecules and hydrogen or hydroxide ions be included. We use this method when the data table doesn’t include the half reactions

Balancing Redox Reactions in Acidic Solutions
In an acidic solution, after writing out the half-reactions of oxidation and reduction, each half reaction must be completed.

Rules for Writing Half-Reactions
1. Write an unbalanced reaction showing formulas for reactants and products 2. Balance all atoms except H and O 3. Balance O by adding H2O(l) where necessary 4. Balance H by adding H+(aq) 5. Balance the charge by adding e- and cancel anything that is the same on both sides (you may have to multiply) 8. Combine the half-reactions to form the net ionic equation 9. If the same entity occurs on both sides subtract the lower number from both sides so that it is only on one side For basic solutions only: 6. Add OH-(aq) to both sides to equal the number of H+(aq) present 7. Combine H+(aq) and OH-(aq) on the same side to form H2O(l). Cancel equal amounts of H2O(l) from both sides.

Practice Given the following reaction occurs in an acidic solution, predict the net equation. Page 76 in SNAP

Practice The redox reaction of thallium and iodate occurs in an acidic environment. Write a balanced redox reaction equation for the reaction Page 50 the key

Practice Page 50 the Key

Practice Copper metal can be oxidized in a basic solution to form copper(I) oxide. What is the half-reaction for this process? Textbook page 565

Practice Chlorine is converted to perchlorate ions in an acidic solution. Write the half- reaction equation. Is this half-reaction an oxidation or reduction? Textbook page 566

Practice Nitrous acid can be reduced in an acidic solution to form nitrogen monoxide gas. What is the reduction half-reaction for nitrous acid?

Practice Aqueous permanganate ions are reduced to solid manganese (IV) oxide in a basic solution. Write the half reaction equation. Is the half reaction an oxidation or a reduction? Textbook page 566

Balance the following redox equation for the reaction that occurs in an acidic solution
Page 79 in SNAP book question 5

Now You Try Complete Practice Question 12a-e on page 566. You must show ALL of your work

Summary 1. Use the information provided to start two half-reaction equations. Using the rules we just learned about half-reactions 2. Balance each half-reaction equation. 3. Multiply each half-reaction by simple whole numbers to balance electrons lost and gained. 4. Add the two half-reaction equations, cancelling the electrons and anything else that is exactly the same on both sides of the equation.

Predicting Redox Reactions by Constructing Half Reactions
Example: A person suspected of being intoxicated blows into this device and the alcohol in the person’s breath reacts with an acidic dichromate ion solution to produce acetic acid (ethanoic acid) and aqueous chromium(III) ions. Predict the balanced redox reaction equation. • Create a skeleton equation from the information provided: Separate the entities into the start of two half-reaction equations Now use the steps you learned for writing half reactions

Practice # 31 on page 581

Answer

Oxidation States An oxidation state is defined as the apparent net electric charge an atom would have if the electron pairs in a covalent bond belonged entirely to the most electronegative atom. An oxidation number is a positive or negative number corresponding to the oxidation state of the atom in a compound. (These are NOT charges! +2 vs 2+) Example: In water, which is the most electronegative atom, H or O? Oxygen, so we act as if the oxygen owns both electrons in the electron pair. Each oxygen atom has 8 p+ and 8 e-. But if the oxygen atom gets to count the two hydrogen electrons (red dots) in the two shared pairs, as its own, then it has 8 p+ but 10 e-, leaving an apparent net charge of -2 Each hydrogen atom has 1 p+, but with no additional electron (since oxygen has already counted it), that leaves hydrogen with an apparent net charge of +1

The sum of the oxidation numbers for a neutral compound = 0
Oxidation States Tip: The sum of the oxidation numbers for a neutral compound = 0 The sum of the oxidation numbers for a polyatomic ion = ion charge ** This method only works if there is only one unknown after referring to the above table

Oxidation States Example: What is the oxidation number of carbon in methane CH4? After referring to Table 1, we assign an oxidation number of +1 to hydrogen So now we have some simple math… Since a methane molecule is electrically neutral, then the oxidation number of the one carbon atom and the four hydrogen atoms 4(+1) must equal zero. x (+1) = So the oxidation number of carbon is = -4 How do we write this?

Oxidation States Example: What is the oxidation number of manganese in a permanganate ion, MnO4- ? After referring to Table 1, we assign an oxidation number of -2 to oxygen Since a permanganate ion has a charge of 1-, then the oxidation number of the one manganese atom and the four oxygen atoms 4(-2) must equal -1. x (-2) = x = So the oxidation number of manganese is = -7 Example: What is the oxidation number of sulfur in sodium sulfate? We know the oxidation numbers of both Na and O, and solve algebraically 2(+1) + x (-2) = 0 x = 0 So the oxidation number of sulfur is +6

Redox in Living Organisms
The ability of carbon to take on different oxidation states is essential to life on Earth. Photosynthesis involves a series of reduction reactions in which the oxidation number of carbon changes from +4 in carbon dioxide to an average of 0 in sugars such as glucose. The smell of a skunk is caused by a thiol compound (R-SH). To deodorize a pet sprayed by a skunk, you need to convert the smelly thiol to an odourless compound. Hydrogen peroxide in a basic solution acts as an oxidizing agent to change the thiol to a disulfide compound (RS-SR), which is odourless.

Assign common oxidation numbers (Table 1 on page 583)
Determining Oxidation Numbers Summary Assign common oxidation numbers (Table 1 on page 583) The total of the oxidation numbers of atoms in a molecule or ion equals the value of the net electric charge of the molecule or ion. The sum of the oxidation numbers for a compound is zero. The sum of the oxidation numbers for a polyatomic ion equals the charge of the ion. Any unknown oxidation number is determined algebraically from the sum of the known oxidation numbers and the net charge on the entity.

Oxidation Numbers and Redox
Although the concept of oxidation states is somewhat arbitrary, because it is based on assigned charges, it is self-consistent and allows predictions of electron transfer. Chemists believe that if the oxidation number of an atom or ion changes during a chemical reaction, then an electron transfer (oxidation-reduction reaction) occurs. Based on oxidation numbers, If the oxidation numbers do not change = no transfer of e-’s = not a redox rxn An increase in the oxidation number = oxidation A decrease in the oxidation number = reduction

REDOX Reactions … the end
Reduction Oxidation Historically, the formation of a metal from its “ore” (or oxide) Example: nickel(II) oxide is reduced by hydrogen gas to nickel metal NiO(s) + H2(g)  Ni(s) + H2O(l) Ni  Nio A gain of electrons occurs (so the entity becomes more negative) Electrons are shown as the reactant in the half- reaction A species undergoing reduction will be responsible for the oxidation of another entity – and is therefore classified as an oxidizing agent (OA) Decrease in oxidation number Historically, reactions with oxygen Example: iron reacts with oxygen to produce iron(III) oxide 4 Fe(s) + O2(g)  Fe2O3(s) Fe  Fe+3 A loss of electrons occurs (so the entity becomes more positive) Electrons are shown as the product in the half- reaction A species undergoing oxidation will be responsible for the reduction of another entity – and is therefore classified as an reducing agent (RA) Increase in oxidation number

Example: Identify the oxidation and reduction in the reaction of zinc metal with hydrochloric acid. First write the chemical equation (as it is not provided) Determine all of the oxidation numbers Now look for the oxidation number of an atom/ion that increases as a result of the reaction and label the change as oxidation. There must also be an atom/ion whose oxidation number decreases. Label this change as reduction.

Example: When natural gas burns in a furnace, carbon dioxide and water form. Identify oxidation and reduction in this reaction. First write the chemical equation (as it is not provided) Determine all of the oxidation numbers Now look for the oxidation number of an atom/ion that increases as a result of the reaction and label the change as oxidation. There must also be an atom/ion whose oxidation number decreases. Label this change as reduction.

Balancing Redox Equations using Oxidation Numbers #1
Example: When hydrogen sulfide is burned in the presence of oxygen, it is converted to sulfur dioxide and water vapour. Use oxidation numbers to balance this equation H2S(g) + O2(g)  SO2(g) + H2O(g) Assign oxidation numbers to all atoms/ions and look for the numbers that change. Highlight these. Notice that a sulfur atom is oxidized from -2 to +4. This is a change of 6 meaning 6 e- have been transferred. An oxygen atom is reduced from 0 to -2. This is a change of 2 or 2e- transferred. Because the substances are molecules, not atoms, we need to specify the change in the number of e-’s per molecule The next step is to determine the simplest whole numbers that will balance the number of electrons transferred for each reactant. The numbers become the coefficients of the reactants 3. The coefficients for the products can be obtained by balancing the atoms whose oxidation numbers have changed and then any other atoms.

Example: Chlorate ions and iodine react in an acidic solution to produce chloride ions and iodate ions. Balance the equation for this reactions. ClO3-(aq) + I2(aq)  Cl-(aq) + IO3-(aq) Assign oxidation numbers to all atoms/ions and look for the numbers that change. Highlight these. Remember to record the change in the number of electrons per atom and per molecule or polyatomic ion. 2. The next step is to determine the simplest whole numbers that will balance the number of electrons transferred for each reactant. The numbers become the coefficients of the reactants. The coefficients for the products can be obtained by balancing the atoms whose oxidation numbers have changed and then any other atoms. 3. Although Cl and I atoms are balanced, oxygen is not. Add H2O(l) molecules to balance the O atoms. 4. Add H+(aq) to balance the hydrogen. The redox equation should now be completely balanced. Check your work by checking the total numbers of each atom/ion on each side and checking the total electric charge, which should also be balanced.

Example: Methanol reacts with permanganate ions in a basic solution. The main reactants and products are shown below. Balance the equation for this reaction. Assign oxidation numbers to all atoms/ions and look for the numbers that change. Highlight these. Remember to record the change in the number of electrons per atom and per molecule or polyatomic ion. Determine the simplest whole numbers that will balance the number of electrons transferred for each reactant. The numbers become the coefficients of the reactants. The coefficients for the products can be obtained by balancing the atoms whose oxidation numbers have changed and then any other atoms. Add H2O(l) to balance the oxygen, add H+(aq) to balance the hydrogen.

Example: Household bleach contains sodium hypochlorite. Some of the hypochlorite ions disproportionate (react with themselves) to produce chloride ions and chlorate ions. Write the balanced redox equation for the disproportionation. For disproportionation reactions, start with two identical entities on the reactant side and follow the usual procedure for balancing equations.

Example: A person suspected of being intoxicated blows into this device and the alcohol in the person’s breath reacts with an acidic dichromate ion solution to produce acetic acid (ethanoic acid) and aqueous chromium(III) ions. Balance the equation for this reaction. Remember to balance the C and Cr first, then add H2O(l) to balance O, H+(aq) to balance H and then stop because this is an acidic solution

Balance the net equation for the reaction of solid copper with nitric acid using oxidation numbers.

Balancing Redox Equations using Oxidation Numbers
Assign oxidation numbers and identify the atoms/ions whose oxidation numbers change Using the change in oxidation numbers, write the number of electrons transferred per atom. Using the chemical formulas, determine the number of electrons transferred per reactant. (Use formula subscripts to do this) Calculate the simplest whole number coefficients for the reactants that will balance the total number of electrons transferred. Balance the reactants and products. Balance the O atoms using H2O(l), and then balance the H atoms using H+(aq)

Predicting Redox Reactions
Now that you know what redox reactions are, you will be responsible for determining if a reaction will occur (is spontaneous) and if so, what the reaction equation will be. How do we do this? The first step is to determine all the entities that are present. Remember: In solutions, molecules and ions behave independently of each other. Example: When a solution of potassium permanganate is slowly poured through acidified iron(II) sulfate solution. Does a redox reaction occur and what is the reaction equation?

The second step is to determine all possible OA’s and RA’s This is a crucial step!! Things to watch out for: Combinations (i.e. MnO4-(aq) is an oxidizing agent only in an acidic solution) To indicate this draw an arc between the permanganate and hydrogen ion Species that can act as both OA and RA Any lower charge multivalent metal i.e. Fe2+, Cu+, Sn2+, Cr2+ Water (H2O(l)) Label both possibilities in your list

Before we move on, let’s practice Step 1 and 2

The third step is to identify the SOA and SRA using the data booklet 4. The fourth step is to show the ½ reactions (from the redox table) and balance SOA equation straight from table. SRA equation read from right to left Are these equations balanced? Do the number of electrons lost = electrons gained If not, multiply one or both equations by a number then add the balanced equations SOA SRA

5. The last step is to predict the spontaneity. Does the net ionic equation represent a spontaneous or non-spontaneous redox reaction? If the SOA above  Spontaneous SRA If the SRA below  Nonspontaneous SOA

Predicting Redox Reactions #2
Could copper pipe be used to transport a hydrochloric acid solution? List all entities 2. Identify all possible OA’s and RA’s 3. Identify the SOA and SRA 4. Show ½ reactions and balance 5. Predict spontaneity Since the reaction is nonspontaneous, it should be possible to use a copper pipe to carry hydrochloric acid

Disproportionation The redox reactions we have covered so far have one reactant (OA) which removes electrons from a second reactant (RA) if a spontaneous reaction is to occur. Although the OA and RA are usually different entities, this is not a requirement. A reaction is which a species is both oxidized and reduced is called disproportionation (aka autoxidation or self oxidation-reduction) Occurs when a substance can act as either as oxidizing or reducing agent Example: Will a spontaneous reaction occur as a result of an electron transfer from one iron(II) ion to another iron (II) ion? No! Using the redox table and spontaneity rule, we see that iron(II) as an oxidizing agent is below iron(II) as a reducing agent, so the reaction is nonspontaneous Disproportionation reactions occur when the same entity is both oxidized and reduced in a redox reaction. Usually, one entity is reduced and the other is oxidized, but when one entity can act as both an oxidizing agent and a reducing agent, it can undergo both oxidation and reduction. In terms of oxidation numbers, this can be described as an entity’s oxidation number both increasing and decreasing over the course of the reaction.

2 Cu+(aq)  Cu2+(aq) + Cu(s)
Disproportionation Example #2: Will a spontaneous reaction occur as a result of an electron transfer from one copper(I) ion to another copper (I) ion? Cu+(aq) + 1 e-  Cu(s) Cu+(aq)  Cu2+(aq) + 1 e- 2 Cu+(aq)  Cu2+(aq) + Cu(s) YES! Using the redox table and spontaneity rule, we see that copper(I) as an oxidizing agent is above copper(I) as a reducing agent. Therefore, an aqueous solution of copper(I) ions will spontaneously, but slowly, disproportionate into copper(II) ions and copper metal. Disproportionation reactions occur when the same entity is both oxidized and reduced in a redox reaction. Usually, one entity is reduced and the other is oxidized, but when one entity can act as both an oxidizing agent and a reducing agent, it can undergo both oxidation and reduction. In terms of oxidation numbers, this can be described as an entity’s oxidation number both increasing and decreasing over the course of the reaction.

Example Reaction of Hydrogen peroxide
Disproportionation is also commonly known as self-oxidation-reduction or autooxidation. - Oxygen has an oxidation number of -1 in water and 0 in O2. Its oxidation number both increases and decreases over the course of the reaction. Because a change in oxidation number indicates a loss or gain of electrons., peroxide is both reduced and oxidized in this reaction.

Summary Redox reactions are characterized by a transfer of electrons from a reducing agent to an oxidizing agent  no electron transfer = no redox reaction Oxidation numbers need to change In disproportionation reactions the same entity both increases and decreases in oxidation number No change in oxidation number = no redox reaction

Try This Determine if the following reaction is a redox reaction
Solid iron reacts with chlorine gas

Titration Review A titration is a quantitative laboratory technique used to determine the concentration of an unknown solution. A reagent, of known concentration, called the titrant is used to react with a solution, called the sample. Using a buret to add the titrant, the volume needed to reach the endpoint can be determined. The endpoint is the point at which the titration is complete, usually determines by an indicator (color change), but not always. This is ideally the same volume as the equivalence point, when stoichiometrically equivalent amounts of each reagent have been added.

Redox Titration In redox titration, no indicator is required because the titrant is a strong oxidizing agent that has a very significant colour change when it undergoes reduction. Common OA’s used in redox titration are MnO4-(aq) and Cr2O72-(aq) , both in acidified solutions. In a redox titration, it is often necessary to standardize the titrant. Due to the reactive nature of the oxidizing agents used as the titrant, they often react with themselves in their storage container. Standardizing involves performing an initial titration with a solution prepared from a solid (so the exact concentration is known) to determine the exact concentration of the titrant.

Titration Procedure Review
An initial reading of the burette is made before any titrant is added to the sample. Then the titrant is added until the reaction is complete; when a final drop of titrant permanently changes the colour of the sample. The final burette reading is then taken. The difference between the readings is the volume of titrant added. Near the endpoint, continuous gentle swirling of the solution is important

Titration Procedure Review
A titration should involve several trials, to improve reliability of the answer. A typical requirement is to repeat titrations until three trials result in volumes within a range of 0.2mL. These three results are then averaged before carrying out the solution stoichiometry calculation; disregard any trial volumes that don’t fall in the range. Remember to read the titrant volume from the bottom of the meniscus. Remember the top of the buret reads 0.0mL, so you will subtract the initial reading from the final reading, to determine the difference or amount of titrant added

Redox Stoichiometry There are many industrial and laboratory applications of redox stoichiometry: Mining engineer must know the concentration of iron in a sample of iron ore to decide whether or not a mine would be profitable. Chemical technicians must monitor the concentration of substances in products (i.e. how much bleach is in a disinfectant) Hospital lab technicians must detect tiny traces of chemicals in human samples. How is this different from Chemistry 20 stoichiometry? We will need to predict the redox equation that will occur, and then we will use the quantities provided to answer the question. The math is the same as Chem 20, we will just be using our knowledge of redox to start the question.

2H+(aq) + Cu(s)  H2(g) + Cu2+(aq)
Redox Stoichiometry Example #1 A strong acid is painted onto a copper sheet to etch a design. If 500 mL of a mol/L solution is used, what mass of copper will react? List entities present, identify SOA and SRA: H+(aq) Cu(s) H2O(l) Write oxidation and reduction half reactions. Balance the number of electrons gained and lost and add the reactions 2H+(aq) + 2e-  H2(g) Cu(s)  Cu2+(aq) + 2e- 2H+(aq) + Cu(s)  H2(g) + Cu2+(aq) V= 500mL m= ???g mol/L 0.500 L x mol H+(aq) x 1 mol Cu(s) x g = 3.97 g Cu(s) L mol H+(aq) mol Cu(s) SOA SRA

3Ni(s) + Cr2O72-(aq) + 14 H+(aq)  3Ni2+(aq + 2Cr3+(aq) + 7H2O(l)
Redox Stoichiometry Example #2 Nickel metal is oxidized to Ni2+(aq) ions by an acidified potassium dichromate solution. If 2.50g of metal is oxidizes by 50.0 mL of solution, what is the concentration of the K2Cr2O7(aq) solution? List entities present, identify SOA and SRA: Ni(s) H+(aq) K+(aq) Cr2O72-(aq) H2O(l) Write oxidation and reduction half reactions. Balance the number of electrons gained and lost and add the reactions 3 [Ni(s)  Ni2+(aq) + 2e- ] Cr2O72-(aq) + 14 H+(aq)+ 6 e-  2Cr3+(aq) + 7H2O(l) 3Ni(s) + Cr2O72-(aq) + 14 H+(aq)  3Ni2+(aq + 2Cr3+(aq) + 7H2O(l) SOA SRA

Redox Stoichiometry A solution of potassium permanganate cannot be directly prepared with a precisely known concentration because the permanganate ion reacts with organic and inorganic impurities in the water and with the water itself. Thus, permanganate is not used as a primary standard. Through an experiment it was found that 16.8mL of permanganate was used to reach the endpoint. What is the concentration of the potassium permanganate solution?

Titration of tin (II) in solution 10. 00 mL of acidified, 0
Titration of tin (II) in solution mL of acidified, mol/L tin (II) chloride solution is reacted with a potassium permanganate solution. Calculate the concentration of potassium permanganate using the following data.

Challenge Question Trial 1 Final Burette reading (mL)
The titration of a 15.0mL sample of mol/L potassium dichromate solution is completed by the addition of mol/L iron (II) chloride solution. Fill in the following titration chart Trial 1 Final Burette reading (mL) Initial Burette reading (mL) 1.0 Volume of iron (II) chloride added

Challenge Question A titration is performed with a 20.0 mL sample of SnCl2 in an acidic solution. The titration requires 15.0mL of M K2Cr2O7. The following balanced equation represents the reaction Cr2O7-2(aq) + 14H+(aq) + 3Sn2+(aq)  2 Cr3+(aq) + 3 Sn2+(aq) + 7 H2O(l) Determine the original concentration of the SnCl2(aq) sample.

Challenge Question An iron nail with a mass of 1.635g is dissolved in an acidic solution yielding Fe2+(aq) which is then titrated with mL of aqueous potassium permanganate. Assume the entire mass of the iron nail is converted to Fe2+(aq) Write the equations representing the half-reactions involved in the given titration reaction Balance the half-reaction equations and write the net balanced redox equation representing the reaction Determine the concentration of the potassium solution

Redox Stoichiometry Complete Lab Exercise 13.C on page 598
Complete questions 3-5 on page 598. You must show ALL of your work Complete Lab Exercise 13.D on page 599 For the following questions you will need to read your textbook to help you answer Questions #2abc-6 and bonus question 7 page 600

Practical Applications of Redox Reactions
In the last chapter we gained experience in the process of electron transfer in redox reactions. We will now at how those electrons can be put to work by converting chemical energy to electrical energy. Remember: If a reaction is spontaneous you can get electricity from chemistry If a reaction is non-spontaneous, you must force in electricity to get chemistry to happen

Electrochemical cells
The main difference between the redox reactions discussed in the last chapter and the voltaic cells of this chapter is that the voltaic cell is a system in which the half-reactions are separated so that the electron transfer must occur through an external circuit (wire) as the chemical change occurs. This flow of electrons is capable of doing work.

Definitions Voltaic Cell: An electrochemical cell that spontaneously reacts to produce electricity (converts chemical energy to electrical energy) Electrolytic Cell: An electrochemical cell that uses electricity to cause a nonspontaneous chemical change (converts electrical energy to chemical energy) Electrode: The solid in the half cell where the wires attach Salt bridge/ porous cup: A barrier that keeps the solutions in a voltaic cell apart, but allows the flow of ions between solutions to prevent electrical charge build up

Definitions The solid metal electrode at which the reduction half reaction occurs. The convention is that this is the ‘+’ electrode in a voltaic cell CATHODE ANODE The solid metal electrode at which the oxidation half reaction occurs. The convention is that this is the ‘-’ electrode in a voltaic cell. CATIONS The positive ions in solution that move toward the cathode to balance the ion charge and maintain charge neutrality in the cell. ANIONS The negative ions in solution that move toward the anode to balance the ion charge and maintain charge neutrality in the cell. Electron Flow Electrons ALWAYS flow from ANODE to CATHODE Electrolyte The solutions in each half-cell

Introduction to Electrochemistry
An electric cell converts chemical energy into electrical energy Alessandro Volta invented the first electric cell but got his inspiration from Luigi Galvani. Galvani’s crucial observation was that two different metals could make the muscles of a frog’s legs twitch. Unfortunately, Galvani thought this was due to some mysterious “animal electricity”. It was Volta who recognized this experiment’s potential. An electric cell produces very little electricity, so Volta came up with a better design: A battery is defined as two or more electric cells connected in series to produce a steady flow of current Volta’s first battery consisted of several bowls of brine (NaCl(aq)) connected by metals that dipped from one bowl to another His revised design, consisted of a sandwich of two metals separated by paper soaked in salt water. Electrochemical cell - any electrochemical reaction in which an external current is made from a chemical reaction or a chemical reaction is made from an electrical current. (generally the same as Voltaic cell and Galvanic cell)

Alessandro Volta’s invention was an immediate technological success because it produced electric current more simply and reliably than methods that depended on static electricity. It also produced a steady electric current –something no other device could do.

A voltmeter is a device that measures the energy difference, per unit charge, between any two points in an electric circuit (called electric potential difference) I.e. A 9V battery releases 6X as much energy compared with the electrons from a 1.5V battery. The voltage of a cell depends mainly on the chemical composition of the reactants in the cell An ammeter is a device that measures the rate of flow of charge past a point in an electrical circuit (called electric current) The larger the electric cell, the greater current that can be produced Electric potential difference is like the potential energy difference between 1kg of water at the top of the dam and 1kg of water at the bottom of the dam. Electric current (or flow of electrons) is like the flow of the water. A larger drain would be like a larger electric cell, allowing more water (or electrons) to flow.

The power of a battery is the rate at which it produces electrical energy. Power is measured in watts (W). Calculated using P = IV (Power = current x potential difference) The energy density is a measure of the quantity of energy stored or supplied per unit mass. Measured in J/kg. Table 1 summarizes some important electrical quantities and their units

Secondary cells can be recharged using electricity, but are expensive
Primary cells cannot be recharged, but are relatively inexpensive Primary - “dry cells” use a paste as the electrolyte which separates the anode and cathode. These cells cannot be recharged. Secondary - can be recharged by using electricity to reverse the chemical reaction. Fuel - produces electricity by the supply of an external fuel. [clip from Alta Inno] Fuel cells produce electricity by the reaction of a fuel that is continuously supplied. More efficient, and used for NASA vehicles, but still too expensive for general or commercial applications

Electric cells are composed of two electrodes – solid electrical conductors and at least one electrolyte (aqueous electrical conductor) In current cells, the electrolyte is often a moist paste (just enough water is added so that the ions can move). Sometimes one electrode is the cell container. The positive electrode is defined as the cathode and the negative electrode is defined as the anode The electrons flow through the external circuit from the anode to the cathode. To test the voltage of a battery, the red(+) lead is connected to the cathode (+ electrode), and the black(-) lead is connected to the anode (- electrode)

Identifying - Anode & Cathode
Oxidation occurs at the anode. (LEO – A or vowels) Anions migrate to the anode (-) Reduction occurs at the cathode. (GER – C or consonants) Cations migrate to the cathode (+) RED CAT and AN OX can be applied to electrochemical cells; “REDuction occurs at the CAThode At the ANode OXidation occurs”

Electron/Ion Flow Anions  “migrate to”  Anode (-)
Cations  “migrate to”  Cathode (+) Electrons flow from the negative terminal (anode) to the positive terminal (cathode) [“Electrons follow the alphabet” – from the anode to the cathode.] Electrons will always travel through the wire – not the solution.

Page 84 in SNAP put in diagram
To predict the reactions that occur in a voltaic cell, the strongest oxidizing agent and the SRA need to be identified. Oxidation occurs at the anode, where the SRA in the cell will lose electrons. The SOA will alsways undergo reduction at the cathode, where it will gain electrons.

Voltaic and Electrochemical Cells
Voltaic Cell Electrolytic Cell Spontaneous Yes No E°net Positive Negative Direction of electron flow Anode to cathode Anode Oxidation Cathode Reduction Direction of ion flow Anions to anode; cations to cathode Salt bridge Required Usually not required

Practice Draw the voltaic cell constructed using magnesium solid and magnesium sulfate, and copper solid and copper (II) sulfate. Determine the SOA and the SRA. Write the half-reactions for the anode and the cathode Indicate the direction of electron flow

Your Task Questions # 1-3, 5abc(that’s all of it) page 621
You also MUST read pages for next class and TAKE YOUR OWN NOTES

Voltaic Cells (aka Galvanic Cell)
Voltaic Cells (aka Galvanic Cell) A device that spontaneously produces electricity by redox Uses chemical substances that will participate in a spontaneous redox reaction. The reduction half-reaction (SOA) will be above the oxidation half-reaction (SRA) in the activity series to ensure a spontaneous reaction. Composed of two half-cells; which each consist of a metal rod or strip immersed in a solution of its own ions or an inert electrolyte. Electrodes: solid conductors connecting the cell to an external circuit Anode: electrode where oxidation occurs (-) Cathode: electrode where reduction occurs (+) The electrons flow from the anode to the cathode (“a before c”) through an electrical circuit rather than passing directly from one substance to another A porous boundary separates the two electrolytes while still allowing ions to flow to maintain cell neutrality Often the porous boundary is a salt bridge, containing an inert aqueous electrolyte (such as Na2SO4(aq) or KNO3(aq)), Or you can use a porous cup containing one electrolyte which sits in a container of a second electrolyte. A voltaic cell is an electrochemical cell that spontaneously produces energy (voltage) It has 2 electrodes (an anode and cathode) which are solid electrical conductors The anode is where oxidation, the loss of electrons, occurs. It is the negative electrode of the cell The cathode electrode where reduction, the gain of electrons, occurs. It is the positive electrode of a voltaic cell Anode releases electrons (oxidation) through a wire to the cathode causing reduction Each electrode is immersed in an aqueous solution that can conduct electricity called an elctrolyte

Draw figure 2 page 622 textbook

Without the presence of a salt bridge the cell would not operate
The Salt Bridge Without the presence of a salt bridge the cell would not operate WHY? The reason is that without the salt bridge there would be a build-up of negatively charged ions (NO3-) as the Cu2+(aq) is converted to Cu(s) at the copper electrode. Similarly, there would be an excess of positive charge around the Zn(s) electrode as the Zn(s) is converted to Zn2+(aq) The salt bridge supplies the ions to neutralize the charge; it allows the cell to maintain charge neutrality. This is accomplished by the NO3- moving out of the salt bridge into the zinc half-cell to balance the charge. Similarly, the K+ (aq) can move into the Cu2+(aq) solution to balance out the excess NO3- charge which exists. The salt bridge allows ions to move freely, without allowing the solutions to mix. The result is a complete circuit.

Voltaic Cells (aka Galvanic Cells)
Voltaic cells can be represented using cell notation: The SOA present in the cell always undergoes reduction at the cathode The SRA present in the cell always undergoes oxidation at the anode The single line represents a phase boundary (electrode to electrolyte) and the double line represents a physical boundary (porous boundary) To predict the reactions that occur in a voltaic cell, the SOA and the SRA need to be identified. The half reactions of the SOA and SRA can then be stoichiometry balanced and combined to predict the net ionic equation RED CAT AN OX

Cell Notation Drawing Cells ALL the time can be tiring!!!
- So Chemists came up with a cell notation to save time!! Cu(s) | Cu2+(aq) || Zn2+(aq) | Zn(s) Indicates a change of state between solid and electrolyte Indicates a porous barrier – will let ions through but prevent mixing of solutions (such as a salt bridge)

Example A silver-copper voltaic cell consists of a copper half-cell with a Cu electrode and a 1.0 mol/L Cu(NO3)2 electrolyte as well as a silver half cell with an Ag electrode and a 1.0mol/L Ag(No3) electrolyte. The two half-cells are connected by a salt bridge containing K(NO3). Describe the reaction that occurs while the cell is operating.

Cell Notation Represent the cell below using cell notation and identify the components

Method for Sketching and Labeling cells
Step 1 List the Species Step 2 Determine S.O.A. And S.R.A Step 3 Write half and net reactions WITH VOLTAGES Step 4 Draw Cell Step 5 Label Cathode/Anode + / - Direction of Electron Flow Direction of Ion Flow Electrolytes

C(s) | Cr2O72-(aq), H+(aq) || Cd2+(aq) | Cd(s)
Consider the cell: C(s) | Cr2O72-(aq), H+(aq) || Cd2+(aq) | Cd(s) SPECIES – see above plus H2O S.O.A E0 = 1.23 V G.E.R - Cathode S.R.A E0 = 0.40 V L.E.O. - Anode Enet = 1.63 V Net Reaction Labeling the Cell Direction of Electron Flow From anode to cathode C(s) + Cathode ANODE – Cd(s) Cations Electrolyte Cd2+(aq) Electrolyte Cr2O72-, H+ Anions

What is happening in this cell?

Recall from your reading that:
The negative terminal is called the anode (oxidation occurs) and anions migrate here. The positive terminal is called the cathode (reduction occurs) and cations migrate here.

e- carry the charge through the wire and move from the anode to the cathode. electrolytes contain ions that carry the charge through solution. e- do not move through solution. (electrons move in the wire to replace electrons being removed by the electrolyte’s anions which are migrating to the anode)

The solid anode is dissolving and releasing electrons (oxidation). The electrons travel through the wire to the cathode where this electrode is gaining electrons and solid metal (reduction). As expected the cations move toward the cathode and the anions move toward the anode.

Half Reactions Labeling the Cell CATHODE + ANODE -
Gain of electrons reduction at the Cathode Loss of electrons oxidation at the Anode Net Reaction Labeling the Cell Direction of Electron Flow From anode to cathode CATHODE + ANODE - Cations Anions Electrolyte Electrolyte

Match the cell notation to the descriptions
Copper placed in a solution of copper(II) chloride and tin metal placed in a solution of tin(II) ions A copper-magnesium cell Magnesium in a solution of magnesium chloride and tin in a solution of tin(II) chloride A tin(IV) ion solution containing tin and a solution of magnesium ions containing magnesium Two tin electrodes in solution of tin(II) chloride and tin (IV) chloride respectively Copper place in a copper(II) solution and tin place in a tin(IV) solution Sn(s) Sn4+(aq) Cu2+(aq) Cu(s) Mg(s) MgCl2(aq) SnCl4(aq) Sn(s) Sn(s) SnCl2(aq) CuCl2(aq) Cu(s) Mg(s) Mg2+(aq) Cu2+(aq) Cu(s) Mg(s) Mg2+(aq) Sn2+(aq) Sn(s) Sn(s) SnCl2(aq) SnCl4(aq) Sn(s)

Anode oxidation; reduction cathode
Voltaic Cells – What is going on? Example: Silver-Copper Cell Cu(s) Cu2+(aq) Ag+(aq) Ag(s) Use the activity series to determine which of the entities is the SOA. The SOA present in the cell always undergoes a reduction at the cathode. Write the reduction half reaction Use the activity series to determine which of the entities is the SRA. The SRA present in the cell always undergoes an oxidation as the anode. Write the oxidation half-reaction Balance the half-reactions and add together to create the net equation. The cathode is the electrode where the strongest oxidizing agent present in the cell reacts. The anode is the electrode where the strongest reducing agent present in the cell reacts. Memory device: “An ox ate a red cat” Anode oxidation; reduction cathode

Voltaic Cells – What is going on?
Example: Silver-Copper Cell Silver ions are the strongest oxidizing agents in the cell, so they undergo a reduction half- reaction at the cathode, creating more Ag(s) Copper atoms are the strongest reducing agents in the cell, so they give up electrons in an oxidation half-reaction and enter the solution (Cu2+ = blue ions) at the anode. Electrons released by the oxidation of copper atoms at the anode travel through the connecting wire to the silver cathode. (Ag+(aq) win in the tug of war for e-’s over Cu2+(aq)) Since the positive silver ions are being removed from solution, you would assume that the solution would become negatively charged. This does not happen. Why? Cations (positively charged ions) move from the salt bridge into the solution in the cathode compartment to maintain an electrically neutral solution. Anions (negatively charged ions) move from the salt bridge into the solution in the anode compartment to maintain an electrically neutral solution.

Voltaic Cell Animation

Voltaic Cell Summary A voltaic cell consists of two-half cells separated by a porous boundary with solid electrodes connected by an external circuit SOA undergoes reduction at the cathode (+ electrode) – cathode increases in mass SRA undergoes oxidation at the anode (- electrode) – anode decreases in mass Electrons always travel in the external circuit from anode to cathode Internally, cations move toward the cathode, anions move toward the anode, keeping the solution neutral

Voltaic Cells with Inert Electrodes
Inert electrodes are needed when the SOA or SRA involved in the reaction is not solid. If this is the case, usually a graphite (C(s)) rod or platinum strip is used as the electrode. Inert (unreactive) electrodes provide a location to connect a wire and a surface on which a half-reaction can occur. Example: a) Write equations for the half-reactions and the overall reaction that occur in the following cell: C(s) Cr2O72-(aq) H+(aq) Cu2+(aq) Cu(s) cathode: Cr2O72-(aq) + 14 H+(aq)+ 6 e-  2Cr3+(aq) + 7H2O(l) anode: 3 [Cu(s) Cu2+(aq) + 2e- ] 3Cu (s) + Cr2O72-(aq) + 14 H+(aq)  3Cu2+(aq + 2Cr3+(aq) + 7H2O(l) b) Draw a diagram of the cell labeling electrodes, electrolytes, the direction of electron flow and the direction of ion movement. For example an acidic dichromate solution is a SOA that reacts spontaneously with copper metal. The copper electrode will decrease in mass and the blue colour of the electrolyte increases (Cu2+), which indicates oxidation at the anode. The carbon electrode remains unchanged, but the orange colour of the dichromate solution becomes less intense and changes to greenish-yellow (Cr3+), evidence that reduction is occurring in this half cell

Example a. Write the equations for the half-reactions and the overall reaction that occurs in the following cell: C(s) Fe2+(aq), Fe3+(aq) Cr2O7-2(aq), H+(aq) C(s) b. Draw a diagram of the cell, labelling electrodes, electrolytes, the direction of electron flow, and the direction of ion movement.

textbook

Practice Draw the voltaic cell constructed using a rod of chromium solid and a chromium (II) nitrate solution and an inert carbon rod and an iron (III) nitrate solution. Determine the SOA and SRA. Write the half-reactions for the anode and the cathode. Indicate the direction of electron flow, which half cell is positive and negative and movement of the salt bridge ions

Standard Cells and Cell Potentials
In chemical cells the voltage recorded depends upon the two half-reactions that make up the cell, the temperature and concentration of the reactants and the pressure in the case of gaseous reactants. Since each half-reaction has its own potential to take electrons scientists needed to make standards. When dealing with voltage, signs cannot be ignored (+5 is different than 5)

A standard cell is a voltaic cell where each ½ cell contains all entities necessary at SATP conditions and all aqueous solutions have a concentration of 1.0 mol/L Standardizing makes comparisons and scientific study easier Standard Cell Potential, E0 cell = the electric potential difference of the cell (voltage) E0 cell = E0r cathode – E0r anode Where E0r is the standard reduction potential, and is a measure of a standard ½ cell’s ability to attract electrons. The higher the E0r , the stronger the OA All standard reduction potentials are based on the standard hydrogen ½ cell being 0.00V. This means that all standard reduction potentials that are positive are stronger OA’s than hydrogen ions and all standard reduction potentials that are negative are weaker. If the E0 cell is positive, the reaction occurring is spontaneous. If the E0 cell is negative, the reaction occurring is non-spontaneous

Rules for Analyzing Standard Cells
Determine which electrode is the cathode. The cathodes is the electrode where the strongest oxidizing agent present in the cell reacts. I.e. The OA that is closet to the top on the left side of the redox table = SOA If required, copy the reduction half-reaction for the strongest oxidizing agent and its reduction potential Determine which electrode is the anode. The anode is the electrode where the strongest reducing agent present in the cell reacts. I.e. The RA that is closet to the bottom on the right side of the redox table = SRA If required, copy the oxidation half-reaction (reverse the half-reaction) Determine the overall cell reaction. Balance the electrons for the two half reactions (but DO NOT change the E0r) and add the half-reaction equations. Determine the standard cell potential, E0cell using the equation: E0 cell = E0r cathode – E0r anode Note that even when the half reactions are multiplied to balance the electrons, THE REDUCATION POTENTIALS WERE NOT ALTERED BY THE FACTORS USED TO BALANCE THE ELECTRONS!!!!

How did ‘THEY’ know that:
E0 = 0.34 V E0 = 0.76 V The answer is that no one knows…. So they decided to set one of the half reaction voltages….. They decided the standard hydrogen reaction would be Volts 2H+ (g) + 2e  H 2(g) Eo = 0.00 Volts This is THE Standard Hydrogen Electrode

Measuring Standard Reduction Potentials
For example: Determine the cell potential of Zn (s) | Zn 2+ (aq) | | Cu 2+ (aq) | Cu (s) The redox table indicates that this is a spontaneous reaction (SOA higher than SRA) and that SOA E o = V and SRA E o = V thus the cell potential ΔE o = V.

An alternative method is to write the E o values with the half reactions;
anode OA SOA cathode Zn (s) | Zn 2+ (aq) | | Cu 2+ (aq) | Cu (s) SRA RA

Cu2+(aq) + 2 e- –> Cu(s) E or = + 0.34
Zn(s) –> Zn2+(aq) + 2 e- E oox = Cu2+(aq) + Zn(s) –> Cu(s) + Zn2+(aq) ΔEo = V Note that just as the half reaction is reversed so is the sign of E o.

Example: What is the standard potential of the cell represented below: Determine the cathode and anode Determine the overall cell reaction Determine the standard cell potential

Example: What is the standard potential of an electrochemical cell made of a cadmium electrode in a 1.0 mol/L cadmium nitrate solution and chromium electrode in a 1.0 mol/L chromium(III) nitrate solution? Cd2+(aq) Cd(s) Cr2+(aq) Cr(s) H2O(l) E0 cell = E0r cathode – E0r anode = (-0.40V) - (-0.91V) = V The E0 cell is positive, therefore the reaction is spontaneous. SOA SRA cathode anode

Example: A standard lead-dichromate cell is constructed. Write the cell notation, label the electrodes, and calculate the standard cell potential. Pb(s) Pb2+(aq) Cr2O72-(aq) H+(aq) Cr3+(aq) C(s) E0 cell = E0r cathode – E0r anode = (+1.23V) - (-0.13V) = V The E0 cell is positive, therefore the reaction is spontaneous. SRA SOA anode cathode Cell Potential Animation

Example: A standard scandium-copper cell is constructed and the cell potential is measured. The voltmeter indicates that copper the copper electrode is positive. Sc(s) Sc3+(aq) Cu2+(aq) Cu(s) E0 cell = +2.36V Write and label the half-reaction and net equations, and calculate the standard reduction potential of the scandium ion. E0 cell = E0r cathode - E0r anode 2.36V = (+0.34V) - (x) E0r anode = V anode cathode

Practice What if solid nickel was the new zero what would hydrogen be?
Fe2+(aq) is often oxidized. The standard electrical potential for Fe2+(aq) is what? If the standard electrical potential of Bromine the element was changed to zero, what would be the standard electrical potential of Ag(s)? What is the net electrical potential of a nickel/magnesium battery? What is the electrical cell potential of a copper/zinc voltaic cell?

Practice Calculate the standard cell potential for the voltaic cell with the reaction: 2I-(aq) + Cl2(g)  I2(aq) + 2Cl-(aq) If the reference half-cell was chosen as Co2+(aq) + 2e-  Co(s) V, what would the new half-cell potential for the half reaction: Fe3+(aq) + e-  Fe2+(aq) be? If the reference half-cell was chosen as Fe3+(aq) + e-  Fe2+(aq) V, what would be the new half-cell potential for the half reaction Mg2+(aq) + 2e-  Mg(s) be?

You have seen that the competition for electrons is much like a tug-of-war. The stronger the oxidizing agent, the higher the reduction potential, so the more easily it can steal electrons from a reducing agent. The difference between the potential energies of the electrons will determine the net cell potential (potential difference). It is from these relative potential differences that the table of redox half-reactions equations was made.

Your Task Questions page 633

Electrochemistry Redox reactions.

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