Electrochemistry

Oxidation-Reduction In the last chapter, we looked at what happened in a reaction where a proton was transferred from one chemical species to another. A proton is not, however, the only part of an atom that can be moved. An electron can also be moved from one chemical species to another.

Demonstration : Aluminum foil in CuCl2 solution.

Where did those electrons go ?

- - In order for the aluminum to lose electrons, there must have been something to gain the electrons. (If this were not the case, the aluminum foil would disintegrate as soon as it was made.) Conversely, in order for the copper ions to turn into copper solid, that is to gain electrons, something must lose the electrons. Clearly, these reactions are dependant on each other. One doesn't happen without the other. Definitions :

The loss of electrons (as the aluminum did above) is known as _____________________. It is called oxidation because this is what happens when a metal oxidizes.

The gaining of electrons (as the copper ions did above) is known as _____________________________. It is called reduction because when ores are mined and refined, a large amount of ore is reduced to a small amount of metal.

The overall reaction that occurred is known as an ___________________________ reaction, sometimes shortened to a REDOX equation.

OXIDIZING AGENT REDUCING AGENT Looking at the above reaction once again, this time in terms of the HALF-REACTIONS, we get :

oxidation

and

reduction

One other thing about the reaction, if we closely examined the solution, we would find that there are NO free electrons floating around. Nature detests a solution that has an electrical charge (a POLARIZED solution) therefore we must look at the reaction once again. If these are the half reactions, one would think that if we simply added them together,

We would get the overall reaction. When we do this:

We get an excess of electrons produced. This will not happen. We will not have polarization occurring. In order for us to account for the extra electron, we must add the two equations together mathematically to eliminate the electrons .

Balancing REDOX reactions (version 1):a) Write the two half -cell reactions:

b) The least common multiple of the electrons in the two half reactions is 6, therefore we multiply each equation in order to make 6 electrons in each :

c) Now we can add together the two equations and eliminate the electrons :

The equation is now balanced.

NOTE : -The total number of atoms of each element is balanced.

- All the charges are balanced.

Easy way to remember what reactions are taking place:

Translation:

Another easy way to remember what is going on:

Translation: Chemistry 12.

Oxidation-Reduction Instant Practice #1

1. Determine which of the following processes are oxidations and which are reductions:

a) Co 2+ becomes Co

b) 2 I- becomes I2c) Fe 3+ becomes Fe 2+d) Sn 2+ becomes Sn 4+2. What is an oxidizing agent? What is a reducing agent?

3. How do you know when a redox reaction is balanced?

4. In the following reactions, indicate the

a) the species oxidized (b) species reduced (c) oxidizing agent (d) reducing agent

a) Hg 2+ Mn ---( Hg + Mn 2+b) H2 + Sn 4+ ---( 2 H+ + Sn 2+c) 2 Li + F2 --( 2 Li + + 2 F-5. When cesium metal is exposed to chlorine gas, a bright flash occurs as the elements react. The product, cesium chloride, is a white solid composed of cesium ions and chloride ions.

a) Write the balanced overall reaction that occurs between chlorine and cesium.

b) Write the half-reactions which occur and identify which half-reaction is the reduction and which is the oxidation.

c) Identify the reducing and oxidizing agents.

6. Balance the following redox reactions. Identify what has been reduced and what is the reducing agent.

a) Na + Cl2 ----( Na+ + Cl b) Cu 2+ + Mg ----( Cu + Mg 2+c) Fe 3+ + Al ----( Fe 2+ + Al 3+

d) Au 3+ + Cd ------( Au + Cd 2+Oxidation Numbers

Another easy way to tell if oxidation or reduction has occurred is to calculate the oxidation number of the element in the compound. The oxidation number is something like the valence of the element. It is assigned to atoms in order to keep track of the redistribution of electrons during a chemical reaction. By keeping account of the oxidation numbers of the reactants and products, it is possible to determine how many electrons are gained or lost in each atom. The oxidation number of an atom in a compound is assigned according to the following rules:

1. The oxidation of any substance in its elemental form is ZERO.

examples:

2. The common oxidation number of hydrogen is +1 except in compounds called hydrides ( here it is -1 )

examples:

3. The common oxidation number of oxygen is -2 except in compounds known as peroxides ( here it is -1)

examples:

4. The oxidation of Alkali metals ( Group I A ) in a compound is always +1.

examples:

5. The oxidation of Alkali Earth ( Group IIA ) in a compound is always +2.

examples:

6. The oxidation number for halogens is usually -1, however, there are exceptions. It can range from -7 to + 7.

examples:

7. The sum of the oxidation numbers of all the atoms in a neutral compound is zero. The sum of the oxidation numbers of the atoms present in a polyatomic ion is equal to the charge of the ion.

examples:

** To determine the oxidation nymber, we use the above rules and the ** electrical charge on the substance

example 1: Determine the oxidation number for Cr in Cr2O7 2-example 2: Determine the oxidation number for C in C3H2O2.

example 3: Determine the oxidation number for Mn in KMnO4.

To determine whether reduction or oxidation has occurred, determine the change in oxidation numbers. If the change is negative, reduction has occurred. If the change is positive, oxidation has occurred.

example: MnO4 - -------> MnO2example: As ------> HAsO2CHEMISTRY 12

OXIDATION NUMBERS

Determine the oxidation numbers of the element underlined in the following formula :

1. SO316. UO331.SO22-46.CH3OH

2. PF317.U3O832.SrSiO347.C4H4O42-3.PCl518.U2O533.S2OCl448.C5H10O

4. Na3P19.K2UO434.S2O3Cl449.C2H3O2-5.S2F1020.MgU2O735.H2SO550.C7H5O2-6.S2O721.MnSO436.SO2ClF

7.NO3-22.MnO237.NH2OH

8.NO223.KMnO438.C2O42-9. N2O324.MnO3+39.WO42-10.NO25.MnCl4-40.NO2-11.N2O26.H2SO341.ClO4-12.N227.H2S2O342.HIO64-13.N3-28.H2S2O743.P2O74-14.N2H5+29.KHSO444.CH415.NH4+30.S445.CH3Cl

ANSWERS :

1. +611. +1 21. +2 31. +2 41. +7

2. +312. 0 22. +4 32. +4 42. +7

3. +513. -1/3 23. +7 33. +3 43. +5

4. -314. -2 24. +7 34. +5 44. -4

5. +515. -3 25. +3 35. +8 45. -2

6. +716. +6 26. +4 36. +6 46. -2

7. +517. +16/3 27. +2 37. -147.+1/2

8. +418. +5 28. +6 38. +3 48. 8/5

9. +319. +6 29. +6 39. +6 49. 0

10. +220. +6 30. 0 40. +3 50. 2/7

Determining if a REDOX reactions will proceed:

Mini Lab: Reactions between metals and metal ions.

Procedures:

Results:

Reactans:

Metal electrode + Metal ion in solutionObservationsReaction? or no Reaction?ION that is stronger oxidizing Agent (stronger attraction to electrons)

Cu + Zn2+

Cu + Mg2+

Cu + Pb2+

Zn + Cu2+

Zn + Mg2+

Zn + Pb2+

Mg + Cu2+

Mg + Zn2+

Mg + Pb2+

Pb + Cu2+

Pb + Zn2+

Pb + Mg2+

Analysis of Results:

1. Arrange the metal ions in terms of increasing the ability to oxidize (from weaker to stronger oxidizing agent).2. Write a balanced REDUCTION Half-Reaction for each metal ion on the table below making sure to arrange the equations in order of decreasing strength of oxidizing agents.

Table of Standard Reduction of Half Cell.

To determine if two species in the table will proceed spontaneously, note the relative placement of each specie.

*** Any oxidizing agent will react with any reducing agent on the right which is LOWER on the list***3. Using the table above, determine if the following reactions will occur spontaneously or if the reaction is non-spontaneous:a) Cu(s) + Mg(NO3)2(aq) -------->

b) Zn(s) + Pb(NO3)2(aq) ---------- >

c) Pb(s) + Cu(NO3)2(s) ---------- >

Standard Reduction Potential Chart

The table you have in front of you is a much more complete list than the one you made in the lab. It will be provided for you on the final exam. It is important that you know how to use this table.

As you can see, there are several reactions that simply involve the transfer of electrons from an ion to an element (as we did in the lab). Also, there are a number of other reactions that have either H+ ions and water in them or OH- ions and water in them. These are the half reactions that will take place in either acidic or basic solutions.

It should be stressed that These are only half of the reaction that will take place. They are all reduction reactions. We have seen that reduction can only take place if oxidation is also occurring. To get an oxidation half reaction all we have to do is to write the reaction in the reverse order. Notice that the arrows are bidirectional - this does not mean that they are in equilibrium but that they can proceed in either direction.

As we said before, in order for a reaction to occur spontaneously, both oxidation and reduction must occur and the reduction reaction must be above the oxidation reaction on the chart.

Assignment: Standard reduction potential ws.

Chemistry 12

Standard Reduction Potential WS

1. Calculate the Eo cell for each reaction and state whether the reaction is expected to be spontaneous.

a) Cr + 3 Ag+ ---( Cr 3+ + 3 Ag

b) Cu + Fe 3+ --( Cu 2+ + Fe 2+

c) Mn 2+ + 2 H2O + I2 --( MnO2 + 4 H+ + 2 I

d) 3 Cu + 2 NO3- + 8 H+ ---( 3 Cu 2+ + 4 H2O + 2 NO

e) 2 Cr 3 + + 7 H2O + 3 Pb 2+ ---( Cr2O7 2- + 14 H+ + 3 Pb

2. When silver metal is placed in 1 M HCl solution under standard conditions, there is no observable reaction. However, when magnesium metal is placed in the same HCl solution, the metal oxidizes and hydrogen gas is produced. Explain these two phenomena.

3. What will happen if an aluminum spoon is used to stir a solution of Fe(NO3)3?

4. Can a 1M solution of Iron III sulfate be stored in a container made of nickel metal? Explain your answer.

Balancing Half-Reactions -

The chart we have is good to use if we have a reaction that is on it (or if we actually have it in our possession at the time we need it). If it isn't on the chart, or if we don't have a chart handy, there is a straight forward, simple way to balance all half reactions :

Rules for balancing half reactions :

1. Balance all elements except hydrogen and oxygen. Do this the same way we would normally balance equations, by changing coefficients.

2. Balance the oxygen atoms by adding water molecules.

3. Balance the hydrogen atoms by adding H+ ions.

4. Balance the electrical charge by adding electrons.

**5. To make the solution basic (if asked for) add the exact same number of OH- ions as H+ ions to both sides of the equation. This will eliminate H+ ions by producing water. Cancel the excess water.

__________________________Examples :

Mn3+ -----> MnO4- acidic conditions

To check, count all the individual atoms on both sides of the equation. As well, the TOTAL CHARGE on both sides of the equation MUST be EQUAL.

example:

I2 -----> IO3- acidic

Example:

MnO4- ----> MnO2 basic

How do we tell whether or not oxidation or reduction has occurred?

Example: FeHPO3 ------- > PO43- + Fe(OH)3 (acidic)

- Assignment :

CHEMISTRY 12 BALANCING HALF-CELL REACTIONS

Balance the following half-cells.

1. Ce4+ ---> Ce2+

2. I2 ---> I-

3. Mn2+ ---> MnO2 (acidic solution)

4. O2 ---> H2O2 (acidic solution)

5. S2O82- ---> HSO4- (acidic solution)

6. MnO4- ---> MnO2 (acidic solution)

7. NO3- ---> NO (acidic solution)

8. ClO3- ---> Cl- (acidic solution)

9. H3AsO4 ---> HAsO2 (acidic solution)

10. H2SeO3 ---> Se (acidic solution)

11. HO2- ---> O2 (basic solution)

12. N2H4 ---> N2 (basic solution)

13. Cr2O72- ---> Cr3+ (acidic solution)

14. HXeO4- ---> HXeO63- (basic solution)

15. Cr(OH)3 ---> CrO42- (basic solution)

16. FeS ---> Fe3+ + SO42- (acidic solution)

17. Cu2S ---> Cu2+ + H2SO3 (acidic solution)

18. FeHPO3 ---> PO43- + Fe(OH)3 (basic solution)

ANSWERS :

1. Ce4+ + 2e- ---> Ce2+

2. I2 + 2e- ---> 2I-

3. Mn2+ + 2H2O ---> MnO2 + 4H+ + 2e-

4. O2 + 2H+ + 2e- ---> H2O2

5. S2O82- + 2H+ + 2e- ---> 2HSO4-

6. MnO4- + 4H+ + 3e- ---> MnO2 + 2H2O

7. NO3- + 4H+ + 3e- ---> NO + 2H2O

8. ClO3- + 6H+ + 6e- ---> Cl- + 3H2O

9. H3AsO4 + 2H+ + 2e- ---> HAsO2 + 2H2O

10. H2SeO3 + 4H+ + 4e- ---> Se + 3H2O

11. HO2- + OH- ---> O2 + H2O + 2e-

12. N2H4 + 4OH- ---> N2 + 4H2O + 4e-

13. Cr2O72- + 14H+ + 6e- ---> 2Cr3+ + 7H2O

14. HXeO4- +4OH- ---> HXeO63- + 2H2O + 2e-

15. Cr(OH)3 + 5OH- ---> CrO42- + 4H2O + 3e-

16. FeS + 4H2O ---> Fe3+ + SO42- 8H+ + 9e-

17. Cu2S + 3H2O ---> 2Cu2+ + H2SO3 + 4H+ + 8e-

18. FeHPO3 + 6OH- ---> PO43- + Fe(OH)3 + 2H2O + 3e-

Balancing Full Reactions

In order to balance overall equations, it is important to do the following:

1. Isolate the two half reactions

2. Balance each half reaction separately

3. Find Least Common Multiple for the electrons

4. Add equations together to get rid of free electrons

Example: Mn3+ + Au 3+ -------> MnO4- + Au

Note: in the overall balanced equation the numbers of atoms of each type are the same on each side of the equation. Also, the electrical charges are equal. Both of these rules are the same as with half reactions but, most importantly, in an overall reaction, there are NO ELECTRONS FREE TO ROAM AROUND!!!

Example: Cr2O7 2- + HXeO4- ----> Cr3+ + HXeO63-example: MnO4-(aq) + Fe2+(aq) ---------- > Fe3+(aq) + Mn2+(aq)Example: Ag(s) + CN-(aq) + O2(aq) -------------- > Ag(CN)2-(aq)

Further Notes on Balancing Redox Eqations

1.Sometimes, there are bizarre circumstances where the reactant undergoes BOTH oxidation and reduction. In order to balance these equations, we must make two equations -- one oxidation and one reduction. From here, we can go through the procedures we did in the previous section.

example: Br2 ----> HBr + HBrO32.Sometimes there are more than one species undergoing either oxidation or reduction ( an even more bizarre case ). In situations such as these, we should try and isolate one element in its own half reaction and place the others in a separate half reaction.

example:Mn3+ + FeSO4 ----> MnO4- + Fe + SO2 ( basic )

example: Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the AuCl4- and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.

example: Chlorine was first prepared in 1774 by C.W. Scheele by oxidizing sodium choloride with manganese (IV) oxide. The reaction is:

NaCl(aq) + H2SO4(aq) + MnO2(s) ---------> Na2SO4(aq) + MnCl2(aq) + H2O(l) + Cl2(aq)Balance this equation:

CHEMISTRY 12

BALANCING OXIDATION-REDUCTION EQUATIONS

1.Cl2 + SO2 ---> Cl- + SO42- (acidic)

2.Cu + NO3- ---> Cu2+ + NO (acidic)

3.S2- + ClO3- ---> Cl- + S (basic)

4.Zn + As2O3 ---> AsH3 + Zn2+ (acidic)

5.Fe2+ + Cr2O72- ---> Cr3+ + Fe3+ (acidic)

6.U4+ + MnO4- ---> Mn2+ + UO22+ (acidic conditions)

7.CN- + IO3- ---> I- + CNO-

8. Mn2+ + HBiO3 ---> Bi3+ + MnO4- (acidic)

9.HSO3- + IO3- ---> I2 + SO42- (acidic)

The following equations involve a species which undergoes BOTH oxidation and reduction at the same time. Use two separate half cells with the same reactant.

10.OCl- ---> Cl- + ClO3-

11.HNO2 ---> HNO3 + NO

12.Br2 ---> Br- + BrO3- (basic)

The following equations involve three changes in oxidation numbers. HINT : One of the species can be isolated. Use this in a half cell by itself. The other two can not be separated. Use a half cell involving both.

13.Sb2S3 + NO3- ---> NO2 + SO42- + Sb2O5 (acidic)

14.As2S3 + NO3- ---> NO + SO42- + H3AsO4 (acidic)

15.FeS + NO3- ---> NO + SO42- + Fe3+ (acidic)

16.FeHPO3 + Cr2O72- ---> Cr3+ + H3PO4 + Fe3+ (acidic)

17.SnS2O3 + MnO4- ---> Mn2+ + SO42- + Sn4+ (acidic)

18. FeHPO3 + OCl- ---> Cl- + PO43- + Fe(OH)3 (acidic)

Answers :

1.Cl2 + SO2 + 2H2O ---> 2Cl- + SO42- + 4H+

2.3Cu + 2NO3- + 8H+ ---> 3Cu2+ + 2NO + 4H2O

3.3S2- + ClO3- + 3H2O ---> 3S + Cl- + 6OH-

4.6Zn + As2O3 + 12H+ ---> 2AsH3 + 6 Zn2+ + 3H2O

5.6Fe2+ + Cr2O72- + 14H+ ---> 2Cr3+ + 6Fe3+ + 7H2O

6.2H2O + 5U4+ + 2MnO4- ---> 2 Mn2+ + 5UO22+ + 4H+

7.3CN- + IO3- ---> I- + 3CNO-

8.2Mn2+ + 5HBiO3 + 9H+ ---> 5Bi3+ + 2MnO4- + 7H2O

9.5HSO3- + 2IO3- ---> I2 + 5SO42- + 3H+ + H2O

10.3OCl- ---> 2 Cl- + ClO3-

11.3HNO2 ---> HNO3 + 2NO + H2O

12.3Br2 + 6OH- ---> 5Br- + BrO3- + 3H2O

13.Sb2S3 + 22H+ + 28NO3- ---> 3SO42- + Sb2O5 + 28NO2 + 11H2O

14.3As2S3 + 4H2O + 10H+ + 28NO3- ---> 9SO42- + 6H3AsO4 + 28NO

15.FeS + 3NO3- + 4H+ ---> SO42- + Fe3+ + 3NO + 2H2O

16.2FeHPO3 + Cr2O72- + 14H+ ---> 2H3PO4 + 2Fe3+ + 2Cr3+ + 5H2O

17.SnS2O3 + 2MnO4- + 6H+ ---> 2SO42- + Sn4+ + 2Mn2+ + 3H2O

18.2FeHPO3 + 5H2O + 3OCl- ---> 2PO43- + 2Fe(OH)3 + 6H+ + 3Cl-

19.6Hg4Fe(CN)6 + 66H+ + 47ClO3- ---> 36NO + 36CO2 + 6Fe3+ +

24Hg2+ + 47Cl- + 33H2O

Stoichiometry and Redox Reactions: Redox Titrations

Similar to acid base titrations.

1. Oxidizing Agents: Acidic Potassium permanganate:

strong oxidizing agent.

MnO4- - purple ----- > Mn2+ - colourless

Other oxidizing agents:

Use standard KMnO4 to determine the concentration of an unknown solution of Fe2+.

Equation:MnO4- + Fe2+ -----( Mn2+ + Fe3+Balanced equation:

Example 1:

Example: A 3.00 g sample of pure iron is dissolved in hydrochloric acid and the resulting solution treated to produce Fe2+. The final volume of the solution is 500.0 mL. What volume of 0.0500 M KMnO4 is required to titrate 25.0 mL sample of Fe 2+ solution? (10.8 mL)

2. Reducing Agents: NaI, or KI (for I-)

Thiosulphate (S2O3 2-)

Sulphite (SO3 2-)

Oxalic acid (H2C2O4)

Example:

Analysis of Bleach can be done by reacting the solution with an excess amount of I-. The reaction:

Titrate a known volume of this sample by reacting the solution with Na2S2O3 and using starch solution as indicator. The solution is a dark blue colour because of the reaction between the starch and the Iodine.

At the endpoint, the iodine reacts completely, forming I-. The endpoint is reached when the blue colour fades.

Example: A 25 mL sample of bleach is reacted with an excess amount of NaI. The solution is then titrated with 0.500M solution of sodium thiosulphate. If 35.00 mL of the thiosulphate solution was required to bring the solution to endpoint, what is the [OCl -] in bleach? (0.35 M)

Assignment:, Hebden p. 213 # 28, 31, 32, 33.

Mini test coming on the first part of electrochemistry coming up on Tuesday

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