Electrochemistry Modified from Scheffler

Electrochemistry Electrochemistry is the study of chemical reactions that generate electrical effects

Oxidation and Reduction Oxidation = loss of electrons. Reduction = gain of electrons. Oxidizing agent = accepts electrons and becomes reduced. Reducing agent = donates electrons and becomes oxidized.

GER You may have heard… LEO says GER Lose Electrons Oxidation Gain Electrons Reduction GER

Or… OILRIG Oxidation Is Loss Reduction Is Gain

Hint… The charge on an ion is written with the number first and then the charge, e.g. Ca2+ The oxidation number is written charge first, then number, e.g. +2

Why are oxidation numbers important? They allow us to track the transfer of electrons through a chemical reaction. For example in the formation of water 2H2 + O2 → 2H2O 0 0 +1 -2 Oxidation number of H goes from 0 to +1 Oxidation number of O goes from 0 to -2

THEREFORE… Oxidation occurs when there is an increase in oxidation number Reduction occurs when there is a REDUCTION in oxidation number Practice Ca + Sn 2+ → Ca 2+ + Sn Which is oxidized?

Rules for assigning oxidation numbers 1. The oxidation number of an atom of any free element is ZERO. 2.The oxidation number of hydrogen in H-containing compound is +1. Exception: in metallic hydrides the oxidation number of hydrogen is -1. a metal hydride has hydrogen bonded to a metal. 3. The oxidation number of oxygen is -2. Exception: In peroxides which contain one extra oxygen, each oxygen is assigned an oxidation number of -1.

Rules Continued 4.The oxidation of a monoatomic ion is equal to its charge. For members of metallic families, the periodic table can be consulted. Also, halides ( a metal + a halogen) will be -1. 5. The sum of the oxidation numbers of all the atoms in a compound is ZERO.

Oxidation-Reduction: A Reaction Consider: Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g). The neutral Ca(s) has lost two e- to 2 H+ to become Ca2+. We say Ca has been oxidized to Ca2+ At the same time 2 electrons are gained by 2 H+ to form H2 . We say H+ is reduced to H2 .

Redox Reaction with Air Consider the reaction of Ca with O2: 2Ca(s) + O2(g)  2CaO(s) Ca is easily oxidized in air. On the left there is shiny Ca metal. On the right we see a white powder – Calcium oxide. Again, Ca(s) loses electrons and is oxidized to Ca+2 And the neutral O2 has gained electrons from the Ca to become O2- in CaO. We say O2 has been reduced to O2-.

Balancing Redox Equations The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution? Write the unbalanced equation for the reaction in ionic form and assign oxidation numbers. Fe2+ + Cr2O72- Fe3+ + Cr3+ Separate the equation into two half-reactions. Fe2+ Fe3+ +2 +3 Oxidation: Cr2O72- Cr3+ +6 +3 Reduction: Balance the atoms other than O and H in each half-reaction. Cr2O72- 2Cr3+

Balancing Redox Equations For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe2+ Fe3+ + 1e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O

Balancing Redox Equations Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. You should also cancel like species. Oxidation: 6Fe2+ 6Fe3+ + 6e- Reduction: 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. You should combine H+ and OH- to make H2O.

1. HNO3(aq) + Cu (s)  Cu(NO3)2 (aq) + NO (g) + H2O (l)      2. H+(aq) + MnO4-(aq) + H2C2O4 (aq)  Mn2+ (aq) + CO2 (g) + H2O (l) 3 H+ + NO3- + I2  HIO3 + NO2 + H2O  4. KMnO4 + HCl  KCl + MnCl2 + H2O + Cl2  5. H3AsO3 + NaBrO3  H3AsO4 + NaBr     6. CrO4- + H+ + Cl-  Cr3+ + Cl2 + H2O

Reactivity Series of Metals Metals act as reducing agents by pushing their electrons on other substances. More reactive metals will lose their electrons more readily which makes them stronger reducing agents than less reactive metals You can test this in solution

Zinc reacts with the CuSO4 solution forming copper metal and Zn2+ Zinc reacts with the CuSO4 solution forming copper metal and Zn2+ . So, Zinc has the reducing strength to “force” copper ions to accept the electrons and be reduced. This is much like the reaction completed in class between Aluminum and Copper (II)Chloride.

By comparing reactions like those between Zinc and Copper II Sulfate, you can build a list of relative strength in terms of reducing agent This is called a reactivity series. You will be required to interpret this and deduce it from given data

Non-Metals We can compare the relative strengths of non-metals similar to the way we did with metals More reactive non-metals are stronger oxidizing agents than less reactive non-metals Non-metals acts as oxidizing agents because they gain electrons (remove them from other substances)

Half-Cells Two half-cells make a voltaic cell Reactions can be separated into half-cells For example a strip of copper metal in a copper solution Two half-cells make a voltaic cell The half-cells are called electrodes

Distinguishing the Anode and Cathode Oxidation occurs at the anode Reduction occurs at the cathode Oxidation (loss of e-) Reduction (gain of e-) 25

Distinguishing the Anode and Cathode Oxidation occurs at the anode Reduction occurs at the cathode Oxidation (loss of e-) Reduction (gain of e-) 26

Electrochemical Cells The voltaic cell on the left has a Potential difference of about 1.1 volts 27

Example 1: The Voltaic Cell The copper electrode is placed in a solution of Cu2+ such as copper (II) sulfate or copper (II) nitrate. The zinc electrode is placed into a solution of Zn2+ ion such as zinc sulfate or zinc nitrate. The two sides are connected with a U tube containing an electrolyte such as KCl or KNO3. This structure is called a salt bridge. 28

Voltaic Cell Cells and Cell Reactions in a Daniel Cell Overall reaction Zn(s) + Cu+2(aq)  Zn+2(aq) + Cu(s) Oxidation half reaction at the anode Zn(s)  Zn+2(aq) + 2 e- Reduction half reaction at the cathode Cu+2(aq) + 2 e-  Cu(s) 29

Shorthand Notation for Electrochemical cells The shorthand representation of an electrochemical cell showing the two half-cells connected by a salt bridge or porous barrier, such as: Zn(s)/ZnSO4(aq)//CuSO4(aq)/Cu(s) anode cathode The electrodes are shown on the ends and the electrolytes for each side are shown in the middle. 31

Information about Batteries NOT REUIRED, but left it in case you are interested.

Batteries Are Applications of Electrochemical Cells device that converts chemical energy into electricity Primary Cells non-reversible electrochemical cell non-rechargeable cell Secondary Cells reversible electrochemical cell rechargeable cell 33

A Common Dry Cell 34

A 9 Volt Dry Cell 35

“Flash Light” Batteries Dry Cell Zn (s) + 2 MnO2 (s) + 2 NH4+ (aq)  Zn+2 (aq) + 2 MnO(OH) (s) + 2 NH3 Alkaline Cell Zn (s) ) + 2 MnO2 (s)  ZnO (s) + Mn2O3 (s) 36

Lead-Acid (Car Battery) 37

Lead-Acid (Car Battery) Overall reaction Pb (s) + PbO2 (s) + 2 H2SO4 (aq) = 2 PbSO4 (s) + 2 H2O (l) E = 2.0- volts per cell Cathode PbO2 (s) + SO42- (aq) + 4H+ (aq) +2e-  PbSO4 (s) + 2 H2O (l) Anode Pb (s) + SO42 (aq)  PbSO4 (s) +2e- 38

Nickel-Cadmium (Ni-Cad) Overall reaction Cd(s) + 2 Ni(OH)3(s) = Cd(OH)2(s) + 2 Ni(OH)2(s) E NiCad = 1.25 v/cell Cathode NiO2 (s) + 2 H2O (l) +2e-  Ni(OH)2 (s) + 2OH- (aq) Anode Cd (s) + 2OH- (aq)  Cd(OH)2 (s) +2e- 39

Electrolysis 40

Electrolysis An electrolysis is the inverse of an electrochemical cell. A non-spontaneous reaction is caused by the passage of an electric current through a solution. By passing a DC current through an electrolyte, the reaction can be made to proceed in the reverse or non-spontaneous direction. 41

Electrolysis The reactions at the anode and cathode depend on the relative reduction potentials of the solute and the solvent. The substance produced at the cathode depends on the cation that has the higher (more positive) reduction potential. The substance produced at the anode depends on the cation that has the lower (more negative) reduction potential. 42

Diagram of a Simple Electrolysis 43

Electrolysis of Molten NaCl If sodium chloride is heated to its melting point, then the resulting liquid contains mobile ions. This is a way of producing sodium metal. 44

Electrolysis of Brine The electrolysis of brine solution results in the reduction of water to hydrogen gas rather than sodium ion to sodium metal 45

Electrolysis Applications Preparation of Aluminum (Hall-Heroult process) The industrial production of aluminum is accomplished by the electrolysis of relatively pure alumina This process was first invented in France in 1886 by Paul Heroult and at almost the same time in the United States by Charles Hall. Adding cryolite, Na3AlF6, to alumina results the mixture can be made to melt at 980° C. rather than the more than 2000oC of alumina alone. It is then electrolyzed using graphite electrodes. 46

Hall-Heroult Process for Aluminum The alumina / cryolite mixture is electrolyzed using graphite electrodes. Aluminum forms at the cathode and oxygen at the anode. The oxygen reacts slowly with the carbon anode to produce carbon dioxide gas. 47

Hall-Heroult Process for Aluminum Chemical reactions in the processing of aluminium Alumina reacts with cryolite: Al2O3 + 4 AlF63-  3 Al2OF62- + 6 F- Cathode: AlF63- + 3 e-  Al + 6 F- Anode: 2 Al2OF62- + 12 F- + C  4 AlF63- + CO2 + 4 e-   The overall cell reaction: Al2O3 + 3 C  4 Al + 3 CO2 48

Electrolysis of Molten NaCl The electrolysis of molten NaCl at high temperatures generates liquid sodium metal and chlorine gas. 49

Industrial Electrolysis of Brine 50

Corrosion Corrosion of metals is a common oxidation-reduction process in nature. The rusting of iron can be thought of as a form of an electrochemical cell. 51

Rusting of Iron 52

Rusting Iron O2(g) + 4 H+(aq) + 4 e-  2 H2O(l) Eo = 1.23V Rusting Process Fe(s)  Fe+2(aq) + 2 e- Eo = 0.44 V O2(g) + 4 H+(aq) + 4 e-  2 H2O(l) Eo = 1.23 V --------------------------------------------------------- -------------- 2 Fe(s) + O2(g) + 4 H+(aq) 2 H2O(l) + Fe+2(aq) Eo = 1.67 V 53

Preventing Corrosion painting galvanizing sacrificial anode 54