V. Electrochemistry. Oxidation Numbers oxidation number: the apparent charge of an atom in a compound/ion if all bonds are considered to be ionic.

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1 V. Electrochemistry Electrochemistry is the branch of chemistry that deals with reactions that involve the transfer of electrons and allow for the interconversion between chemical and electrical energy. Oxidation Numbers oxidation number: the apparent charge of an atom in a compound/ion if all bonds are considered to be ionic. Rules for Assigning Oxidation Numbers (1) The oxidation number for an atom in an element is zero. ex. Na, O 2, He etc. all have an oxidation number equal to zero (2) The oxidation number of an ion is equal to its charge. ex. for NaCl, Na + has an oxidation number of +1 and Cl has an oxidation number of 1 (3) In a compound, hydrogen usually has an oxidation number of +1. ex. HBr, H + has an oxidation number of +1 and Br has an oxidation number of 1. The exception is for metal hydrides, where hydrogen has an oxidation number of 1. ex. LiH, Li + has an oxidation number of +1 and hydrogen has an oxidation number of 1. (4) In a compound, oxygen usually has an oxidation number of 2. ex. K 2O, K + has an oxidation number of +1 and O 2 has an oxidation number of 2. The exception is for peroxides, where oxygen has an oxidation number of 1. ex. hydrogen peroxide, H 2O 2, H + has an oxidation number of +1 and oxygen has an oxidation number of 1 (5) In neutral compounds, the oxidation numbers of atoms must add up to zero. In an ion, the oxidation numbers of all atoms must add up to the total charge. ex. Assign oxidation numbers to each element in the following compounds/ions. (1) SO 3 2 (2) Cr 2O 3 (3) KMnO 4 (4) HPO 4 2 (5) Fe 3O 4 S: O: Cr: O: K: Mn: O: H: P: O: Fe: O: * fractional oxidation numbers are possible! In this case, the oxidation number represents an average of all of that the atoms of the substance. Reduction Oxidation (Redox) Reactions A Redox reaction is one that involves a change of oxidation number of two elements: one element must increase in oxidation number (one element must be oxidixed) and one element must decrease in oxidation number (on element must be reduced). Oxidation: a chemical reaction in which the oxidation number of an atom increases as a result of losing electrons. Reduction: a chemical reaction in which the oxidation number of an atom decreases as a result of gaining electrons. Oxidizing agent: the substance that is reduced in a redox reaction. Reducing agent: the substance that is oxidized in a redox reaction. Just remember: LEO says GER (losing electrons oxidation/gaining electrons reduction)

2 ex. Assign oxidation numbers to each element. Determine if the reaction is a redox reaction. (1) 2Na + Cl 2 2NaCl Na: Cl: Na: Cl: (2) Na 2CO 3 + CaCl 2 2NaCl + CaCO 3 Na: Ca: Na: Ca: C: Cl: Cl: C: O: O: (3) CuCl 2+ 2Ag 2AgCl + Cu Cu: Ag: Ag: Cu: Cl: Cl: Typically, decomposition, synthesis, single replacement, and combustion reactions are redox reactions. Double replacement and neutralizations reactions are not redox reactions. ex. Determine if each half-reaction represents oxidation or reduction. (1) Co 2+ Co (2) 2Cl Cl 2 (3) Fe 3+ Fe 2+ (4) Sn 2+ Sn 4+ ex. Assign oxidation numbers to each atom and determine which element is being oxidized and which element is being reduced. Identify the oxidizing agent and the reducing agent. ex. SO I + 8H + S 2 + 4I 2 + 4H 2O S: I: H: S: I: H: O: O: ex. 2NO 3 + 8H + + 3Cu 2NO + 4H 2O + 3Cu 2+ N: H: Cu: N: H: Cu: O: O: O: Balancing Redox Reactions Half Reaction Method (1) Write separate equations for the oxidation and reduction half reactions. (2) For each half reaction: balance all the elements balance the charge by adding electrons (3) Add the two half reactions together. If necessary multiply one or both of the reactions by a number in order to equalize the number of electrons in the two half reactions. Cancel identical chemicals. ex. Balance each Redox equation using the half reaction method. (1) Ag + + Mg Ag + Mg 2+ (2) Cl 2 + Na Cl + Na + (3) Al + F 2 Al 3+ + F

3 Balancing Redox Equations in Acidic Solution (1) Write separate equations for the oxidation and reduction half reactions. (2) For each half reaction: balance all the elements except oxygen and hydrogen. balance the oxygen atoms by adding H 2O balance the hydrogen atoms by adding H + balance the charge by adding electrons (3) Add the two half reactions together. If necessary multiply one or both of the reactions by a number in order to equalize the number of electrons in the two half reactions. Cancel identical chemicals. ex. Balance each Redox equation occurring in Acidic Solution. (1) MnO 2 + Cl Mn 2+ + Cl 2 Half Reactions MnO 2 Mn 2+ Cl Cl 2 Balance Elements (preliminary balancing) Balance Oxygen (add H2O) Balance Hydrogen (add H + ) Balance Charge (add e ) MnO 2 Mn 2+ (Mn is already balanced) MnO 2 Mn H2O (add 2H 2O on the right to balance the 2 oxygen atoms on the left) MnO 2 + 4H + Mn H 2O (add 4H + on the left to balance the 4 hydrogen atoms on the right) MnO 2 + 4H + + 2e Mn H 2O (the total charge on the left is +4, the total charge on the right is 2+, add 2e on the left to balance the charge) 2Cl Cl 2 (add a coefficient of 2 for chlorine on the left) 2Cl Cl 2 (no oxygen to balance) 2Cl Cl 2 (no hydrogen to balance) 2Cl Cl 2 + 2e (the total charge on the left is 2, the total charge on the right is 0, add 2e on the right to balance the charge) Since the two reactions have the same number of electrons, add the reactions to get the overall reaction and cancel the electrons MnO 2 + 4H + + 2e Mn H 2O 2Cl Cl 2 + 2e MnO 2 + 2Cl + 4H + Mn 2+ + Cl 2 + 2H 2O (2) Cr 2O SO 2 Cr 3+ + SO 4 2 (3) Os + IO 3 OsO 4 + I 2

4 Balancing Redox Equations in Basic Solution (1) Write separate equations for the oxidation and reduction half reactions. (2) For each half reaction: balance all the elements except oxygen and hydrogen. balance the oxygen atoms by adding H 2O balance the hydrogen atoms by adding H + balance the charge by adding electrons (3) Add the two half reactions together. If necessary multiply one or both of the reactions by a number in order to equalize the number of electrons in the two half reactions. Cancel identical chemicals. (4) For a basic solution, add the same number of OH as there are H + ions to both sides of the reaction. (H + will be eliminated from one side of the reaction by forming H 2O and OH will remain on the other side of the reaction) ex. Balance each Redox equation occurring in Basic Solution. (1) HClO 2 + I Cl 2 + HIO Half Reactions HClO 2 Cl 2 I HIO Balance Elements (preliminary balancing) Balance Oxygen (add H2O) Balance Hydrogen (add H + ) Balance Charge (add e ) 2HClO 2 Cl 2 (add a coefficient of 2 for chlorine on the left) 2HClO 2 Cl 2 + 4H2O (add 4H 2O on the right to balance the 4 oxygen atoms on the left) 2HClO 2 + 6H + Cl 2 + 4H 2O (add 6H + on the left to balance the 8 hydrogen atoms on the right with the two on the left) 2HClO 2 + 6H + + 6e Cl 2 + 4H 2O (the total charge on the left is +6, the total charge on the right is zero, add 6e on the left to balance the charge) I HIO (I is already balanced) I + H2O HIO (add one H 2O on the right to balance the one oxygen atom on the left) I + H 2O HIO + H + (add one H + on the left to balance the 2 hydrogen atoms on the left with the one on the right) I + H 2O HIO + H + + 2e (the total charge on the left is 1, the total charge on the right is +1, add 2e on the right to balance the charge) Since the two reactions have different numbers of electrons, multiply the reaction on the right by three in order to balance with the six electrons on the left. Cancel identical chemicals and add the reactions (i.e. 3H + on the right and 6H + on the left reduces to 3H + on the left and 3H 2O on the left and 4 H 2O on the right reduced to one H 2O on the right). 2HClO 2 + 6H + + 6e Cl 2 + 4H 2O 3I + 3H 2O 3HIO + 3H + + 6e 2HClO 2 + 3I + 3H + Cl 2 + H 2O + 3HIO Since the reaction is occurring in base, add the same number of OH as there are H + ions to both sides of the reaction. In this case, since there are 3H +, add 3OH on both sides of the reaction. 2HClO 2 + 3I + 3H + Cl 2 + H 2O + 3HIO +3OH +3OH = 3H 2O This yield 3H 2O on the left which can be reduced with one H 2O on the right to give 2H 2O on the left in the overall reaction. 2HClO 2 + 3I + 2H 2O Cl 2 + 3HIO + 3OH

5 (2) MnO 4 + NO 2 MnO 2 + NO 3 (3) ClO 3 + N 2H 4 NO 3 + Cl Disproportionation Reactions In a disproportionation reaction, a single substance is both oxidized and reduced. Begin each half reaction with the given reactant. Simplify the overall reaction if possible. ex. Balance the following reaction in acidic solution: I 2 I + IO ex. Balance the following reaction in basic solution: Cl 2 ClO 3 + Cl

6 Reduction Oxidation Titrations Reduction Oxidation titrations involve redox reactions. Titration Calculations Determine the balanced chemical equation for the redox reaction. Use the concentration and volume given to determine the moles of titrant used in the experiment. Determine the moles of titrate/analyte using the mole ratio in the reaction. Determine the concentration/volume/mass of the titrate/analyte from the volume. ex. In acidic solution, MnO 4 reacts with Fe 2+ to produce Mn 2+ and Fe 3+. Write the balanced chemical equation for this reaction. What volume of M KMnO 4 solution is required to titrate ml of M Fe(NO 3) 2? moles (mol) mass (g) volume (L) Molarity (M) molar mass (g/mol) MnO 4 + 5Fe H + Mn Fe H 2O 8.736x Fe(NO 3) 2 Fe NO M M [Fe 2+ ] = M n= CV = ( M) ( L) = mol Fe 2+ 1mol MnO mol Fe x x10 mol MnO x10 mol KMnO 5mol Fe V = n C x10 mol L of KMnO 4 solution M ex. Tin is often mined as the ore "cassiterite" which contains tin and other impurities. A sample of ore weighing g is dissolved in acid to release Sn 2+ ions. The solution was then titrated with 8.08 ml of M KMnO 4 solution. In acidic solution, Sn 2+ reacts with MnO 4 to produce Sn 4+ and Mn 2+. Write the balanced chemical equation for this reaction. Determine the mass of pure tin in the sample of tin ore. Determine the percent tin in the sample of tin ore.

7 Predicting Redox Reactions Common Reduction Reactions: Common Oxidation Reactions: Reactants (oxidizing agent) permanganate: MnO4 (acidic solution) MnO4 (basic or neutral solution) MnO2 Products formed Mn 2+ MnO2 Mn 2+ Reactants (reducing agent) halide ions (ex. Cl ) metals (ex. Cu) metallous ions (lower oxidation number ex. ferrous ion Fe 2+ ) Products formed halogens (ex. Cl2) metal ions (ex. Cu 2+ ) metallic ions (higher oxidation number ex. ferric ionfe 3+ ) dichromate: Cr2O7 2 (acidic solution) Cr 3+ H2O O2 + H + halogens (ex. Cl2) metal ions (ex. Cu 2+ ) metallic ions (higher oxidation number ex. ferric ion Fe 3+ ) H2O halide ions (ex. Cl ) metals (ex. Cu) metallous ions (lower oxidation number ex. ferrous ion Fe 2+ ) H2 + OH Common multivalent metals Fe 3+ /Fe 2+ Cu 2+ /Cu + Pb 4+ /Pb 2+ Sn 4+ /Sn 2+ Co 3+ /Co 2+ Cr 3+ /Cr 2+ Ni 3+ /Ni 2+ Metals with only one common charge Ag + Zn 2+ Cd 2+ Al 3+ Complete and balance the following reactions: (1) A strip of lead metal is added to a solution of silver nitrate. (2) An acidic solution of potassium dichromate is added to a solution of iron (II) nitrate (3) An alkaline (basic) solution of potassium permanganate and tin (II) nitrate are mixed (4) Liquid bromine is added to a solution of potassium iodide. (5) Cadmium nitrate is mixed with chromium (II) nitrate

8 Standard Reduction Potentials In order to determine if a redox reaction will occur spontaneously, each reduction reaction is assigned a standard reduction potential (E ) measured in volts (V). Standard reduction potentials given are for standard conditions: at a temperature of 25 ºC, all gases at 1.00 atm and all solutions having a concentration equal to 1.0 M. Standard reduction potentials are relative to the reduction of hydrogen, which is assigned a voltage of zero. The higher the reduction potential, the more easily the element/compound is reduced. The lower the reduction potential, the more easily the element/compound is oxidized. Note: If the reduction reaction is reversed and is written as an oxidation, the sign on the voltage is reversed. Multiplying the reaction by a number does NOT change the voltage. The overall voltage for a reaction can be determined by adding the voltages for the reduction half reaction and the oxidation half reaction. If the overall voltage is positive, the reaction is spontaneous. If the overall voltage is negative the reaction is non spontaneous. ex. Which of the following substances is the strongest oxidizing agent? F 2, Cd 2+, or Na + ex. Which of the following substances is the strongest reducing agent? Rb, Cl, or Sn 2+ ex. Determine the voltage for each halfreaction. Determine the overall voltage. Is the reaction spontaneous or non spontaneous? (1) Cl 2 + Cd 2Cl + Cd 2+ (2) 2I + Zn 2+ I 2 + Zn Metal Reactivity A ranking according to reactivity is called an activity series. An activity series of metals places the most reactive (i.e. the most easily oxidized) metal at the top. If a solid piece of metal is combined with a solution of a different metal ion and a reaction takes place, the solid metal is considered to be the more reactive metal since it undergoes oxidation. If a solid piece of metal is combined with a solution of a different metal ion and no reaction takes place, the metal in solution is considered to be the more reactive metal. The relative reactivity metals can be found on an activity series. In addition, the solid metal that is most easily oxidized will have the metal ion that is most difficult to reduce and the solid metal that is the most difficult to oxidize will have the metal ion that is most easily reduced. Activity Series of Metals lithium potassium calcium sodium magnesium aluminum zinc chromium iron nickel tin lead hydrogen copper silver mercury platinum gold A reaction between a metal that is higher on the activity series with a metal ion lower on the activity series will be spontaneous and have a positive overall voltage. ex. The following: reaction Mg + 2Ag + Mg Ag is a SPONTANEOUS reaction because magnesium is a more reactive metal and is higher on the activity series than silver. Note: The reduction potential for Ag + + e Ag (0.80 V) is higher than the reduction potential for Mg e Mg (2.37 V) The overall voltage for the reaction is 3.17 V. The less reactive metal will have the more easily reduced metal ion. The more reactive metal will have the less easily reduced metal ion.

9 Example: An experiment was conducted by combining solid metals with solutions of metal ions and the following data was collected. Ti 3+ Ga 3+ In 3+ Ti reaction reaction Ga In no reaction no reaction reaction no reaction Rank the metals from most reactive to least reactive. Rank the metal ions from most easily reduced/highest standard reduction potential to least easily reduced/lowest standard reduction potential. Electrochemical Cells Electrochemical or galvanic cells (also called batteries) convert the chemical energy of redox reactions into electrical energy. In an electrochemical cell, the reduction half reaction and the oxidation half reaction are separated into two half cells. Each half cell consists of an electrode and a solution. There may also be a liquid or a gas that is part of the redox reaction. The electrons transferred during the redox reaction create a current that moves through a wire connecting the two half cells. In order to complete the circuit, a salt bridge also connects the two half cells. A salt bridge is a glass tube filled with an electrolyte (i.e. KNO 3). The voltage produced by the redox reaction (E cell) is measured using a voltmeter. The overall redox reaction in an electrochemical cell is always spontaneous (always has a positive overall voltage). The half cell where oxidation occurs is called the anode and the half cell where reduction occurs is called the cathode. Electrons are produced at the anode and travel through the wire towards the cathode where they are consumed. The metal electrode of the anode will generally decrease in mass as the metal is oxidized and the metal electrode of the cathode will generally increase in mass as the metal ions in the solution are reduced. In addition to completing the circuit in an electrochemical cell, the salt bridge also works to maintain electrical neutrality of each half cell. In the salt bridge, ions migrate in order to keep the half cells neutral (uncharged). As the electrochemical cell operates, the half cell where reduction occurs become more negative and the half cell where oxidation occurs becomes more positive. The cation in the salt bridge (i.e. K + ) migrates towards the cathode and the anion in the salt bridge (i.e. NO 3 ) migrates towards the anode in order to balance out the effect of these charges building up in the each half cell. Line Notation Electrochemical cells can be represented using line notation. Anode components are listed on the left and cathode components are listed on the right. The half cells are separated by a double vertical line: " ". Phase boundaries are indicated by a single vertical line: " ". (The order of anode components and cathode components is not set).

10 ex. Consider the following electrochemical cell. (a) Write the cathode half reaction. Label the cathode. (b) Write the anode half reaction. Label the anode. (c) Write the overall reaction. (d) Calculate the cell voltage. (Eº cell) (e) On the diagram, show electron flow and ion migration. (f) Which electrode loses mass and which electrode gains mass? (g) Give the line notation for this electrochemical cell. ex. Consider the following electrochemical cell. (a) Write the cathode half reaction. Label the cathode. (b) Write the anode half reaction. Label the anode. (c) Write the overall reaction. (d) Calculate the cell voltage. (Eº cell) (e) On the diagram, show electron flow and ion migration. (f) Give the line notation for this electrochemical cell.

11 The Effect of Half-Cell Concentrations on Standard Cell Potential (Eºcell) For the given electrochemical cell Cathode Reaction: Pb e Pb (0.13 V) Anode Reaction: Al Al e (1.66 V) Overall Reaction: 3Pb Al 3Pb + 2Al 3+ Eº cell = 0.13 V V = 1.53 V In order to produce a cell potential of 1.53 V, the cell must be under standard conditions (at 25 ºC and with the concentration of each half cell equal to 1.0 M). If the concentration of solutions in one or both of the half cells is changed, the cell potential will not be equal to the standard cell potential of 1.53 V. In an electrochemical cell, standard conditions can be thought of as the cell in a state of equilibrium. An equilibrium expression can be written for the overall reaction (which includes ions since they are aqueous, but does not include metals since they are solids). For the above electrochemical cell, the equilibrium expression is written as follows: K = [Al3+ ] 2 [Pb 2+ ] 3 Under standard conditions, since [Pb 2+ ] = [Al 3+ ] = 1.0 M K = (1.0)2 (1.0) 3 = 1.0 When the concentrations are not equal to 1.0 M, a reaction quotient, Q, must be calculated. - If Q<1, the cell potential, E cell, will be GREATER than Eº cell - If Q>1, the cell potential, E cell, will be LESS than Eº cell For example, Calculate Q and predict the effect on cell potential for the following concentrations of Pb 2+ and Al 3+ (1) [Pb 2+ ] = [Al 3+ ] = 2.0 M Q = (2.0)2 (2.0) 3 = 0.50 Since Q<1, the cell potential, Ecell, will be GREATER than Eºcell (1.53 V) (2) [Pb 2+ ] = [Al 3+ ] = 0.10 M Q = (0.10)2 (0.10) 3 = 10 Since Q>1, the cell potential, Ecell, will be LESS than Eºcell (1.53 V) (3) [Pb 2+ ] = 1.0 M and [Al 3+ ] = 0.80 M Q = (0.80)2 (1.0) 3 = 0.64 Since Q<1, the cell potential, Ecell, will be GREATER than Eºcell (4) [Pb 2+ ] = 0.60 M and [Al 3+ ] = 1.0 M Q = (1.0)2 (0.60) 3 = 4.6 Since Q>1, the cell potential, Ecell, will be LESS than Eºcell Operation of an Electrochemical Cell and the Effect on Cell Potential As an electrochemical cell operates, over time the concentration of the product will increase and concentration of the reactant will decrease. The result is that Q (products/reactants) will become greater than 1 and the cell potential will decrease. The longer the cell operates, the more the cell potential will decrease. Eventually, the concentration of the products will increase and the concentration of the reactants will decrease to such an extent that the cell potential will be equal to zero. This is why batteries (which are electrochemical cells) eventually "die". example: The cell operates until the [Al 3+ ] = 1.24 M. Determine [Pb 2+ ]. Calculate Q and predict the effect on cell potential. To determine [Pb 2+ ], use an ICE table with the initial concentration of each ion equal to 1.0 M. Since the final concentration of Al 3+ is known, the change in concentration can be determined and the final concentration of Pb 2+ can be found. From the final concentrations, a Q value can be calculated. 3Pb Al 3Pb + 2Al 3+ I C 3 (0.24) E Q = (1.24)2 (0.64) 3 = 5.9 Since Q>1, the cell potential will be LESS than 1.53 V (As the cell operates, [Al 3+ ] increases and [Pb 2+ ] decreases, and as a result, the cell potential will decrease)

12 Electrolytic Cells Electrolytic cells convert electrical energy into chemical energy in order to enable a nonspontaneous reaction to occur. In an electrolytic cell, electrical energy is supplied by a power source (i.e. a battery). In order to determine the reaction taking place at the cathode, list all possible reduction reactions and the voltage of each. The reduction reaction that takes place is the one with the highest voltage. In order to determine the reaction taking place at the anode list all possible oxidation reactions and the voltage of each. The oxidation reaction that takes place is the one with the highest voltage. The power supply required to operate the electrolytic cell will be equal to the overall voltage for the electrolytic cell. There are three basic types of electrolytic cells. Type I: Inert electrodes (i.e. platinum, graphite, etc.), molten electrolyte ex. Consider the following electrolytic cell. (a) Write the cathode half reaction. (b) Write the anode half reaction. (c) Write the overall reaction. (d) Calculate the power that must be supplied to operate the cell. Type II: Inert electrodes electrolyte solution (ionic compound in water) ex. Consider the following electrolytic cell. (a) Write the cathode half reaction. (b) Write the anode half reaction. (c) Write the overall reaction. (d) Calculate the power that must be supplied to operate the cell.

13 Type III: Reactive electrodes, electrolyte solution ex. Consider the following electrolytic cell. (a) Write the cathode half reaction. (b) Write the anode half reaction. (c) Write the overall reaction. (d) Calculate the power that must be supplied to operate the cell. Current Calculations In an electrolytic cell, the current supplied by the power source for a given amount of time can be related to the mass change or charge of the metal of an electrode. Current is calculated according to the following equation: where: I = Current (Amperes or amps, A) 1 amp = 1 Coulomb/second q q = charge (coulombs, C) 1 mole of electrons has a charge of 1 Faraday; 1 Faraday = Coulombs I therefore, 1 mol e t = Coulombs t = time (seconds) ex. Consider the half reaction: Cu e Cu. What mass of copper is produced if 10.0 A is applied over 30.0 minutes? 60 s 30.0 minx 1 min = 1.80x103 s I = q q = It = ( 10.0C ) (1.80x10 3 s) = 1.80x10 4 C t s 1.80x mol e 1 mol Cu g Cu C x x x = 5.93 g Cu C 2 mol e 1 mol Cu ex. Consider the half reaction: Ag + + e Ag. How long (in minutes) must a 5.00 A current be applied to produce 10.5 g of silver metal? ex. Consider the half reaction: Tl e Tl. What current must be applied in order to produce 2.04 g of thallium in 45.0 min? ex. The reduction of an ion of Titanium is carried out in an electrolytic cell. What is the charge on the ion if a 16.0 A current is applied for 20.0 min and produces 3.18 g of titanium metal. Give the balanced reduction reaction.