# Unit 14.  Oxidation States  Oxidation versus Reduction  Redox reactions Identifying Half Reactions Balancing  Voltaic (Galvanic) Cells Cell potential.

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Unit 14

 Oxidation States  Oxidation versus Reduction  Redox reactions Identifying Half Reactions Balancing  Voltaic (Galvanic) Cells Cell potential Standard reduction potentials Standard hydrogen electrode  Strength of Agents  Nernst Equation Calculations Equilibrium Gibbs Free Energy  Electrolytic cells Electroplating Stoichiometry

 Actual charge of a monatomic ion  Rules for assigning oxidation numbers 1. For an atom in its elemental form (uncombined with other elements) the oxidation number is always 0.  Example: H 2 = 0 2. For any monatomic ion, the oxidation number equals the charge of the ion.  Alkali metals always = +1  Alkaline earth metals always = +2  Aluminum always = +3

3. Nonmetals usually have negative oxidation numbers (although they can be positive)  Oxygen is usually -2 (exception: peroxide)  Hydrogen is usually +1 (-1 when bonded to metals)  Fluorine is -1.  Other halogens have oxidation number of -1 in binary compounds.  In oxyanions, halogens usually have positive oxidation states. 4. The sum of oxidation numbers in neutral compounds is zero. The sum of oxidation numbers in polyatomic ions equals the charge of the ion.

 What is the oxidation state of S? H 2 S = -2 SCl 2 = +2 S 8 = 0 Na 2 SO 3 = +4 SO 4 -2 = +6

 Oxidation  Oxidation – A process in which an element attains a more positive oxidation state (loses electrons) Na(s)  Na + + e -  Reduction  Reduction – A process in which an element attains a more negative oxidation state (gains electrons) Cl 2 + 2e -  2Cl -

LEO says GER G ain E lectrons = R eduction L ose E lectrons = O xidation

 OR…

 Redox Reaction:  Redox Reaction: reaction in which oxidation and reduction occur Oxidation and reduction occur spontaneously Mg + S → Mg 2+ + S 2- The magnesium atom (which has zero charge) changes to a magnesium ion by losing 2 electrons, and is oxidized to Mg 2+ The sulfur atom (which has no charge) is changed to a sulfide ion by gaining 2 electrons, and is reduced to S 2- (MgS)

Each sodium atom loses one electron (oxidation): Each chlorine atom gains one electron (reduction):

 Active metals: Lose electrons easily Are easily oxidized Are strong reducing agents   Active nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents

 A common approach for listing species that undergo REDOX is as half-reactions. 2Fe 3+ + Zn o (s) = 2Fe 2+ + Zn 2+ For 2Fe 3+ + Zn o (s) = 2Fe 2+ + Zn 2+ Fe 3+ + e -  Fe 2+ (reduction) Zn o (s)  Zn 2+ + 2e - (oxidation)  They are individual equations showing just the oxidation or just the reduction.

 The amount of each element must be the same on both sides (like normal)  BUT the gain/loss of electrons must also be balanced If an element loses a certain amount of electrons, another element must gain the same amount

#1 Divide the equation into two half reactions #2Balance each half reaction  Balance all elements other than H and O  Balance the O atoms by adding H 2 O  Balance the H atoms by adding H +  Balance the charge by adding e -

#3 Multiply the half reactions by integers so that the electrons lost in one half- reaction equals electrons gained in the other reaction #4Add the two half-reactions together  Simplify by cancelling out species appearing on both sides of the equation

Balancing Redox Reactions 19.1 Balance the following reaction in acidic solution. Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr 3+ 1. Separate the equation into two half-reactions. Oxidation: Cr 2 O 7 2- Cr 3+ +6+3 Reduction: Fe 2+ Fe 3+ +2+3 2. Balance the atoms other than O and H in each half- reaction. Cr 2 O 7 2- 2Cr 3+

Balancing Redox Reactions 3. For reactions in acid, add H 2 O to balance O atoms and H + to balance H atoms. Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 4. Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe 2+ Fe 3+ + 1e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 5. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 6Fe 2+ 6Fe 3+ + 6e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 19.1

7.Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 6Fe 2+ 6Fe 3+ + 6e - Oxidation: Reduction: 14H + + Cr 2 O 7 2- + 6Fe 2+ 6Fe 3+ + 2Cr 3+ + 7H 2 O 19.1

 Cu + NO 3 -  Cu +2 + NO 2 Answer: Cu + 4H + + 2NO 3 -  Cu +2 + 2NO 2 + 2H 2 O  Cr 2 O 7 -2 + Cl -  Cr +3 + Cl 2 Answer: 14H + + Cr 2 O 7 -2 + 6Cl -  2Cr +3 + 7H 2 O + 3Cl 2

 If redox reaction occurs in “basic solution”… Follow all steps as usual Before half-reactions are combined, add 1 OH - to BOTH sides of the equation for every 1 H + in the equation Combine H + and OH + to form H 2 O when they appear on the same side of the equation

 The following reactions occur in basic solutions  CN - + MnO 4 -  CNO - + MnO 2 Answer: 3CN - + H 2 O + 2MnO 4 -  3CNO - + 2MnO 2 + 2OH -  NO 2 - + Al  NH 3 + Al(OH) 4 - Answer: NO 2 - + 2Al + 5H 2 O + OH -  NH 3 + 2Al(OH) 4 -

 There are two general ways to conduct an oxidation-reduction reaction #1 Mixing oxidant and reductant together Cu 2+ + Zn (s) Cu (s) + Zn 2+ This approach does not allow for control of the reaction.

 Electrochemical cells Each half reaction is put in a separate ‘half cell.’ They can then be connected electrically. Called a “voltaic” or “galvanic” cell This permits better control over the system.

Cu 2+ + Zn (s) Cu (s) + Zn 2+ Zn Cu Cu 2+ Zn 2+ e-e- e-e- Electrons are transferred from one half-cell to the other using an external metal conductor. Electrons are transferred from one half-cell to the other using an external metal conductor.

 Electrodes Two solid metals that are connected by the external circuit  Anode The electrode at which oxidation occurs  Cathode The electrode at which reduction occurs

Just remember the… Red Cat Red “Red uction occurs cat at the cat hode”

 Oxidation occurs at the anode  Reduction occurs at the cathode electrons flow from the anode to the cathode through the external circuit  This means that electrons flow from the anode to the cathode through the external circuit

e-e- e-e- To complete the circuit, a salt bridge is used. To complete the circuit, a salt bridge is used. salt bridge

 Salt bridge Allows ions to migrate without mixing electrolytes (keeps solutions neutral). It can be a simple porous disk or a gel saturated with a non-interfering, strong electrolyte like KCl. KCl Cl - K+K+ Cl - is released to Zn side as Zn is converted to Zn +2 K + is released as Cu +2 is converted to Cu

 Salt bridge Consists of a U-shaped tube that contains the electrolyte solution (ex. NaNO 3 )  Usually solution is a paste or gel Ions of electrolyte solution will not react with other ions in the cell or with the electrode materials  Ions migrate to neutralize the cells  Anions migrate toward the anode  Cations migrate toward the cathode

 Rather than drawing an entire cell, a type of shorthand can be used. For our copper - zinc cell, it would be: Zn | Zn 2+ (1M) || Cu 2+ (1M) | Cu  The anode is always on the left. | = boundaries between phases || = salt bridge  Other conditions like concentration are listed just after each species.

 Electromotive Force (EMF):  Electromotive Force (EMF): Driving force that pushes electrons through the external circuit Measures how willing a species is to gain or lose electrons Also referred to as “cell potential” or “cell voltage” E cell Denoted E cell Spontaneous cell reaction = positive (+) E cell Measured in volts (1 V = 1Joule/Coulomb)

 EMF under standard state conditions (E o cell ) All soluble species are at 1 M Slightly soluble species must be at saturation. Any gas is constantly introduced at 1 atm Any metal must be in electrical contact Other solids must also be present and in contact.

 E o cell is determined from the reduction potentials of the two half reactions E o cell = E o red(cathode) – E o red(anode)  Reduction potentials can be looked up for all half reactions (table given on AP test)

Half reaction E o, V F 2 (g) + 2H + + e - 2HF (aq) 3.053 Ce 4+ + e - Ce 3+ (in 1M HCl) 1.28 O 2 (g) + 4H + + 4e - 2H 2 O (l) 1.229 Ag + + e - Ag (s) 0.7991 2H + + 2e - H 2 (g) 0.000 Fe 2+ + 2e - Fe (s) -0.44 Zn 2+ + 2e - Zn (s) -0.763 Al 3+ + 3e - Al (s) -1.676 Li + + e - Li (s) -3.040

 The table of standard reduction potentials relates to the activity series of metals Flip the table upside down and only look at the metals (solids), and it is the activity series list The AP test does not give you an activity series list so you have to use the standard reduction potentials to predict products of reactions

 E 0 is for the reaction as written  The more positive E 0 the greater the tendency for the substance to be reduced  The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed (do this if you need the “oxidation potential” of an element)  Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0

 Reversing the sign  For the reduction of Zn +2 + 2e -  Zn E red = -0.76 The reduction potential of Zn is -0.76  If we are looking at the reverse reaction and want the oxidizing potential, the sign is reversed Zn  of Zn +2 + 2e - E ox = +0.76  We will do redox calculations using only “reduction potentials”

 Standard Hydrogen Electrode (SHE) We can’t measure the standard reduction potential of a half reaction directly SHE used as a reference point  Reduction potential = 0 V  Consists of… Platinum wire connected to piece of platinum foil Electrode encased in a glass tube so that hydrogen at standard conditions can bubble over the platinum

 To solve for the E cell, subtract the reduction potential for the half-reaction at the cathode from the reduction potential for the half-reaction at the anode. E o cell = E o cathode – E o anode

 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? 2Cr (s) + 3Cd 2+ (1 M) 3Cd (s) + 2Cr 3+ (1 M) Cd 2+ (aq) + 2e - Cd (s) E 0 = -0.40 V Cr 3+ (aq) + 3e - Cr (s) E 0 = -0.74 V E 0 = E 0 cathode - E 0 anode cel l E 0 = -0.40 – (-0.74) cell E 0 = 0.34 V cell

 Given that the standard reduction potential of Zn +2 to Zn(s) is -0.76 V, calculate the E 0 red for the reduction of Cu +2 to Cu(s). Zn(s) + Cu 2+ (aq,1 M) Cu(s) + Zn +2 (aq,1 M) E 0 cell = 1.10 V E 0 = E 0 cathode - E 0 anode cel l 1.10 = E 0 cathode - (-0.76) E 0 = 0.34 V cathode

 The more positive the value of E 0 red, the greater the driving force for reduction under standard conditions Meaning…the reaction at the cathode has a more positive value of E 0 red than the reaction at the anode

 A voltaic cell is based on the following two standard half reactions. Determine which half-reaction occurs at the cathode and which occurs at the anode as well as the standard cell potential. E 0 red of (Cd +2 /Cd) is -0.403 and E 0 red of (Sn +2 /Sn) is -0.136 Reaction of Sn +2 /Sn is more positive so it occurs at the cathode Cd 2+ (aq) + 2e - Cd (s) Sn 2+ (aq) + 2e - Sn (s)

 Cathode reaction is reduction/Anode reaction is oxidation Cathode:Sn 2+ (aq) + 2e - Sn (s) E 0 red = -0.136 Anode : Cd (s) Cd 2+ (aq) + 2e - E 0 red = -0.403 Cd 2+ (aq) + 2e - Cd (s) Sn 2+ (aq) + 2e - Sn (s) E 0 = E 0 cathode - E 0 anode cel l E 0 = -0.136 – (-0.403) cell E 0 = 0.267 V cell

 The more readily an ion is reduced (the more positive its E 0 red value), the stronger it is as an oxidizing agent.  The half-reaction with the smallest reduction potential (most negative E 0 red ) is most easily reversed as an oxidation Smallest reduction potential is strongest reducing agent  The stronger the oxidizing agent, the weaker it acts as a reducing agent (and vice versa)

 E versus E 0 E means not at standard conditions  E versus E cell E doesn’t have to occur in a voltaic cell so there is no cathode and anode E = E reduction - E oxidation

 Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M. R = 8.31 J/(mol  K) T = Temperature in K n = moles of electrons transferred in balanced redox equation F = Faraday constant = 96,500 coulombs/mol e - (Charge of one mole of electrons) (Charge of one mole of electrons) Q = reaction quotient

 At room temperature (25  C or 298 K) the Nernst equation can be simplified  As the concentration of products increases in a redox reaction, voltage decreases  As the concentration of reactants in a redox equation increases, voltage increases

 Determine the potential of a Pt indicator electrode if dipped in a solution containing 0.1M Sn 4+ and 0.01M Sn 2+. Sn 4+ + 2e - Sn 2+ E o = 0.15V E = 0.15V - log = 0.18 V 0.0592 2 0.01 M 0.1M

 At equilibrium, the forward and reverse reactions are equal The cell potential is 0 V The battery is “dead”  Modifying the Nernst equation for equilibrium (at 25°C)gives us…

 Calculate the equilibrium constant, K, for the following reaction. E 0 Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V

 Any reaction that can occur in a voltaic cell must be spontaneous Not all redox reactions are spontaneous  A positive value of E indicates a spontaneous process A negative value means it is not spontaneous

 Both ∆G and E can be used to tell whether or not a reaction is spontaneous Negative ∆G or positive E means spontaneous  The relationship between the two is  G 0 = -nFE 0

 Calculate ∆G for the following reaction: E 0 Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V

 As reactants are converted to products, Q increases and E decreases until E = 0  Using the Nernst equation, when E = 0, G = 0 When G = 0, the system is at equilibrium No net reaction is occurring

 Earlier, we explained that  G and the equilibrium constant can be related. Since E cell is also related to K, we know the following. Q  G E Forward change, spontaneous< K -+ At equilibrium= K 00 Reverse change, spontaneous> K + -

Greater voltages are achieved by connecting multiple voltaic cells Cathode is labeled with a (+) sign Anode is labeled with a (-) sign Life of battery depends on quantities of substances packed in the battery to oxidize and reduce at the anode and cathode  Portable, self-contained power source that contains one or more voltaic cells

 An outside source is used to force a nonspontaneous redox reaction to take place Two electrodes are in a molten salt or a solution Outside source example is a battery  E 0 cell is negative and ∆G is positive (not spontaneous)

Example: Electrolysis of Water

In acidic solution Anode rxn: Cathode rxn: -1.23 V -0.83 V -2.06 V

 Uses electrolysis to deposit a thin layer of one metal onto another metal to improve beauty or to resist corrosion  We will sometimes be asked find how much metal “plates out” when electroplating occurs.

#1Given the current used (amps) and time, calculate the charge (coulombs) #2Solve for moles of electrons involved using Faraday’s constant of 96,500 C/mol I = qtqt I = current (amperes, A) q = charge (coulombs, C) t = time (sec) ___ Coulombs ___ Mole of e- 96,500 C 1 mole e- =

#3Identify the balanced half reaction that you are working with (with electrons transferred) #4Use stoichiometry to find moles of metal, then grams of metal Example: Au +3 + 3e-  Au(s) e- transferred  moles of metal  grams of metal

How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of 0.452 A is passed through the cell for 1.5 hours? Step 1: determine the charge, C I = qtqt 0.452 = q 5400 sec q = 2440.8 C Ca 2+ (l) + 2Cl - (l) Ca (s) + Cl 2 (g)

How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of 0.452 A is passed through the cell for 1.5 hours? Step 2: determine the moles of e- involved Mole of e- = 0.026 2440.8 Coulombs ___ Mole of e- = 96,500 C 1 mole e- Ca 2+ (l) + 2Cl - (l) Ca (s) + Cl 2 (g)

How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of 0.452 A is passed through the cell for 1.5 hours? Step 3: Determine the balanced half reaction Ca 2+ (l) + 2Cl - (l) Ca (s) + Cl 2 (g) Cathode: Ca 2+ (l) + 2e - Ca (s)

How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of 0.452 A is passed through the cell for 1.5 hours? Step 4: Stoichiometry time Ca 2+ (l) + 2Cl - (l) Ca (s) + Cl 2 (g) Cathode: Ca 2+ (l) + 2e - Ca (s) 0.026 mol e- = 1 mol Ca(s) 2 mol e- 1 mol Ca 40.08 g Ca(s) x = 0.52 g Ca(s) 0.52 grams of Ca will “plate out” or be produced

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