Electrochemistry

ElectrochemistryUnit 14OverviewOxidation StatesOxidation versus ReductionRedox reactionsIdentifyingHalf ReactionsBalancingVoltaic (Galvanic) CellsCell potentialStandard reduction potentialsStandard hydrogen electrode

Strength of AgentsNernst EquationCalculationsEquilibriumGibbs Free EnergyElectrolytic cellsElectroplatingStoichiometry

Oxidation States (Numbers)Actual charge of a monatomic ionRules for assigning oxidation numbers

For an atom in its elemental form (uncombined with other elements) the oxidation number is always 0.Example: H2 = 0

For any monatomic ion, the oxidation number equals the charge of the ion.Alkali metals always = +1Alkaline earth metals always = +2Aluminum always = +3

Oxidation States (Numbers)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.

The sum of oxidation numbers in neutral compounds is zero. The sum of oxidation numbers in polyatomic ions equals the charge of the ion.Examples (Oxidation States)What is the oxidation state of S?

H2S = -2

SCl2 = +2

S8 = 0

Na2SO3 = +4

SO4-2 = +6

Oxidation - ReductionOxidation A process in which an element attains a more positive oxidation state (loses electrons) Na(s) Na+ + e-

Reduction A process in which an element attains a more negative oxidation state (gains electrons)

Cl2 + 2e- 2Cl-

Leo the Lion (Oxidation Reduction)

LEO says GERGain Electrons = ReductionLose Electrons = OxidationOxidation - ReductionOR

Redox ReactionsRedox Reaction: reaction in which oxidation and reduction occurOxidation and reduction occur spontaneouslyMg + S Mg2+ + S2-The magnesium atom (which has zero charge) changes to a magnesium ion by losing 2 electrons, and is oxidized to Mg2+

The sulfur atom (which has no charge) is changed to a sulfide ion by gaining 2 electrons, and is reduced to S2-(MgS)Redox Reactions

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

10TrendsActive metals: Lose electrons easily Are easily oxidized Are strong reducing agentsActive nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents

Half ReactionsA common approach for listing species that undergo REDOX is as half-reactions.

For 2Fe3+ + Zno(s) = 2Fe2+ + Zn2+

Fe3+ + e- Fe2+(reduction)Zno(s) Zn2+ + 2e- (oxidation)

They are individual equations showing just the oxidation or just the reduction.

Balancing RedoxThe amount of each element must be the same on both sides (like normal)

BUT the gain/loss of electrons must also be balancedIf an element loses a certain amount of electrons, another element must gain the same amountBalancing Redox#1 Divide the equation into two half reactions

#2Balance each half reactionBalance all elements other than H and OBalance the O atoms by adding H2OBalance the H atoms by adding H+Balance the charge by adding e-Balancing Redox#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 togetherSimplify by cancelling out species appearing on both sides of the equationBalancing Redox Reactions19.1Balance the following reaction in acidic solution.Fe2+ + Cr2O72- Fe3+ + Cr3+1. Separate the equation into two half-reactions.Oxidation:Cr2O72- Cr3++6+3Reduction:Fe2+ Fe3++2+32. Balance the atoms other than O and H in each half-reaction.Cr2O72- 2Cr3+Balancing Redox Reactions3. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.Cr2O72- 2Cr3+ + 7H2O14H+ + Cr2O72- 2Cr3+ + 7H2O4. Add electrons to one side of each half-reaction to balance the charges on the half-reaction.Fe2+ Fe3+ + 1e-6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O5. 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+ + 7H2O19.1Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O6Fe2+ 6Fe3+ + 6e-Oxidation:Reduction:14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O19.1Balancing Redox ReactionsMore ExamplesCu + NO3- Cu+2 + NO2

Answer: Cu + 4H+ + 2NO3- Cu+2 + 2NO2 + 2H2O

Cr2O7-2 + Cl- Cr+3 + Cl2

Answer: 14H+ + Cr2O7-2 + 6Cl- 2Cr+3 + 7H2O + 3Cl2

Balancing RedoxIf redox reaction occurs in basic solutionFollow all steps as usualBefore half-reactions are combined, add 1 OH- to BOTH sides of the equation for every 1 H+ in the equationCombine H+ and OH+ to form H2O when they appear on the same side of the equationExamplesThe following reactions occur in basic solutions

CN- + MnO4- CNO- + MnO2

Answer: 3CN- + H2O + 2MnO4- 3CNO- + 2MnO2 + 2OH-

NO2- + Al NH3 + Al(OH)4-

Answer: NO2- + 2Al + 5H2O + OH- NH3 + 2Al(OH)4-

Electrochemical CellsThere are two general ways to conduct an oxidation-reduction reaction

#1 Mixing oxidant and reductant together

Cu2+ + Zn(s) Cu(s) + Zn2+

This approach doesnot allow forcontrol of the reaction.

Electrochemical cellsElectrochemical cellsEach 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.

Voltaic cells

Cu2+ + Zn(s) Cu(s) + Zn2+ZnCuCu2+Zn2+e-e-Electrons aretransferred fromone half-cell tothe other usingan external metalconductor.Voltaic CellsElectrodesTwo solid metals that are connected by the external circuit

AnodeThe electrode at which oxidation occurs

CathodeThe electrode at which reduction occurs

Voltaic CellsJust remember the

Red Cat

Reduction occursat the cathode

Voltaic CellsOxidation occurs at the anodeReduction occurs at the cathode

This means that electrons flow from the anode to the cathode through the external circuitVoltaic cells

e-e-To complete thecircuit, a saltbridge is used.salt bridgeVoltaic CellsSalt bridgeAllows 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.KClCl-K+Cl- is releasedto Zn side as Zn is converted to Zn+2K+ is releasedas Cu+2 isconverted to CuVoltaic CellsSalt bridgeConsists of a U-shaped tube that contains the electrolyte solution (ex. NaNO3)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 cellsAnions migrate toward the anodeCations migrate toward the cathodeVoltaic Cell

Cell DiagramsRather than drawing an entire cell, a type of shorthand can be used.

For our copper - zinc cell, it would be:

Zn | Zn2+ (1M) || Cu2+ (1M) | Cu

The anode is always on the left.| = boundaries between phases|| = salt bridge

Other conditions like concentration are listed just after each species.Cell PotentialElectromotive Force (EMF): Driving force that pushes electrons through the external circuitMeasures how willing a species is to gain or lose electronsAlso referred to as cell potential or cell voltageDenoted EcellSpontaneous cell reaction = positive (+) EcellMeasured in volts (1 V = 1Joule/Coulomb)Standard Cell PotentialEMF under standard state conditions (Eocell)All soluble species are at 1 MSlightly soluble species must be at saturation.Any gas is constantly introduced at 1 atmAny metal must be in electrical contactOther solids must also be present and in contact.

Standard Reduction PotentialsEocell is determined from the reduction potentials of the two half reactions

Eocell = Eored(cathode) Eored(anode)

Reduction potentials can be looked up for all half reactions (table given on AP test)Standard Reduction PotentialsHalf reaction Eo, VF2 (g) + 2H+ + e- 2HF (aq) 3.053Ce4+ + e- Ce3+ (in 1M HCl)1.28O2 (g) + 4H+ + 4e- 2H2O (l) 1.229Ag+ + e- Ag (s) 0.79912H+ + 2e- H2 (g)0.000Fe2+ + 2e- Fe (s) -0.44Zn2+ + 2e- Zn (s) -0.763Al3+ + 3e- Al (s) -1.676Li+ + e- Li (s) -3.040NoteThe table of standard reduction potentials relates to the activity series of metalsFlip the table upside down and only look at the metals (solids), and it is the activity series listThe AP test does not give you an activity series list so you have to use the standard reduction potentials to predict products of reactions

Standard Reduction PotentialsE0 is for the reaction as writtenThe more positive E0 the greater the tendency for the substance to be reducedThe half-cell reactions are reversibleThe sign of E0 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 E0

Standard Reduction PotentialsReversing the sign

For the reduction of Zn+2 + 2e- Zn Ered = -0.76The reduction potential of Zn is -0.76

If we are looking at the reverse reaction and want the oxidizing potential, the sign is reversedZn of Zn+2 + 2e- Eox = +0.76

We will do redox calculations using only reduction potentialsStandard Hydrogen ElectrodeStandard Hydrogen Electrode (SHE)We cant measure the standard reduction potential of a half reaction directlySHE 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

Standard Cell PotentialTo solve for the Ecell, subtract the reduction potential for the half-reaction at the cathode from the reduction potential for the half-reaction at the anode.

Eocell = Eocathode Eoanode

Standard Cell Potential (Example 1)What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?

2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 VCr3+ (aq) + 3e- Cr (s) E0 = -0.74 VE0 = E0 cathode - E0 anodecel lE0 = -0.40 (-0.74) cellE0 = 0.34 V cellStandard Cell Potential (Example 2)Given that the standard reduction potential of Zn+2 to Zn(s) is -0.76 V, calculate the E0red for the reduction of Cu+2 to Cu(s).

Zn(s) + Cu2+ (aq,1 M) Cu(s) + Zn+2(aq,1 M) E0cell = 1.10 VE0 = E0 cathode - E0 anodecel l1.10 = E0 cathode - (-0.76) E0 = 0.34 V cathodeStandard Cell PotentialThe more positive the value of E0red, the greater the driving force for reduction under standard conditionsMeaningthe reaction at the cathode has a more positive value of E0red than the reaction at the anodeStandard Cell Potential (Example 3)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.

E0red of (Cd+2/Cd) is -0.403 and E0red of (Sn+2/Sn) is -0.136

Reaction of Sn+2/Sn is more positive so it occurs at the cathode

Cd2+ (aq) + 2e- Cd (s)Sn2+ (aq) + 2e- Sn (s)Standard Cell Potential (Example 3)Cathode reaction is reduction/Anode reaction is oxidation

Cathode:Sn2+ (aq) + 2e- Sn (s) E0red = -0.136Anode: Cd (s) Cd2+ (aq) + 2e- E0red = -0.403

Cd2+ (aq) + 2e- Cd (s)Sn2+ (aq) + 2e- Sn (s)E0 = E0 cathode - E0 anodecel lE0 = -0.136 (-0.403) cellE0 = 0.267 V cellStrengths of Oxidizing AgentsThe more readily an ion is reduced (the more positive its E0red value), the stronger it is as an oxidizing agent.

The half-reaction with the smallest reduction potential (most negative E0red) is most easily reversed as an oxidationSmallest reduction potential is strongest reducing agent

The stronger the oxidizing agent, the weaker it acts as a reducing agent (and vice versa)Strengths of Oxidizing Agents

E versusE versus E0E means not at standard conditions

E versus EcellE doesnt have to occur in a voltaic cell so there is no cathode and anodeE = Ereduction - EoxidationNernst EquationStandard 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/(molK) T = Temperature in Kn = moles of electrons transferred in balanced redox equationF = Faraday constant = 96,500 coulombs/mol e- (Charge of one mole of electrons)Q = reaction quotientNernst EquationAt room temperature (25 C or 298 K) the Nernst equation can be simplified

As the concentration of products increases in a redox reaction, voltage decreasesAs the concentration of reactants in a redox equation increases, voltage increasesNernst Equation (Example 1)Determine the potential of a Pt indicator electrode if dipped in a solution containing 0.1M Sn4+ and 0.01M Sn2+.

Sn4+ + 2e- Sn2+Eo = 0.15V

E = 0.15V - log

= 0.18 V0.059220.01 M0.1MEquilibrium and NernstAt equilibrium, the forward and reverse reactions are equalThe cell potential is 0 VThe battery is dead

Modifying the Nernst equation for equilibrium (at 25C)gives usEquilibrium and NernstCalculate the equilibrium constant, K, for the following reaction.Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Spontaneous ProcessesAny reaction that can occur in a voltaic cell must be spontaneousNot all redox reactions are spontaneous

A positive value of E indicates a spontaneous processA negative value means it is not spontaneousGibbs free energy (G) and EMFBoth G and E can be used to tell whether or not a reaction is spontaneousNegative G or positive E means spontaneous

The relationship between the two is

G0 = -nFE0Gibbs free energy (G) and EMFCalculate G for the following reaction:

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Gibbs free energy (G) and EMFAs reactants are converted to products, Q increases and E decreases until E = 0

Using the Nernst equation, when E = 0, G = 0When G = 0, the system is at equilibriumNo net reaction is occurringSummaryEarlier, we explained that DG and the equilibrium constant can be related. Since Ecell is also related to K, we know the following.

Q DG EForward change, spontaneous< K -+

At equilibrium= K 00

Reverse change, spontaneous> K + -BatteriesGreater voltages are achieved by connecting multiple voltaic cellsCathode is labeled with a (+) signAnode is labeled with a (-) signLife 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 cellsElectrolytic CellsAn outside source is used to force a nonspontaneous redox reaction to take placeTwo electrodes are in a molten salt or a solutionOutside source example is a batteryE0cell is negative and G is positive (not spontaneous)

Example: Electrolysis of Water

Electrolysis of Water

In acidic solutionAnode rxn:Cathode rxn:-1.23 V-0.83 V-2.06 V

63ElectroplatingUses 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.Electroplating Stoichiometry!#1Given the current used (amps) and time, calculate the charge (coulombs)

#2Solve for moles of electrons involved using Faradays constant of 96,500 C/mol

I = qtI = current (amperes, A)q = charge (coulombs, C)t = time (sec)___ Coulombs ___ Mole of e-96,500 C1 mole e-=Electroplating Stoichiometry!#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 metalExample (stoichiometry)How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours?

Step 1: determine the charge, C

I = qt0.452 = q5400 secq = 2440.8 C Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)Example (stoichiometry)How much Ca will be produced in an electrolytic cell of molten CaCl2 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 C1 mole e-Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)Example (stoichiometry)How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours?

Step 3: Determine the balanced half reaction

Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)Cathode:Ca2+ (l) + 2e- Ca (s)Example (stoichiometry)How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours?

Step 4: Stoichiometry time

Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)Cathode:Ca2+ (l) + 2e- Ca (s)0.026 mol e- =1 mol Ca(s)2 mol e-1 mol Ca40.08 g Ca(s)x = 0.52 g Ca(s)0.52 grams of Ca will plate out or be produced