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Electrochemistry

Electrochemistry Terminology Oxidation A process in which an element attains a more positive oxidation state Na(s) Na+ + e- Reduction A process in which an element attains a more negative oxidation state Cl2 + 2e- 2Cl- Oxidizing agent -The substance that is reduced is the oxidizing agent Reducing agent - The substance that is oxidized is the reducing agentLEO says GERElectrochemistry Terminology Anode -The electrode where oxidation occurs Cathode - The electrode where reduction occurs

Reduction at the CathodeLeo is aBalancing EquationsConsider the reduction of Ag+ ions with copper metal.

Cu + Ag+ ----> Cu2+ + Ag

Balancing EquationsStep 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction.OxCu ---> Cu2+Red Ag+ ---> AgStep 2: Balance each element for mass. Already done in this case.Step 3: Balance each half-reaction for charge by adding electrons.Ox Cu ---> Cu2+ + 2e-Red Ag+ + e- ---> AgCu + Ag+ ----> Cu2+ + AgBalancing EquationsStep 4:Multiply each half-reaction by a factor so that the reducing agent supplies as many electrons as the oxidizing agent requires.Reducing agent Cu ---> Cu2+ + 2e-Oxidizing agent 2 Ag+ + 2 e- ---> 2 AgStep 5:Add half-reactions to give the overall equation.Cu + 2 Ag+ ---> Cu2+ + 2AgThe equation is now balanced for both charge and mass.Balancing EquationsBalance the following in acid solution VO2+ + Zn ---> VO2+ + Zn2+Step 1:Write the half-reactionsOxZn ---> Zn2+RedVO2+ ---> VO2+Step 2:Balance each half-reaction for mass.OxZn ---> Zn2+RedVO2+ ---> VO2+ + H2O2 H+ +Add H2O on O-deficient side and add H+ on other side for H-balance.Balancing EquationsStep 3:Balance half-reactions for charge.OxZn ---> Zn2+ + 2e-Rede- + 2 H+ + VO2+ --> VO2+ + H2OStep 4:Multiply by an appropriate factor.OxZn ---> Zn2+ + 2e-Red 2e- + 4 H+ + 2 VO2+ ---> 2 VO2+ + 2 H2OStep 5:Add balanced half-reactionsZn + 4 H+ + 2 VO2+ ---> Zn2+ + 2 VO2+ + 2 H2O

Balancing Equations for Redox ReactionsA great example of a thermodynamically spontaneous reaction is the thermite reaction. Here, iron oxide (Fe2O3 = rust) and aluminum metal powder undergo a redox (reduction-oxidation) reaction to form iron metal and aluminum oxide (Al2O3 = alumina): Fe2O3(s) + Al(s) Al2O3(s) + Fe(l)Al = 0Fe = +3Fe = 0Al = +3

Tips on Balancing EquationsNever add O2, O atoms, or O2- to balance oxygen.Never add H2 or H atoms to balance hydrogen.Be sure to write the correct charges on all the ions.Check your work at the end to make sure mass and charge are balanced.

How many electrons are transferred in the following reaction?2ClO3 + 12H+ + 10I 5I2 + Cl2 + 6H2O

1252301000of30Which of the following reactions is possible at the anode of a galvanic cell?

Zn Zn2+ + 2eZn2+ + 2e ZnZn2+ + Cu Zn + Cu2+ Zn + Cu2+ Zn2+ + Cutwo of these0of3010SecondsRemainingWhich of the following species cannot function as an oxidizing agent?

S(s)NO3(ag)Cr2O72(aq)I (aq)MnO4 (aq)0of3015Electrochemical CellsAn apparatus that allows a redox reaction to occur by transferring electrons through an external connector.Product favored reaction ---> voltaic or galvanic cell ----> electric currentReactant favored reaction ---> electrolytic cell ---> electric current used to cause chemical change.

Batteries are voltaic cellsAnodeCathodeBasic Concepts of Electrochemical Cells

Terms Used for Voltaic Cells

CELL POTENTIAL, EFor Zn/Cu cell, potential is +1.10 V at 25 C and when [Zn2+] and [Cu2+] = 1.0 M.This is the STANDARD CELL POTENTIAL, Eo- a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 C.

Calculating Cell VoltageBalanced half-reactions can be added together to get overall, balanced equation. Zn(s) ---> Zn2+(aq) + 2e-Cu2+(aq) + 2e- ---> Cu(s)--------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)If we know Eo for each half-reaction, we could get Eo for net reaction.Measuring Standard Electrode Potential

Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.Table of Reduction Potentials

Measured against the StandardHydrogenElectrodeTABLE OF STANDARD REDUCTION POTENTIALS 2 Eo (V)Cu2+ + 2e- Cu+0.342 H+ + 2e- H0.00Zn2+ + 2e- Zn-0.76oxidizingability of ionreducing abilityof elementTo determine an oxidation from a reduction table, just take the opposite sign of the reduction!Zn/Cu Electrochemical CellZn(s) ---> Zn2+(aq) + 2e-Eo = +0.76 VCu2+(aq) + 2e- ---> Cu(s)Eo = +0.34 V---------------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s) Eo = +1.10 V

Cathode, positive, sink for electronsAnode, negative, source of electrons

+

Cd --> Cd2+ + 2e-orCd2+ + 2e- --> CdFe --> Fe2+ + 2e-orFe2+ + 2e- --> FeEo for a Voltaic CellAll ingredients are present. Which way does reaction proceed?

From the table, you see Fe is a better reducing agent than CdCd2+ is a better oxidizing agent than Fe2+

Eo for a Voltaic Cell

More About Calculating Cell VoltageAssume I- ion can reduce water. 2 H2O + 2e- ---> H2 + 2 OH-2 I- ---> I2 + 2e--------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2Assuming reaction occurs as written, E = Ecat+ Ean= (-0.828 V) + (- 0.535 V) = -1.363 VNegative E means rxn. occurs in opposite direction (the connection is backwards or you are recharging the battery)CathodeAnodeGalvanic (Electrochemical) Cells Spontaneous redox processes have:

A positive cell potential, E0 A negative free energy change, (-G)

Zn - Cu Galvanic CellZn2+ + 2e- Zn E = -0.76V Cu2+ + 2e- Cu E = +0.34V

From a table of reduction potentials:Zn - Cu Galvanic Cell Cu2+ + 2e- Cu E = +0.34V

The less positive, or more negative reduction potential becomes the oxidationZn Zn2+ + 2e- E = +0.76VZn + Cu2+ Zn2+ + Cu E0 = + 1.10 V Line Notation

Zn(s) | Zn2+(aq) (1.0M) || Cu2+(aq) (1.0M) | Cu(s)An abbreviated representation of an electrochemical cellAnode solutionAnode materialCathode solutionCathode material||||Line notation is cool, just like AC

Zn(s) |Zn2+(aq) (1.0M) ||H+(aq) (1.0M)|H2(g) (1.00 atm) |Pt(s)

Calculating G0 for a Cell

G0 = -nFE0n = moles of electrons in balanced redox equationF = Faraday constant = 96,485 coulombs/mol e-Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Calculate DG for the following reaction:Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq)

Fe+2(aq) + 2e- Fe(s)E = 0.44 VCu+2(aq) +2e- Cu(s) E = 0.34 VTry this oneDay 3 (dahditdadahditahhCharge!)

The Nernst Equation

Standard potentials assume a concentration of 1.0 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 in balanced redox equationF = Faraday constant = 96,485 coulombs/mol e-Nernst Equation Simplified

At 25 C (298 K) the Nernst Equation is simplified this way:Equilibrium Constants and Cell PotentialAt equilibrium, forward and reverse reactions occur at equal rates, therefore: The battery is dead The cell potential, E, is zero volts

Modifying the Nernst Equation (at 25 C):Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V Calculating an Equilibrium Constant from a Cell Potential

Concentration Cell

Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.Both sides have the same components but at different concentrations.???Concentration Cell

Both sides have the same components but at different concentrations.The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentrationZn2+ (1.0M) + 2e- Zn (reduction) Zn Zn2+ (0.10M) + 2e- (oxidation) ???CathodeAnodeZn2+ (1.0M) Zn2+ (0.10M) Concentration Cell

Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C).Both sides have the same components but at different concentrations.???CathodeAnodeZn2+ (1.0M) Zn2+ (0.10M) Concentration Cell

Nernst CalculationsZn2+ (1.0M) Zn2+ (0.10M)

Charging a Battery

When you charge a battery, you are forcing the electrons backwards (from the + to the -). To do this, you will need a higher voltage backwards than forwards. This is why the ammeter in your car often goes slightly higher while your battery is charging, and then returns to normal.

In your car, the battery charger is called an alternator. If you have a dead battery, it could be the battery needs to be replaced OR the alternator is not charging the battery properly.Dry Cell BatteryAnode (-)Zn ---> Zn2+ + 2e-Cathode (+)2 NH4+ + 2e- ---> 2 NH3 + H2

Alkaline BatteryNearly same reactions as in common dry cell, but under basic conditions.

Anode (-): Zn + 2 OH- ---> ZnO + H2O + 2e-Cathode (+): 2 MnO2 + H2O + 2e- ---> Mn2O3 + 2 OH-

Mercury BatteryAnode:Zn is reducing agent under basic conditionsCathode:HgO + H2O + 2e- ---> Hg + 2 OH-

Lead Storage BatteryAnode (-) Eo = +0.36 VPb + HSO4- ---> PbSO4 + H+ + 2e-Cathode (+) Eo = +1.68 VPbO2 + HSO4- + 3 H+ + 2e- ---> PbSO4 + 2 H2O

Ni-Cad BatteryAnode (-)Cd + 2 OH- ---> Cd(OH)2 + 2e-Cathode (+) NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-

The positive electrode is made of Lithium cobalt oxide, or LiCoO2The negative electrode is made of carbon.H2 as a Fuel

Cars can use electricity generated by H2/O2 fuel cells.H2 carried in tanks or generated from hydrocarbons

The Electrochemical Corrosion of Iron53Preventing CorrosionCoating to keep out air and water.Galvanizing - Putting on a zinc coatHas a lower reduction potential, so it is more. easily oxidized.Alloying with metals that form oxide coats.Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.54Cathodic Protection of an Underground Pipe

Electrolytic ProcessesA negative cell potential, (-E0) A positive free energy change, (+G)

Electrolytic processes are NOT spontaneous. They have:

Electrolysis of Water

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

Electroplating of Silver

Anode reaction:Ag Ag+ + e- Electroplating requirements:1. Solution of the plating metal3. Cathode with the object to be plated2. Anode made of the plating metal4. Source of currentCathode reaction:Ag+ + e- Ag Step 1 convert current and time to quantity of charge in coulombsa. Amps x time = total charge transferred in coulombs(Coulomb/sec) x sec = coulombsStep 2 convert quantity of charge in coulombs to moles of electronscoulombs /(96,485 coulombs/mol e-) = mol e-Step 3 Convert moles of electrons to moles of substancemol e- x (mole substance/mol e-) = mol substanceStep 4 Convert moles of substance to grams of substancemol substance x formula mass of substance = mass of substanceCalculating plating1 mol PbSuppose that in starting a car on a cold morning a current of 125 amperes is drawn for 15.0 seconds from a cell of the type described above. How many grams of Pb would be consumed? (The atomic weight of Pb is 207.19.)125 C15 sec1 mol e-1 mol Pb96 485 C2 mol e-207.19 g1 secPb2+ + 2e- Pb 2.01 g PbSolving an Electroplating ProblemQ: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO3 using a 20.0 Ampere current?5.0 gAg+ + e- Ag 1 mol Ag107.87 g1 mol e-1 mol Ag96 485 C1 mol e-1 s20.0 C= 2.2 x 102 s