ELECTROCHEMISTRY

1. ELECTROCHEMISTRY

3. OXIDATION- REDUCTION Iron ores are minerals with a high iron content. One of them hematite, Fe2O3 is chemically very similar to iron rust. The reaction which iron metal is produced from hematite in a blast furnace is described as: Fe2O3(s) + 3CO(g) 2 Fe(l) + 3 CO2(g) We can think of the CO(g) as taking O atoms away from Fe2O3 to produce CO2(g) and the free element iron. A commonly used term to describe a reaction in which a substance loses O atoms, is « reduction» and gains O atoms is «oxidation». An oxidation and reduction must always occur together, and the reaction in which they do is called an oxidation-reduction reaction.

4. Oxidation State (O.S) Changes +4 -2 0 +3 +2 -2 -2 Fe2O3(s) + 3CO(g) 2 Fe(l) + 3 CO2(g) The O.S of oxygen is -2 everywhere it appears. That of iron(shwon in red) changes. It decreases from + 3 in Fe2O3 to 0 in the free element, Fe. The O.S of carbon also changes. It increases from +2 in CO to +4 in CO2 . In terms of oxidation state changes, in an oxidation process the O.S of some element increases and in reduction the O.S of an element decreases.

5. Identifying Oxidation- Reduction Reactions • Indicate whether each of the following is an oxidation-reduction reaction • a) MnO2(k) + 4 H+(aq) + 2 Cl-(aq) Mn2+(aq) + 2 H2O(s) + Cl2(g) • b) H2PO4-(aq) + OH-(aq) HPO42-(aq) + H2O(s) • Solution: • The O.S of Mn in MnO2 decreases from +4 to +2 in Mn2+. MnO2 is reducedto Mn2+. The O.S of O remains --2 throughout the reaction, and that of H at +1. The O.S of Clincreases from -1 in Cl- to 0 in Cl2. Cl- is oxidized to Cl2 .The reaction is an oxidation-reduction reaction. • The O.S of H is +1 on both sides of the equation. The O.S of O remains -2 throughout the reaction. The O.S of phosphorus is +5 in both H2PO4- and HPO42-. There are no changes in O.S . This is not an oxidation-reduction reaction. It is infact an acid-base reaction.

6. Oxidation and Reduction Half reactions The reaction illustrated below is an oxidation-reduction reaction Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) We can show this by evaluating changes in the O.S states. But we may think of the reaction as involving two half-reactions occuring at the same time. The overall or net reaction is the sum of these two reactions: Oxidation: Zn(k) Zn2+(aq) + 2e- Reduction: Cu2+(aq) + 2e- Cu(k) Net: Zn(k) + Cu2+(aq) Zn2+(aq) + Cu(k) In the half-equation the O.S of Zn increases from 0 to +2, corresponding to a loss of two electrons by each Zinc atom. In the second half-equation the O.S of Copper decreases from +2 to 0 corresponding to the gain of 2e- by each Cu2+(aq) . To summarize;

7. Oxidation and Reduction Half reactions Oxidation, is a process in which the O.S of some element increases, and in which electrons appear on the right in a half-equation Reduction, is a process in which the O.S of some element decreases, and in which the electrons appear on the left in a half equation. Oxidation and reduction half-reactions must always occur together, and the total number of electrons associated with the oxidation must equal to the total number associated with the reduction.

8. BALANCING OXIDATION- REDUCTION EQUATIONS • The same principles of equation balancing apply to oxidation-reduction (redox equations are balance for numbers of atoms and balance for electric charge.Several methods are possible but we emphasize the one described below: • The-Half reaction(Ion-Electron Method) • In this method of balancing a redox equation we: • write and balance seperate half-equations for oxidation and reduction. • adjust coefficients in the two-half equations so that the same number of electrons appears in each half equation • add together the two half-equations to obtain the balanced net equation.

9. Balancing the Equation for a Redox Reaction in Acidic Solution • Example: Write thebalancedequationforthereaction: • SO32-(aq) + MnO4-(aq) SO42-(aq) + Mn2+(aq) • Step 1.Write the «skeleton» half-equationsbased on thespeciesundergoingoxidationandreduction. The O.S of sulfurincreasesfrom +4 in SO32-to +6 in SO42-. The O.S. Of Mn decreasesfrom +7 in MnO4-to +2 in Mn2+. Theskeletonhalf-equationsare: • SO32-(aq) SO42-(aq) • MnO4-(aq) Mn2+(aq) • Step 2.Balanceeachhalf-equationatomically in thisorder • Atomsotherthan H and O • O atomsbyaddingH2O withtheappropriatecoefficient • H atomsbyadding H+withtheappropriatecoefficient • Theotheratoms( S and Mn) arealreadybalanced in theskeletonhalf-equations. Tobalance O atomsweaddone H2O moleculetotheleft of thefirsthalf-equationandfourtotheright of thesecond.

10. BalancingtheEquationfor a RedoxReaction in Acidic Solution SO32-(aq) + H2O(l) SO42-(aq) MnO4-(aq) Mn2+(aq) + 4 H2O(l) To balance H atoms we add two H+ to the right of the first half-equation and eight to the left of the second SO32-(aq) + H2O(l) SO42-(aq) + 2H+(aq) MnO4-(aq) + 8 H+(aq) Mn2+(aq) + 4 H2O(l) Step 3. Balance each half-equation electrically. Add the number of electrons necessary to get the same electric charge on both sides of each half equation.The half-equation in which the electrons appear on the right side is the oxidation half reaction. The other half equation with electrons on the left is the reduction half-equation. Oxidation: SO32-(aq) + H2O(l) SO42-(aq) + 2H+(aq) + 2e- (net charge on each side, -2) Reduction: MnO4-(aq) + 8 H+(aq) + 5e- Mn2+(aq) + 4 H2O(l) (net charge on each side, +2)

11. BalancingtheEquationfor a RedoxReaction in Acidic Solution Step 4. Obtain the net redox reaction by combining the half-equations. Multiply through the oxidation half-equation by 5 and through the reduction half-equation by 2. This results in 10 e- on each side of the net equation: 5 SO32-(aq) + 5 H2O(l)5 SO42-(aq) + 10 H+(aq) + 10e- 2MnO4-(aq) + 16 H+(aq) + 10e- 2 Mn2+(aq) + 8 H2O(l) 5 SO32-(aq) + 2 MnO4-(aq) + 5 H2O(l) + 16 H+(aq) 5 SO42-(aq) + 2 Mn2+(aq) + 8 H2O(l) + 10 H+(aq) Step 5. Simplify. The net equation should not contain the same species on both sides. Subtract 5 H2O from each side of the equation in step 4. This leaves 3 H2O on the right. Also subtract 10 H+ from each side, leaving 6 on the left. 5 SO32-(aq) + 2 MnO4-(aq) + 6 H+(aq) 5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(s) Step 6. Verify. Check the final net equation to ensure that is balanced both atomically and electrically.e.g the net charge on each side of the equation is: (5x2-) +(2 x 1-) + (6 x 1+) = (5 x 2-) + (2 x 2+) = -6.

12. BalancingtheEquationfor a RedoxReaction in Basic Solution • Balancetheequationforthereaction in whichchromateionoxidizessulfideion in basicsolutiontoproducefreesulfurandchromium(III) hydroxide: • CrO42- (aq) + S2-(aq) + OH- Cr(OH)3(s) + S(s) +H2O • Solution: Initially, wetreatthehalf-reactions as iftheyoccur in acidicsolution, andthenweadjustthemfor a basissolution: • STEP 1. writethetwoskeletonhalf-equationsandbalancethemfor Cr and S atoms • CrO42- (aq) Cr(OH)3(s) • S2-(aq) S(s) • STEP 2. Balanceeachhalf-equationfor H and O atoms. Notethatthesecondhalf-equation has no H or O atoms • CrO42- (aq) + 5 H+(aq) Cr(OH)3(s) + H2O • S2-(aq) S(s)

13. BalancingtheEquationfor a RedoxReaction in Basic Solution STEP 3. Balance the half-equations for electric charge by adding the appropriate numbers of electrons Reduction: CrO42- (aq) + 5 H+(aq) + 3 e- Cr(OH)3(s) + H2O Oxidation: S2-(aq) S(s) + 2 e- STEP 4. Change from an acidic to a basic medium by adding OH- ions and eliminating H+. The oxidation half equation is unaffected because it has no H+ ions. Add 5 OH- ions to each side of the reductionhalf equation; combine H+ and OH- into H2O; eliminate H2O from the right side of the half-equation. CrO42- (aq) + 5 H+(aq) + 5OH- (aq) + 3 e- Cr(OH)3(s) + H2O + 5OH- (aq) CrO42- (aq) + 5 H2O + 3 e- Cr(OH)3(s) + H2O + 5OH- (aq) CrO42- (aq) + 4 H2O + 3 e- Cr(OH)3(s) + 5OH- (aq)

14. BalancingtheEquationfor a RedoxReaction in Basic Solution STEP 5. Combine the half-equations to obtain the net redox equation. (multiply the reduction half-equation by 2 and the oxidation half reaction by 3) 2 CrO42- (aq) + 8 H2O + 6 e- 2 Cr(OH)3(s) + 10OH- (aq) 3 S2-(aq) 3 S(s) + 6 e- 2 CrO42- (aq) + 8 H2O + 3 S2-(aq) 2 Cr(OH)3(s) + 10 OH- (aq) + 3 S(s)

15. OXIDIZING AND REDUCING AGENTS • In a redoxreaction, thesubstancethatmakes it possibleforsomeothersubstancesto be oxidized is calledoxidizingagentoroxidant. Indoingso, theoxidizingagent is itselfreduced. Similarly, thesubstancethatcausessomeothersubstancesto be reduced is calledthereducingagentorreductant. Inthereactionthereducingagentitself is oxidized. • An oxidizingagent(oxidant): • contains an element whoseO.S. decreasesin a redoxreaction • «gains» electrons(electronsarefound on theleftside of thehalf-equation • A reducingagent(reductant): • contains an element whoseO.S. increases in a redoxreaction • «loses» electrons(electronsarefound on therightside of thehalf-equation

16. OXIDIZING AND REDUCING AGENTS Example: Hydrogen peroxide, H2O2 is a versatile chemical. Its uses include bleaching wood pulp and fabrics and substituting for chlorine in water purification. One reason for its versatility is that it can be either an oxidizing or reducing agent. For the following reactions, identify whether hydrogenperoxide is an oxidizing or reducing agent. a) H2O2 + 2 Fe2+(aq) + 2 H+(aq) 2 H2O + 2 Fe3+ b) 5 H2O2(aq) + 2 MnO4- (aq) + 6 H+ 8 H2O +2 Mn2+(aq) + 5 O2(g) Solution: a) Fe2+ is oxidized to Fe3+ and H2O2 makes it possible; H2O2 is an oxidizing agent. b) MnO4- is reduced to Mn2+ and H2O2 makes this possible; H2O2is a reducing agent

17. ELECTRODE POTENTIALS AND THEIR MEASUREMENT Twokinds of interactionsarepossiblebetween metal atoms on theelectrodeand metal ions in solution A metal ion Mn+maycollidewiththeelectrode, gain n electronsand be convertedto a metal atom M, theion is reduced A metal atom M on theelectrodemaylose n electronsandenterthesolution as theion Mn+. The metal atom is oxidized.

18. ELECTRODE POTENTIALS AND THEIR MEASUREMENT Electrode potential is a property proportional to the density of negative electric charge. To measure a difference in potential, we need to connect two half-cells in a special way. The flow of electric current is in the form of migration of ions from the greater negative charge to the other electrode.(here from Cu to Ag)

19. ELECTRODE POTENTIALS AND THEIR MEASUREMENT The net reaction that occurs as electric current flows through the electrochemical cell is: Anode:Oxidation Cathode:Reduction Net • Thereading on thevoltmeter is (0,463 V) significant. It is thepotentialdifferencebetweentwohalfcells. Becausethispotentialdifference is the « drivingforce» forelectrons, it is oftencalledtheelectromotiveforce(emf) of thecellorthecellpotential . • Whydoescopper not displace Zn2+fromsolution? • Ifwe set up an electrochemicalcellZn(s) / Zn2+halfcelland Cu2+/Cu(s) halfcell, wefindthatelectronsflowfromtheZntothe Cu. Thespontaneous net reaction in theelectrochemicalcell is • Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) • This is a spontaneousreactionandthereverse of thisreactiondoes not occurspontaneously.

20. CELL DIAGRAMS

21. CELL DIAGRAMS • A celldiagramshowsthecomponents of an electrochemicalcell in a symbolicway. Wewillusethefollowingconventions in writingcelldiagrams: • Theanode, theelectrode at whichtheoxidationoccurs, is placed at theleftside of thediagram • Thecathode, theelectrode at whichreductionoccurs, is placed at therightside of thediagram. • A boundarybetweendifferentphases(e.g, an electrodeand a solution) is representedby a singleverticalline(/) • Theboundarybetweenhalf-cellcomponents, usually a salt bridge, is representedby a doubleverticalline(//) • Theelectrochemicalcellswhichproduceelectricity as a result of chemicalreactionsarecalledvoltaicorgalvaniccells

22. CELL DIAGRAMS Representing a Redoxreaction Through a Cell Diagram-Aluminium metal displaceszinc(II) ionfromaqueoussolution Write oxidationandreductionhalfequationsand a net equationforthisredoxreaction Write a celldiagramfor a voltaiccell in whichthisreactionoccurs. Solution: a) Theterm « displaces» meanstheAluminiumgoesintosolution as Al3+and Zn2+comesout of thesolution as zinc metal. b) Al(s) is oxidized to Al3+ in the anode half-cell(left) and Zn2+ (aq) is reduced to Zn(s) in the cathode half-cell(right)

23. STANDARD ELECTRODE POTENTIALS To do a calculation of cell voltages, we choose a particular half-cell to which we assign a potential of zero. We then compare other half cells to this reference. The commonly accepted reference is the standard hydrogen electrode

24. STANDARD ELECTRODE POTENTIALS • Standard electrode potential, E˚, measures the tendency for a reduction process to occur at an electrode. • To determine the value of E˚ for an electrode, we compare it with a standard hydrogen electrode(SHE). In the voltaic cell indicated below the measured potential difference is 0,337 V, with electrons flowing from H2 to the Cu electrode.

25. STANDARD ELECTRODE POTENTIALS A standard cell potential, Ecell˚, is the potential difference or voltage of a cell formed from two standard electrodes. The net reaction that occurs in the voltaic cell is The cell reaction indicates that Cu2+ (1M) is more easily reduced than is H+(1M). The standard electrode potential representing the reduction of Cu2+ (aq) to Cu(s) is assigned a positive value.

26. STANDARD ELECTRODE POTENTIALS When a standard hydrogen electrode is combined with a standard zinc electrode, electrons flow in the opposite direction, that is from the zinc to the hydrogen electrode.

27. STANDARD ELECTRODE POTENTIALS Insummary, thepotential of thestandardhydrogenelectrode is set at 0. Anyelectrode at which a reductionhalf-reactionshows a greatertendencytooccurthandoesthereduction of H+(1M) to H2(g) has a positivevalueforitsreductionpotential,E˚. Anyelectrode at which a reductionhalf-reactionshows a lessertendencytooccurthandoesthereaction of H+(1M) to H2(g) has a negativevalueforitsstandardreductionpotantial, E˚. Thevalue of E˚ does not depend on theamount of substancesinvolved. E˚ valuesareunaffectedbymultiplyinghalf-equationsbyconstantcoefficients.

28. SOME STANDARD ELECTRODE(REDUCTION) POTENTIALS AT 25˚C

29. SOME STANDARD ELECTRODE(REDUCTION) POTENTIALS AT 25˚C

30. SOME STANDARD ELECTRODE(REDUCTION) POTENTIALS AT 25˚C

31. CombiningE˚ValuesintoEcell˚ for a Reaction A new battery system currently under study for possible use in electric vehicles is the zinc-chlorine battery. The net reaction producing electricity in this cell is What is theEcell˚ of thisvoltaiccell?

32. Determining an Unknown E˚ fromEcell˚ Measurement Cadmium is found in smallquantitieswhereverzinc is found. Unlikezinc, which in traceamounts is an essential element, cadmium is an environmentalpoison. Todeterminecadmiumionconcentrationsbyelectricalmeasurementsthestandardreductionpotentialforthe Cd2+/Cdelectrode is needed. Thvoltage of thefollowingvoltaiccell is measured: Cd(s) / Cd2+(1 M) // Cu2+(1 M) / Cu(s) Ecell˚ = 0,740 V What is thestandardreductionpotentialforthe Cd2+/Cdelectrode? Solution: =ECd2+/Cd

33. ECELLand SPONTANEOUS CHANGE Determining a Free Energy Change from A Cell Potential. Determine ΔG˚ for the reaction

34. n mol e- = electric current(ampere) x time(s) / 96485 C For the reaction Cu2+(aq) + 2 e- Cu(s) The mass of Cu accumulated on the electrode= n/2 x MA(atomic mass of Cu)

35. SpontaneousChange in OxidationReductionReactions Our main criterion for spontaneous change is that ΔG<0 . However, according to the equation if ΔG< 0, then ΔG< 0 Ecell>0. If Ecell>0 , a reaction occurs spontaneously in the forward direction If Ecell<0, the reaction occurs spontaneously in the reverse direction. If Ecell=0, a reaction is at equilibrium. If a cell reaction is reversed, Ecell changes sign

36. SpontaneousChange in OxidationReductionReactions

37. Making Qualitative Predictions with Electrode Potential Data Solution Because the E˚value for the reduction of S2O82-(aq) is larger than that for the reduction of Cr2O72-(aq) , S2O82-(aq) should be the better oxidizing agent.

38. TheBehavior of MetalsTowardsAcids Example:

39. RELATIONSHIP BETWEEN Ecell˚ and Keq R=8,314 Jmol-1K-1 at 25˚C Example: What is the value of the equilibrium constant Keq for the reaction between copper metal and iron(III) ions in aqueous solution at 25˚C?

40. APPLICATION FIELDS OF Ecell˚ IN COMBINATION WITH Keq AND ΔG

41. ECELL AS A FUNCTION OF CONCENTRATION

42. NERNST EQUATION Example : What is the value of Ecell for the voltaic cell pictured below? Ecell= 0,011 V

43. PredictingSpontaneousReactionsForNonstandardConditions Example: Will the cell reaction proceed spontaneously as written for the following cell? Solution:

44. CONCENTRATION CELLS aq

45. CONCENTRATION CELLS Any voltaic cell in which the net cell reaction involves only a change in the concentration of some species( here H+) is called a concentration cell. A concentration cell consists of two half-cells with identical electrodes but different ion concentrations. Because the electrodes are identical, the standard half-cell potentials are numerically equal and opposite in sign. This makes Ecell=0. However, because the ion concentrations differ, there is a potential difference between the two half-cells. Spontaneous change in a concentration cell always occurs in the direction that produces a more dilute solution

46. CONCENTRATION CELLS Constructing and using a hydrogen electrode is difficult. A better approach is to replace the SHE by a different reference electrode and the other hydrogen electrode by a glass electrode. A glass electrode is a thin glass membrane enclosing HCl(aq) and a silver wire coated with AgCl(s) The potential of the glass electrode depends on the hydrogen ion concentration of the solution being tested. The difference in potential between the glass electrode and the reference electrode is converted to a pH reading by the meter

47. BATTERIES:PRODUCING ELECTRICITY THROUGH CHEMICAL REACTIONS A device that stores chemical energy for later release as electricity is called a battery Primary batteries: The cell reaction is not reversible. When the reactants have been mostly converted to products, no more electricity is produced and the battery is dead. Secondary batteries: The cell reaction can be reversed by passing electricity through the battery(charging). This means that a battery can be used several times by charging. Flow batteries: Materials(Reactants, products, electrolytes) pass through the battery, which is the conversion of chemical energy into electrical energy

48. LECLANCHE(DRY) CELL In this cell oxidation occurs at a zinc anode and reduction at an inert carbon cathode. The electrolyte is a moist paste of MnO2, ZnCl2, NH4Cl and carbon black. Oxidation: Zn(s) Zn2+(aq) +2 e- Reduction: 2 MnO2(s) +H2O + 2e- Mn2O3(s) + 2 OH-(aq) An acid-base reaction occurs between NH4+(aq) + OH-(aq) NH3(g) + H2O(l). The Leclanche cell is a primary cell. It is cheap to make, but it has some drawbacks. When current is rapidly drawn from the cell, products build up on the electrodes(e.g NH3) and this causes the voltage to drop. Also because the electrolyte medium is acidic, zinc metal slowly dissolves. In order to overcome this problem, an alkaline form of this battery can be produced . The advantage of the alkaline battery are that zinc does not dissolve in a base and the battery does a better job in maintaining the voltage.

49. LEAD-ACID STORAGE BATTERY The most common secondary battery is the automobile storage battery (see below) . The electrodes are lead-antimony alloy. The anodes are impregnated with lead metal and the cathodes with red-brown leaddioxide. The electrolyte is dilute sulfuric acid. When the cell is allowed to discharge, following reactions occur: Oxidation: Pb(s) + SO42-(aq) PbSO4(s) + 2 e- Reduction: PbO2(s) + 4 H+(aq) + SO42-(aq)+ 2 e- PbSO4(s) + 2 H2O Net: Pb(s) + PbO2(s)+ 4 H+(aq)+SO42-(aq) 2 PbSO4(s)+2 H2O To recharge it,electrons are forced in the opposite direction by connecting the battery to an external electric source. The reverse of reactions occurs when the battery is recharged

50. BUTTON CELL BATTERIES The cell-diagram of a silver-zinc cell is Zn(s)/,ZnO(s)/ KOH(satd)/Ag2O(s),Ag(s) Oxidation: Zn(s) + 2 OH-(aq) ZnO(s) +H2O+ 2 e- Reduction: Ag2O(s)+H2O+ 2 e- 2 Ag(s)+ 2 OH-(aq) Net: Zn(s) + Ag2O(s) ZnO(s) + 2 Ag(s) Because no solution species is involved in the net reaction, the quantity of the electrode is very small, and the electrodes can be maintained very close together. The storage capacity of a silver-zinc cell is about six times as great as a lead-acid cell of the same size. These batteries are used in watches, electronic calculators, hearing aids and cameras