1 9 & 19. ELECTROCHEMISTRY 1

2 Electron Transfer Reactions 1. Electron transfer reactions are redox reactions. 2. Results in the generation of an electric current (electricity) or be caused by imposing an electric current. 3. Therefore, this field of chemistry is often called ELECTROCHEMISTRY.

3 Two industrial applications: I. Voltaic Cells ____ Produces energy from a spontaneous chemical reaction. II. Electrolytic Cells____ Uses energy in order to promote a non spontaneous chemical reaction.

4 Voltaic Cells/ Electrochemical Cells Voltaic Cells/ Electrochemical Cells A device that obtains electrical energy from a spontaneous chemical reaction Batteries are voltaic cells

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6 Terms Used for Voltaic Cells

7 Electrons travel thru external wire. Salt bridge allows anions and cations to move between electrode compartments.Salt bridge allows anions and cations to move between electrode compartments. Electrons travel thru external wire. Salt bridge allows anions and cations to move between electrode compartments.Salt bridge allows anions and cations to move between electrode compartments. Zn --> Zn e- Cu e- --> Cu <--AnionsCations--> OxidationAnodeNegativeOxidationAnodeNegative Reduction Cathode Positive Reduction Cathode Positive RED CAT

8 AN OX chases a RED CAT

9 Zn/Cu Electrochemical Cell Zn(s) ---> Zn 2+ (aq) + 2e-E o = V Cu 2+ (aq) + 2e- ---> Cu(s)E o = V Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) E o = V E o = V Cathode, positive, sink for electrons Anode, negative, source of electrons +

10 Cd --> Cd e- or Cd e- --> Cd Fe --> Fe e- or Fe e- --> Fe E o for a Voltaic Cell All ingredients are present. Which way does reaction proceed?

11 CHEMICAL CHANGE ---> ELECTRIC CURRENT With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” Zn is oxidized and is the reducing agent Zn(s) ---> Zn 2+ (aq) + 2e-Zn is oxidized and is the reducing agent Zn(s) ---> Zn 2+ (aq) + 2e- Cu 2+ is reduced and is the oxidizing agent Cu 2+ (aq) + 2e- ---> Cu(s)Cu 2+ is reduced and is the oxidizing agent Cu 2+ (aq) + 2e- ---> Cu(s)

12 More About Calculating Cell Voltage Assume I - ion can reduce water. 2 H 2 O + 2e- ---> H OH - Cathode 2 I - ---> I 2 + 2e- Anode I H 2 O --> I OH - + H 2 2 H 2 O + 2e- ---> H OH - Cathode 2 I - ---> I 2 + 2e- Anode I H 2 O --> I OH - + H 2 Assuming reaction occurs as written, E˚ = E˚ cat + E˚ an = ( V) - ( V) = V Minus E˚ means rxn. occurs in opposite direction (the connection is backwards or you are recharging the battery)

13 Write the equation for the spontaneous reaction that will occur when a magnesium half cell is connected to na aluminum half cell.

14 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.

15 Dry Cell Battery Anode (-) Zn ---> Zn e- Cathode (+) 2 NH e- ---> 2 NH 3 + H 2

16 Alkaline Battery Nearly same reactions as in common dry cell, but under basic conditions. Anode (-): Zn + 2 OH - ---> ZnO + H 2 O + 2e- Cathode (+): 2 MnO 2 + H 2 O + 2e- ---> Mn 2 O OH -

17 Mercury Battery Anode: Zn is reducing agent under basic conditions Cathode: HgO + H 2 O + 2e- ---> Hg + 2 OH -

18 Lead Storage Battery Anode (-) E o = V Pb + HSO > PbSO 4 + H + + 2e- Cathode (+) E o = V PbO 2 + HSO H + + 2e- ---> PbSO H 2 O

19 Ni-Cad Battery Anode (-) Cd + 2 OH - ---> Cd(OH) 2 + 2e- Cathode (+) NiO(OH) + H 2 O + e- ---> Ni(OH) 2 + OH -

20 H 2 as a Fuel Cars can use electricity generated by H 2 /O 2 fuel cells. H 2 carried in tanks or generated from hydrocarbons

21 Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques. MnO Fe H + ---> Mn Fe H 2 O Mn = +7 Fe = +2 Fe = +3 Mn = +2

22 Balancing Equations Consider the reduction of Ag + ions with copper metal. Cu + Ag + --give--> Cu 2+ + Ag

23 Balancing Equations Step 1:Divide the reaction into half-reactions, one for oxidation and the other for reduction. OxCu ---> Cu 2+ Red Ag + ---> Ag Step 2:Balance each element for mass. Already done in this case. Step 3:Balance each half-reaction for charge by adding electrons. Ox Cu ---> Cu e- Red Ag + + e- ---> Ag

24 Balancing Equations Step 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 ---> Cu e- Oxidizing agent 2 Ag e- ---> 2 Ag Step 5:Add half-reactions to give the overall equation. Cu + 2 Ag + ---> Cu Ag The equation is now balanced for both charge and mass.

25 Balancing Equations Balance the following in acid solution— VO Zn ---> VO 2+ + Zn 2+ VO Zn ---> VO 2+ + Zn 2+ Step 1:Write the half-reactions OxZn ---> Zn 2+ RedVO > VO 2+ Step 2:Balance each half-reaction for mass. OxZn ---> Zn 2+ Red VO > VO 2+ + H 2 O 2 H + + Add H 2 O on O-deficient side and add H + on other side for H-balance.

26 Balancing Equations Step 3:Balance half-reactions for charge. OxZn ---> Zn e- Rede- + 2 H + + VO > VO 2+ + H 2 O Step 4:Multiply by an appropriate factor. OxZn ---> Zn e- Red 2e- + 4 H VO > 2 VO H 2 O Step 5:Add balanced half-reactions Zn + 4 H VO > Zn VO H 2 O

27 Tips on Balancing Equations Never add O 2, O atoms, or O 2- to balance oxygen.Never add O 2, O atoms, or O 2- to balance oxygen. Never add H 2 or H atoms to balance hydrogen.Never add H 2 or H atoms to balance hydrogen. Be sure to write the correct charges on all the ions.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.Check your work at the end to make sure mass and charge are balanced. PRACTICE!PRACTICE!

28 1. Can an acidified aqueous solution of potassium dichromate spontaneously oxidize a solution of bromide ions to bromine?

29 2. Can a solution of tin II ions reduce a solution of iron III ions? If so, are the iron III ions reduced to iron II or to iron metal?

30 3. What will happen when copper I sulfate(s) dissolves in water?

31 II. Electrolysis hp Electrolysis is the situation when redox cells are forced to run in reverse by attaching an electricity source to overcome the potential difference.

32 Salgema-Maceio/Brasil

33 Some applications:

34 a)Electrolysis of molten NaCl: NaCl(l) Species present: Na + and Cl - electrolyte

35 Electrolysis of molten NaCl: NaCl(l) Species present: Na + and Cl - Positive electrode=> oxidation Anode 2 Cl - (l) => Cl 2 (g) + 2e Negative electrode=> reduction Cathode Na + (l) + e => Na(l ) Overall: 2 Cl - (l)+ 2 Na + (l) => Cl 2 (g) + 2 Na(l) electrolyte

36 b)Electrolysis of aqueous NaCl, NaCl(aq) Species: Na +1, H +1, Cl -1, H 2 O This time,H + will be reduced instead of Na + Cathode (-): Reduction 2 H 2 O(l) + 2 e - =>H 2 (g) + 2 OH - (aq) Anode (+): Oxidation 2 Cl - => Cl 2 (g)+ 2 e - Overall: 2 NaCl(aq) + 2 H 2 O(l) => 2 Na + (aq) + 2 OH - (aq) + H 2 (g) + Cl 2 (g)

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38 Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially. Electrolysis of an aqueous NaCl solution has two other advantages: It produces H 2 gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

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40 Electrolysis Al

How electric current is conducted. 7WAhttp:// 7WA In an electrolytic cell, current is conducted by electrons in the wire and by ions in the electrolyte.

42 Hodder Q5

43 II. Electrolytic Cell The ions that are successfully released at the electrodes depend on three factors 1)The position of the ion in the electrochemical series. As a rule of thumb, if the metal appears below hydrogen in the electrochemical series then it will be preferentially deposited. 2)The concentration of the ion in the solution. 3)The nature of the electrode: Platinum, graphite

Deduce the products of the electrolysis of any molten salt, PbBr 2 M&feature=relatedhttp:// M&feature=related Hodder Q1b and Q2

45 The ions that are successfully released at the electrodes depend on three factors 1)The position of the ion in the electrochemical series. As a rule of thumb, if the metal appears below hydrogen in the electrochemical series then it will be preferentially deposited. 2)The concentration of the ion in the solution. 3)The nature of the electrode: Platinum, graphite

46 IB Questions

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48 raday.php Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially. Electrolysis of an aqueous NaCl solution has two other advantages. It produces H 2 gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

49 Solid sodium chloride doesn't conduct electricity, because there are no electrons which are free to move. When it melts, sodium chloride undergoes electrolysis, which involves conduction of electricity because of the movement and discharge of the ions. In the process, sodium and chlorine are produced. This is a chemical change rather than a physical process.

50 Predict the products of electrolysis of strong calcium chloride solution At the cathode Species present Ca2+ and H+. Ca2+ is higher in the reactivity seriers than hydrogen and therefore cannot be released. The reaction is therefore: 2H + (aq) + 2e H 2 (g) At the anode Species present OH - and Cl -. The chloride concentration is strong and so it is preferentially oxidised and the reaction is: 2Cl - (aq) Cl 2(g) + 2e Species remaining in solution: Calcium ions and hydroxide ions

51 Quantities produced by electrolysis Faraday's law states that the mass of product produced will be proportional to the charge passed. The charge (Coulombs)= current (amps) x time (seconds) Q=It F = Q/96500 Faraday's law may also be restated as...the number of faradays required to discharge 1 mol of an ion at an electrode equals the number of charges on that ion. 1 Faraday = Coulombs # Faradays, F = Q / 96500

52 A metallic object to be plated with copper is placed in a solution of CuSO 4. a) To which electrode of a direct current power supply should the object be connected? b) What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours? Solution:

53 A metallic object to be plated with copper is placed in a solution of CuSO 4. a) To which electrode of a direct current power supply should the object be connected? b) What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours? Solution: a)Since Cu 2+ ions are being reduced, the object acts as a cathode and must be connected to the negative terminal (where the electrons come from!) b) The amount of charge passing through the cell is Q = I x t = (0.22 amp) × (5400 sec) = 1200 c F = Q / = (1200 ) ÷ (96500 ) = F Since the reduction of one mole of Cu 2+ ion requires the addition of two moles of electrons, the mass of Cu deposited will be n =0.012/2 = moles m = n x MM = (63.54 g mol –1 ) (0.006) = 0.39 g of copper

54 Example: Calculate the mass of copper released by a current of 10A passing for 200 seconds through a Copper II sulphate solution.

55 Example: Calculate the mass of copper released by a current of 10A passing for 200 seconds through a Copper II sulphate solution. Q = I t =Charge =10 x 200 = 2000 coulombs Number of Faradays = 2000/96500 = 0,0207 Faradays The reaction occurring at the cathode is: Cu 2+ (aq) + 2e - Cu(s) Therefore: 2 Faradays will produce 1 mole of copper 0,0207 Faradays will produce 0,0207/2 moles of copper = 0,0104 moles Therefore the mass of copper produced = 0,0104 x 63,5 = 0,658g

56 Calculate the number of moles of hydrogen released when 5 amps of current passes for 3000 seconds through a solution of sulphuric acid Q = It Q = 5 x 3000 = C Q = 15000/96500 Faradays Q = F

57 Factors affecting the relative amounts of products formed during electrolysis 1. Charge on the ions Na + (l) + e => Na(l) 1 mol of e required to produce 1 mol of Na atoms Cu 2+ (aq) + 2e => Cu(s) 2 mol of e to produce 1 mol Cu atoms 2. Quantity of electrons The quantity of electrons depends on the current and time. If two electrolytic cells are connected in series, the same current will flow through both and for the same length of time.

58 In the electrolysis the hydrogen gas is released at the cathode as follows: 2H + + 2e => H 2 2 moles of electrons release 1 mole of gas 2 Faradays of charge are needed to release 1 mole of hydrogen F releases 0.155/2 moles of hydrogen moles of hydrogen released = moles

59 Electrolysis of CuSO4 (aq) With graphite electrodesWith Cu electrodes

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61 Electroplating Use of electrolysis in electroplating: The object to be electroplated is made the negative electrode, and it is placed in a solution of the ions of the metal used to plate it. Electroplating different from electrolysis in that the metal deposited from electrolysis plates out on the surface of another metal. The electrolyte contains the plating metal in the form of dissolved ions and the anode usually is made of the plating metal. The object to be plated is the cathode.

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