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Chapter 21 Electrochemistry

Chapter 21 Electrochemistry. AP Chemistry Milam. Overview. In physics electricity typically deals with electrons flowing through metals, the flow of charge is called current

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Chapter 21 Electrochemistry

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  1. Chapter 21 Electrochemistry AP Chemistry Milam

  2. Overview • In physics electricity typically deals with electrons flowing through metals, the flow of charge is called current • In chemistry we look at what causes electrons to flow, construction of batteries and which chemicals will work best for a specific function such as a battery • Some concepts from both will be present and some unique to chemistry will be there as well

  3. Overview • An electrolytic cell is one where external energy is used to cause a nonspontaneous chemical reaction to occur • A voltaic cell is where a spontaneous chemical reaction (usually redox) produces electricity and can supply it to something external

  4. 21-1 Electrical Conduction • As mentioned earlier, charge is the key thing that must move for electrical currents. In physics charge can be an electron, and in chemistry we will see electron flow, as well as ion flow. • Both of these occur in typical cells (batteries)

  5. 21-2 Electrodes • Electrodes are surfaces where reduction or oxidation reactions occur • In a battery, a redox reaction occurs, but typically the reduction/cathode is separated from the oxidation/anode by a wire and a salt solution called a salt bridge • The separation allows for electron and ion flow between the two ½ reactions giving us current

  6. 21-2 • The cathode has the reduction occur, so it gains electrons and likewise the anode (oxidation) loses electrons. Therefore it only makes sense that electrons will flow from the anode to the cathode. • The directionality of electron flow is typically 1 point on an electrochemistry question on the AP exams.

  7. 21-3 The Electrolysis of Molten Sodium Chloride (The Downs Cell) • This is a good example to show the basics of redox reactions as well as provide a good visual example • Here we have molten NaCl and by running electrical current through it, we can produce sodium metal and chlorine gas • Na+ + e- Na (l)

  8. 21-3 • Since the sodium ion picks up an electrons or Gains an Electron it is Reduced (GER) • 2Cl- 2e-+Cl2(g) • Since the chloride Loses Electrons it is Oxidized (LEO) • Always remember that LEO the lion goes GER

  9. 21-3 • To combine the two ½ reactions, you need to have the same number of electrons so that they cancel, thus we multiply the Na ½ reaction by 2 • 2x [Na+ + e- Na (l)] • + 2Cl- 2e-+Cl2(g) • 2Na+ + 2Cl- 2Na(l) +Cl2(g) • Since Na is reduced, it is formed at the cathode and Chlorine is formed at the anode

  10. 21-4 The Electrolysis of Aqueous Sodium Chloride • This reaction is similar, but occurs in solution rather than molten sodium chloride • The redox reaction is the same, but after Na(s) is produced, it immediately begins to react with water to make Na+ and H2 and OH-

  11. 21-5 The Electrolysis of Aqueous Sodium Sulfate • This is actually just an electrolysis of water to produce hydrogen and oxygen gas • While it appears to be a simple reaction, it might be a good example to analyze by splitting up the net reaction into ½ reactions • H2O(l)  H2(g) + O2(g) • The hydrogen in water is a +1 oxidation state, it goes to 0 so it is reduced and oxygen is therefore oxidized

  12. 21-5 • 2H2O  O2 + 4H+ + 4e- • We can easily now see the loss of electrons for this oxidation ½ reaction • 2H2O + 2e-  H2 + 2OH- • Additionally we can see the hydrogen is reduced and gains electrons • To combine these reactions, we would multiply the 2nd reaction by 2, and the OH- and H+ would combine to form H2O and cancel with 4 waters from the reactant side

  13. 21-6 Counting Electrons: Coulometry and Faraday’s Law of Electrolysis • In physics, current is measured in Amps, or a Couloumb/second, a Couloumb (C) is a measure of charge • In chemistry we can use Amps to determine how much electrolysis will occur, similar to stoichiometry, using ½ reactions with the # of electrons as mol ratios

  14. 21-6 • A mol of electrons is called a Faraday and has 96, 485 C of charge. • Therefore we can convert between time and amount (mass/mol/volume/particles) or material by using time, amps, Faraday’s constant, # of electrons and the ratios between # of electrons and amount of substance

  15. 21-7 Commercial Applications of Electrolytic Cells • Plating of metals is very common, but also expensive so typically it is only done as a surface and sometimes very thin surfaces. The chromium surface of bumpers is only 0.0002 mm thick. • How many atoms thick is that?

  16. 21-8 The Construction of Simple Voltaic Cells • Voltaic (galvanic) cells have spontaneous redox reactions occuring, separated into two halves. The half-cells are separated and connected by a wire and a salt bridge to maintain neutrality of charge and allow charge to flow and be extracted. • A standard cell has all solutions as 1M and all gases at 1atm

  17. 21-9 The Zinc-Copper Cell • In this cell we have a zinc electrode in contact with 1M Zinc Sulfate solution and the other ½ has a copper electrode in 1M copper sulfate solution connected to the zinc by a wire and a salt bridge • The reaction is: • Cu+2(aq) + Zn(s)  Cu(s) + Zn+2 (aq)

  18. 21-9 • If a statement like the electrode loses mass, or gains mass is said, this tells you whether reduction or oxidation is occurring at that electrode. If metal is plating the electrode to make it gain mass, then it is more than likely reduction and the cathode • In the zinc copper cell the copper electrode will gain mass and the current flows from the zinc electrode to the copper electrode via the wire

  19. 21-9 • The equation for this cell can be simplified to Zn|Zn+2 (1M) ||Cu+2 (1M)|Cu • The Double line in the middle indicates a salt bridge, the single lines separate the electrodes and the ions being reduced or oxidized on those electrodes as well as the concentrations. • The anode is written on the left

  20. 21-9 • If you drop a piece of metal zinc into copper sulfate solution, the exact same reaction will occur. The point of the salt bridge and the separate ½ reactions is to cause a flow of electrons from one electrode to another. It is the separation that allows us to capture the electrical potential energy in these chemicals.

  21. 21-10 The Copper-Silver Cell • This is another specific example of a cell, we won’t get into details on this until after we go through some thermodynamics and how to find cell potentials (voltages) • There is a reference here back to activity series of metals for single displacement reactions way back in chapter 4, this is how that list is developed.

  22. 21-11 The Standard Hydrogen Electrode • We cannot determine the potential of a single electrode, both reduction and oxidation occur at the same time so you cannot just measure one, you can only measure one relative to another • For this reason, we set the standard hydrogen electrode (SHE) to be 0 Volts • From there we can compare electrodes to this one and have a set of values to use

  23. 21-11 • The SHE converts protons (H+) into hydrogen gas (H2) and vice versa. Both reactions are assigned a potential of 0 V

  24. 21-12 The Zinc-SHE Cell • This means that we can compare a zinc electrode (Zn|Zn+2 (1M)) in conjunction with the SHE • This produces a cell with a potential (voltage) of 0.763 V and we can therefore assign this voltage to zinc • The voltage of a cell is determined by adding the voltage of the anode and cathode

  25. 21-12 • Since the SHE is 0V and the total was 0.763V, the zinc electrode must also be 0.763V. • Zn+2 (aq)+ 2e- Zn(s) Eored = -0.763V • Zn(s)  Zn+2 (aq)+ 2e-Eoox = +0.763V • A spontaneous cell will have a + Ecell and so in this case the zinc metal will oxidize and the hydrogen will be reduced to produce hydrogen gas

  26. 21-13 The Copper-SHE Cell • Likewise with copper, we can arrange a cell and measure the potential difference • Cu+2(aq) + 2e- Cu(s) Eored = +.337 V • Cu(s)  Cu+2(aq) + 2e- Eoox = -.337 V • The reverse will always produce the same potential with the opposite sign

  27. 21-14 Standard Electrode Potentials • Now we can use the SHE against all kinds of electrode and end up with a list of potentials of all electrodes, in addition we can now compare 2 electrodes where neither is a SHE • If the electrode has a + Eored then it will undergo reduction with the SHE and if it is – then it will undergo oxidation with the SHE

  28. 21-14 • The more positive the value of Eored, the more likely that substance is to undergo reduction at the cathode in a galvanic cell or the better an oxidizing agent it will be • The reverse is true with large negative values for Eored

  29. 21-15 Uses of Standard Electrode Potentials • The uses of these potentials includes determining which direction in a cell will occur spontaneously as well as determining the potential of that given cell. • The more positive Eored will be the reaction at the cathode being reduced in the spontaneous reaction, with zinc and copper we will see copper reduced

  30. 21-15 • Cu+2(aq) + Zn(s)  Cu(s) + Zn+2 (aq) • Cu+2(aq) + 2e- Cu(s) Eored = +.337 V • Zn(s)  Zn+2 (aq)+ 2e-Eoox = +0.763V • Then we can also calculate the Eocell • Eocell = Eored +Eoox • Eocell = +.337V + (+0.763V) = 1.100 V • So we will get 1.100 V and electrons will flow from the Zn electrode to the Cu electrode

  31. 21-16 Standard Electrode Potentials for Other Half-Reactions • A summary of everything you should know by now: • Labeling oxidation, reduction, oxidizing agents, reducing agents, cathode, anode, direction of flow of electrons • Calculating the potential of a cell given the net reaction • Going from ½ reactions to a full balanced reaction and vice versa

  32. 21-16 • Making calculations involving Faraday’s Constant 96485 C and amount of substance deposited during electrolysis • Predicting which metal will be reduced in a galvanic cell based on potentials • Determine spontaneity based on overall cell potential • Describe metals as good oxidizing or reducing agents based on reduction potentials

  33. 21-16 • Standard conditions (1M, 1atm, 25 °C) • Describe how cell potentials can be determined using a SHE and also know that potentials are relative and arbitrarily assigned • Be familiar with common examples such as decomposition of water, Zn/Cu cells and so forth.

  34. 21-17 Corrosion • Corrosion occurs when metals react with oxygen, typically with moisture present • Corrosion is a big problem because a metal surface becomes ionic where it can crumble and not have the protection that metals offer • For iron the reaction is 4Fe(s) + 3O2 (g) + xH2O (l)  Fe2O3*xH2O (s)

  35. 21-18 Corrosion Protection • To protect from corrosion there are several things you can do, interestingly you can add a layer of metal on the surface of the iron, and the metal can either be a more reactive or less reactive metal. • The less reactive metal will do nothing, but if it breaks, the iron will corrode very quickly because of the subsequent reaction available

  36. 21-18 • Zinc is more reactive than iron, so galvanized steel is often used. The zinc will not rust, and if the iron is exposed, the iron will not react with oxygen until there is no longer any zinc metal in contact.

  37. 21-19 The Nernst Equation • The Nernst Equation is used for cells that do not have standard amounts, for example if I create a battery where the concentrations are much higher than 1M • E = Eo – 0.0592/n * logQ • E is the new potential, n is the number of electrons transferred, Q is the reaction quotient (from equilibrium), Eo is the standard reduction potential

  38. 21-20 Using Electrochemical Cells to Determine Concentrations • The Nernst Equation can be used to calculate potentials using concentrations, or if the potentials are measured, we can measure concentration. • Common examples of this include pH meters which use voltage to determine pH in a solution. • You do not need to know about saturated calomel electrodes not glass electrodes for pH meters

  39. 21-20 • The only thing you need to be able to do with the Nernst Equation is the calculations of concentration and potential

  40. 21-21 The Relationship of Eocell to ΔGo and K • Eocell = RTlnK/(nF) • lnK = nFEocell/(RT) • R is 8.314 J/(mol*K), n is the # of electrons transferred, F is Faraday’s constant (9.65x104 J) • If you know the cell potential you can calculate the equilibrium constant and vice versa

  41. 21-21 • It’s important to be able to calculate the cell potential and equilibrium constant, but it’s also important to be able to analyze relationships between Eocell, ΔGo, and K • For example, if ΔGo is -, which means the reaction is spontaneous in the forward direction, what type of ΔEo would also provide a spontaneous forward reaction?

  42. 21-22 Dry Cells – 21-25 The Hydrogen-Oxygen Fuel Cell • These sections go into specific types of batteries and show how some can be recharged and others cannot. If you’re curious about what chemicals are in car batteries or Duracell’s read on. It won’t be on the AP test and we won’t have the time to cover it.

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