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Chapter 18

Chapter 18. Electrochemistry. Oxidation-Reduction. REDOX reactions are at the heart of all batteries. Take a simple single replacement reaction like: Zn (s) + CuCl 2( aq )  ZnCl 2( aq ) + Cu (s)

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Chapter 18

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  1. Chapter 18 Electrochemistry

  2. Oxidation-Reduction • REDOX reactions are at the heart of all batteries. • Take a simple single replacement reaction like: • Zn(s) + CuCl2(aq) ZnCl2(aq) + Cu(s) • Reducing this to a net ionic equation and then to half-reactions reveals the transfer of the electrons. • Oxidation = loss of electron(s). • Reduction = gain of electron(s).

  3. Oxidation-Reduction • In Chapter 4, the activity series was used to predict the outcome of a single replacement reaction. • LEP #1

  4. Oxidation-Reduction • The substance being oxidized can also be referred to as the reducing agent and the substance being reduced as the oxidizing agent. • 2 Zn(s) + O2(g) 2 ZnO(s) • Co(s) + NiCl2(aq) CoCl2(aq) + Ni(s) • LEP #2

  5. Balancing REDOX Reactions • Reactions are split into half-reactions – • Oxidation half-reaction • Reduction half-reaction • These are then balanced separately. • Electrons are added to show the net change in oxidation number of the REDOX half-reactions. • Each half reaction is then multiplied by the LCM to cancel out the electrons. • LEP #3

  6. Voltaic Cells • In a spontaneous reaction, electrons are transferred from the more active metal to the less active one. • If we can put these in separate compartments, then the electron transfer can be harnessed.

  7. Voltaic Cells • A voltaic cell is one that produces a positive voltage. • A complete pathway must be present for the electrons to travel.

  8. Voltaic Cells • A standard cell looks like this. • Oxidation takes place at the anode. • Reduction takes place at the cathode. • The flow of electrons is from the anode to the cathode.

  9. Voltaic Cells • In a molecular view of the reaction…

  10. Electromotive force (emf) • Water flows spontaneously from higher to lower levels. • Electrons do the same – they flow from higher electron pressure to lower electron pressure.

  11. Emf • This difference in pressure is called the emf – aka the voltage. Note: 1 V = 1 J / 1 C • Emf’s are relative to a position, so you must have a point of reference. • S.H.E. = standard hydrogen electrode • Thus, 2H+(aq) + 2e- H2(g) ;Eocell= 0.00V

  12. Emf • Cell voltages are for 1 M solutions, 1 atm, 25oC • For a Zn / Cu cell (LEP#4): • Zn+2(aq) + 2e- Zn(s) ; Eo = -0.76V • Cu+2(aq) + 2e- Cu(s) ; Eo = +0.34V • Because Zn metal is oxidized, that equation is reversed and its sign is changed. • Or: Eocell= Ecathode- Eanode • LEP #5

  13. Oxidizing / Reducing Agents • If placed in order of most positive to most negative voltages (handout), then top left has strongest oxidizing agent and bottom right has strongest reducing agent.

  14. Free Energy and REDOX • Can use reduction potentials to predict whether a REDOX reaction is spontaneous. • Eo = +, then spontaneous. • Eo = -, then non-spontaneous. • I2(s) + 5 Cu+2(aq) + 6 H2O(l) 2 IO3-(aq) + 5 Cu(s) + 12 H+(aq) • H2SO3(aq) + 2 Mn(s) + 4 H+(aq)  S(s) + 2 Mn+2(aq) + 3 H2O(l)

  15. Free Energy • The relationship between Free Energy (DG) and cell emf is: DGo = -nFE0 ; where n is the net number of electrons transferred (balanced reaction) and F is the Faraday constant 96,485 C/mol. • From Chapter 19, we also know that: DGo = -RT lnK. • Thus, -nFEo = -RT lnK. • And, ln K = nFEo / RT. • The constants can be combined and the original equations used log instead of the natural log. • Log K = nEo / (0.0592). • LEP #6

  16. Nernst Equation • What if concentrations are not 1 M? • What happens to a battery in use over time? • Recall from Chapter 19: • DG = DGo + RT lnQ • Since DG = -nFE, we can replace this for both DG and DGo.

  17. Nernst Equation • Suppose [Cu+2] = 0.10M and [Zn+2] = 0.10M, will we need to use the Nernst Equation? • Net: Zn(s) + Cu+2(aq) Zn+2(aq) + Cu(s) • What if the reaction replaced the Cu with Ag? • Net: Zn(s) + 2 Ag+1(aq) Zn+2(aq) + 2 Ag(s) • LEP #7 and #8

  18. Concentration Cells • All voltaic cells so far involved two different metals. • Can we produce a voltage when the same metal is used for both half-cells? • Yes!

  19. Concentration Cells • Since both half-reactions are the same, but opposite reactions, then Eo = 0.00 V (always!). • Ni+2(aq) + 2e- Ni(s) ; Eored = -0.28V • Ni(s)  Ni+2(aq) + 2e- ; Eoox = +0.28V • LEP #9

  20. Importance of Concentration Cells • Pacemaker cells in the heart are a concentration cell. • Potassium ion concentrations inside and outside of the cell are different – ICF = 135 mM and ECF = 4 mM. • Generates an electrical pulse of about 94 mV. • This can be measured with an EKG.

  21. Electrolysis • Can use electrical energy to force a non-spontaneous reaction to occur. • This is called and electrolytic cell. • Two types of electrolysis: • Aqueous – the ions are present in a solution of water and requires only the energy for the electolysis. • Molten Salt – the ionic compound is heated until it liquefies. This requires energy (lots) before the electrolysis can take place.

  22. Electrolysis • Suppose we have a solution of NaCl(aq) and try to electrolyze this solution. • At the cathode, reduction of the metal is possible: • Na+(aq) + 1e- Na(s) ; Eo = -2.71 V • However, there is a second competing reaction: • 2 H2O(l) + 2e-  H2(g) + 2 OH-(aq) ; Eo = -0.83 V • Only the easier one will occur!!!

  23. Electrolysis • Likewise, at the anode where oxidation occurs, the oxidation of the non-metal is possible: • 2 Cl-(aq) Cl2(g) + 2e- ; Eo = -1.36 V • Once again, there is a competing reaction: • 2 H2O(l)  O2(g) + 4 H+(aq) + 4e- ; Eo = -1.23 V • As before, only the easier one will occur!!! • Therefore, electrolysis on NaCl(aq) will yield:

  24. Electrolysis • The ONLY way to produce sodium metal is with molten salt electrolysis.

  25. Predicting the Products • We will use our electrode potentials to predict the products of an aqueous electrolysis. • Key reactions to highlight are: • Cathode: • 2 H2O(l) + 2e- H2(g) + 2 OH-(aq) ; Eo = -0.83 V • Anode: • 2 H2O(l) O2(g) + 4 H+(aq) + 4e- ; Eo = -1.23 V • LEP #11 and #12

  26. Applications of Electrolysis • Besides producing certain metals there are other uses of electrolysis like electroplating and purifying certain metals.

  27. Quantitative Electrolysis • Current is measured in Amperes or amps for short. • 1 Amp = 1 C / 1 s • wmax = -nFE • Problems are set up using standard dimensional analysis. • Will require a conversion between moles of electrons and moles of metal reduced. • LEP #13 and #14

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