Transcript Slide 1
Electrochemistry 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). Oxidation-Reduction In Chapter 4, the activity series was used to predict the outcome of a single replacement reaction. LEP #1 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 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 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. Voltaic Cells A voltaic cell is one that produces a positive voltage. A complete pathway must be present for the electrons to travel. 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. Voltaic Cells In a molecular view of the reaction… 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. Emf This difference in pressure is called the emf – aka the voltage. Note: 1 V =1J/1C Emf’s are relative to a position, so you must have a point of reference. S.H.E. = standard hydrogen electrode Thus, 2H+(aq) + 2eH2(g) ; Eocell = 0.00V 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 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. 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) 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 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. 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 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! 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 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. 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. 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!!! 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: Electrolysis The ONLY way to produce sodium metal is with molten salt electrolysis. 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 Applications of Electrolysis Besides producing certain metals there are other uses of electrolysis like electroplating and purifying certain metals. 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