Electrochemistry Ron RobertsonElectrochemistry Slide 1 I. Applications of Redox Reactions A. Terminology Zn (s) + Cu+2(aq) → Zn+2 (aq) + Cu (s) This reaction is redox because electrons have been transferred. Oxidation - loss of electrons, ox # increases, Zn → Zn+2 Reduction - gain of electrons, ox # decreases, Cu+2 → Cu Reducing agent - agent that allows reduction to occur, it is the substance that is oxidized and can provide electrons Oxidizing agent - agent that allows oxidation to occur, it is the substance that is reduced and takes the electrons provided. B. Examples Corrosion Combustion Metabolism Voltaic cells (batteries) ElectrolysisElectrochemistry Slide 2 B. Balancing Redox Reactions Oxidation reduction reactions are sometimes very difficult to balance. Mass, charge, and # of electrons must be balanced. The following method for balancing is slightly different from the book. 1. Write reaction in ionic form and assign oxidation numbers 2. Separate into half cells 3. Balance # of atoms of oxidized and reduced species 4. Balance # of electrons of oxidized and reduced species 5. Make sure electrons lost in the oxidation = electrons gained in the reduction 6. Balance charge in each half cell using H+ if in acid and OH- if basic 7. Balance # of H and O using water molecules.Electrochemistry Slide 3 II. Voltaic cells (Batteries) A. Theory of operation Zn(s) + Cu+2(aq) → Zn+2(aq) + Cu(s) This does not look like a battery because the electrons are being transferred directly between Cu+2 and the Zn strip. If we separate the reactions and force the electrons to travel through a circuit we could use these electrons as they try to get from the Zn to the Cu+2.Electrochemistry Slide 4 The push of the electrons to go from Zn to Cu+2 is called the voltage. Voltage = work/charge V = W/Q Volt [=] Joule/coulomb This push can be measured with a voltmeter. Why do we need the salt bridge? In the figure above the electrons from Zn → Zn+2 will move through the wire and over to the other side to the Cu+2. Almost immediately the flow stops because of a charge buildup on both sides.Electrochemistry Slide 5 Zn+2 ions are accumulating in the left compartment (+) and Cu+2 ions are leaving on the right side (-). We must allow ions to flow to equalize the charge. This is the function of the salt bridge. With the salt bridge the circle is complete. Electrons flow through the wire from Zn to Cu+2 and anions move from the right side to the left and cations move from left side to the right to equalize the charge. This total movement of charge, ions in the internal circuit and electrons in the external, completes the voltaic cell. B. The notation for the cell is M (electrode)M+(solution)N+(solution) N(electrode) anode cathodeElectrochemistry Slide 6 C. Thermodynamics and the Cell Potential Standard Potential - When all reactants and products present are pure solid or in solution at 1.0 M (or gases at 1.0 Atm) the conditions are standard and the measured voltage is called the standard potential E°. This E° is a measure of the tendency of pure reactants to become products, sort of like ∆Gο. There is a relationship between the two: Under standard conditions ∆G° = -nF E° Under any conditions ∆G= -nF E Since spontaneous reactions have a positive E° there must be a negative sign in the proportionality. n represents the number of moles of electrons (also called Faradays) transferred in the balanced equation. F is a conversion between Joules, Volts and moles of electrons. It is called Faraday’s constant. F = 96,500 coulombs/moleElectrochemistry Slide 7 D. Half Cell Potentials We can measure the push of electrons (called the voltage) through the external wire for a voltaic cell. This voltage can be envisioned as a sum of 2 pushes - one for the oxidation and one for the reduction. Ecell = Ered + Eox The tendency for a metal to lose or gain electrons can be measured by these pushes - called the oxidation and reduction potentials. The larger the value the greater the tendency for that reaction to occur - or put another way the more positive the oxidation or reduction potential is the greater the tendency for that oxidation or reduction to occur. oxidation potential (Eox) M+x⇒ M+y + ze- reduction potential (Ered) M+y + ze-⇒ M+x Eox= - EredElectrochemistry Slide 8 The problem is that we cannot measure each individual push so we cannot measure the absolute cell potential. So we define a cell potential to be zero and measure all others against it. This allows the calculation of all other potentials. The actual zero potential that has been agreed upon is the H2 ⇒ 2H+ + 2e-. E. Rules of thumb for using half cell potentials 1. E° values are normally tabulated as reduction potentials 2. If you need to find the oxidation potential simply reverse the algebraic sign of E°. All reactions are reversible 3. The more positive the value of E° is the more likely the process is to occur. Since the reduction process is the agent of oxidation, the elements at the top of the reduction table are the best ox agents and the species at the bottom (for the reverse oxidation reaction) are the best reducing agents.Electrochemistry Slide 9 4. Under standard conditions (1M or 1Atm) any substance will spontaneously oxidize any other substance underneath it in the reduction table. 5. Potentials are intensive properties. Changing the stoichiometric coefficients for a half-rxn does not change the value of E. 6. To predict the spontaneity of a rxn, add E° of the red rxn and E° of the ox rxn together. If E° is positive the rxn is spontaneous under standard conditions. F. Example problems and questions using half cell potentials • Will aluminum dissolve in a solution of Sn+4? • Why do people who have dental fillings feel pain when chewing a piece of aluminum foil? • Predict the voltage of an aluminum/zinc battery. • Which is spontaneous, the reaction of iron with the cupric ion or the reaction of copper metal with ferrous ion?Electrochemistry Slide 10 G. Voltaic cells at nonstandard conditions What if the concentrations are not 1 M for all ions? The Nernst Equation relates the cell potential to the concentrations of the ions. E = E° - [RT/nF] ln Q Q is the reaction quotient, the ratio of product concentrations divided by reactant concentrations. Since this equation is a holdover from the days when