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  1. Updates • Assignment 07 is due Fri., March 30 (in class) • Prepare well for the final exam; a good score can compensate for low midterm marks!

  2. Free Energy and Equilibrium Note: at equilibrium, there is no net change in [reactants] and [products] so ΔG = 0. This is true for any reaction. To compare free energies of different reactions, it is useful to tabulate ΔGo values under identical conditions (1 M or 1 atm), and we call these conditions standard conditions (ΔGo). Standard conditions does not imply the reactions are at equilibrium; therefore ΔGo will be positive or negative and indicative of whether a reaction is spontaneous. We can use this number to calculate the actual ΔG under any set of conditions. Under standard conditions, ΔG = ΔGo Under equilibrium conditions,ΔG = 0 Under nonequilibrium, nonstandard conditions, ΔG = something other than 0 or ΔGo Under any conditions, the free energy change can be found this way: G = G + RT lnQ (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out and G = G)

  3. Free Energy and Equilibrium • At equilibrium, Q = K, and G = 0. • The equation becomes 0 = G + RT lnK • Rearranging, this becomes G = RT lnK or, K = eG/RT

  4. Redox Reactions and Electrochemistry Chapter 19

  5. Oxidation and Reduction • What is reduced is the oxidizing agent. • H+ oxidizes Zn by taking electrons from it. • What is oxidized is the reducing agent. • Zn reduces H+ by giving it electrons.

  6. Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. • Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 • In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 • The oxidation number of oxygen is usually–2. In H2O2 and O22- it is –1. 19.1

  7. Oxidation numbers of all the atoms in HCO3- ? • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. • Group IA metals are +1, IIA metals are +2 and fluorine is always –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. HCO3- O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4 19.1

  8. Fe2+ + Cr2O72- Fe3+ + Cr3+ +2 +3 Fe2+ Fe3+ +6 +3 Cr2O72- Cr3+ Cr2O72- 2Cr3+ Balancing Redox Equations The oxidation of Fe2+ to Fe3+ by Cr2O72- (where dichromate ions are reduced to Cr3+ ions) in acid solution? • Write the unbalanced equation for the reaction in ionic form. • Separate the equation into two half-reactions. Oxidation: Reduction: • Balance the atoms other than O and H in each half-reaction. 19.1

  9. Fe2+ Fe3+ + 1e- 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O Cr2O72- 2Cr3+ + 7H2O Balancing Redox Equations • For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. • Add electrons to one side of each half-reaction to balance the charges on the half-reaction. • If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 19.1

  10. 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O Balancing Redox Equations • Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. Oxidation: Reduction: • Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 • For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. 19.1

  11. Step 2: We write the incomplete, unbalanced half-reactions: Step 3: We balance each half-reaction as if it took place in acidic solution: SAMPLE EXERCISE Balancing Redox Equations in Basic Solution Complete and balance this equation for a redox reaction that takes place in basic solution: Solution Solve:Step 1: We assign oxidation states. This is a tricky one! Mn goes from 7+ to 4+. The sum of the oxidation states of C and N in CN– must be –1, the overall charge of the ion. In CNO–, if oxygen has an oxidation state of –2 as usual, the sum of the oxidation states of C and N must be +1. So, overall, CN– is oxidized by two electrons.

  12. Now we need to take into account that the reaction occurs in basic solution, adding OH– to both sides of both half-reactions to neutralize H+: We now “neutralize” H+ and OH– by forming H2O when they are on the same side of either half-reaction: Next, we cancel water molecules that appear as both reactants and products: SAMPLE EXERCISE continued Both half-reactions are now balanced—you can check the atoms and the overall charge.

  13. Step 4: Now we multiply the cyanide half-reaction through by 3, which will give six electrons on the product side; and multiply the permanganate half-reaction through by 2, which will give six electrons on the reactant side: Steps 6 and 7: Check that the atoms and charges are balanced. There are 3 C, 3 N, 2 H, 9 O, 2 Mn, and a charge of 5– on both sides of the equation. Step 5: Now we can add the two half-reactions together and simplify by canceling species that appear as both reactants and products: SAMPLE EXERCISE continued

  14. Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

  15. Voltaic Cells • We can use that energy to do work if we make the electrons flow through an external device. • We call such a setup a voltaic cell.

  16. Voltaic Cells • A typical cell looks like this. • The oxidation occurs at the anode. • The reduction occurs at the cathode.

  17. Voltaic Cells Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

  18. Voltaic Cells • Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. • Cations move toward the cathode. • Anions move toward the anode.

  19. Voltaic Cells • In the cell, then, electrons leave the anode and flow through the wire to the cathode. • As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

  20. Voltaic Cells • As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. • The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

  21. Electromotive Force (emf) • Water only spontaneously flows one way in a waterfall. • Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

  22. Electromotive Force (emf) • The potential difference between the anode and cathode in a cell is called the electromotive force (emf). • It is also called the cell potential, and is designated Ecell.

  23. Cell Potential J C 1 V = 1 Cell potential is measured in volts (V).

  24. Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated.