Chapter 20Electrochemistry• This chapter deviates from any other recent material to cover oxidation-reduction reactions and electrochemistry. • The main feature of these reactions is that electrons are transferred between compounds that result in the change in their oxidation state.Overview:• Review – Oxidation and reduction– Calculating oxidation states and identifying redox reactions– Balancing redox reactions• Introduce voltaic cells– Define standard reduction potential– Calculate the cell potential as: Ecell= Ereduction− Eoxidation– Relate Gibb’s free energy to cell potential: ΔG = −nFE– Use the Nernst Equation for calculating cell potential at non-standard conditions: Ecell= E° − (RT/nF)ln(Q)• Define and apply electrolysis– Basically forcing a non-spontaneous reaction to proceed by adding electricity.Definitions:• There are two reactions that take place at the same time. They are oxidation and reduction.Oxidation represents the loss of electrons from a atom.Reduction represents the gain of electrons by a atom.• These reaction always go in pairs since the electrons given up by one compound must be gained by some other compound.– They are called oxidation-reduction reactions or “redox” for short.Definitions• Another mnemonic besides LEO goes GER is: OIL RIG:– Oxidation Involves Loss– Reduction Involves GainExamples• The following are oxidation reactions for the first chemical listed in the reaction:Na(s) + H2O(l) = Na++ OH−+ ½H2(g)2Fe(s) + 3/2O2(g) = Fe2O3(s)CH4(g) + O2(g) = CO2(g) + 2H2O(g)• The following are reduction reactions for the first chemical listed in the reaction:NO + 5/2H2= NH3+ H2OCO2+ H2O    C6H12O6(glucose)Fe2O3+ 2Al = Fe(s) + Al2O3Oxidation States• Oxidation reactions that change the charge of an ion are easy to identify, but many redox reactions are a little more difficult to identify whether electrons have been transferred.• The means of determining whether a redox reaction has occurred uses the oxidation state (or oxidation number) of the atoms of interest in a molecule.• If the oxidation state has changed, then a redox reaction has occurred.Oxidation States• The rules for calculating oxidation states of an atom are as follows. – The rules are ranked in order of precedence, to higher number rules take precedence over lower number rules for assigning oxidation numbers to an atom.– Note that oxidation states are calculated for individual atoms and not as whole molecules.– Skipping steps will come back to haunt you…Oxidation States#1) The oxidation state of an atom in its elemental form is always 0.– The oxygens in O2have a value of 0.– The carbons in graphite have a value of 0.– Pure metals have a value of zero.Elements with multiple forms (e.g. Cgraphitevs Cdiamond) still have a value of zero for both forms.Oxidation States#2) The oxidation state of a monoatomic ion is equal to its charge. A monoatomic ion is a single atom that is in an ionic form.The oxidation state of F−is −1.The oxidation state of Al3+is +3.The oxidation state of Ca2+is +2.Atoms in an ionic substance (e.g. Fe2O3) count because they are individual atoms held together by ionic charges.Molecular ions (where the atoms are covalently bound together) do NOT count as monoatomic. (e.g. NO3−, SO42−, HCO3−, PO43−, etc.)Oxidation States#3) The oxidation state of oxygen in molecules is generally −2 (except for peroxides, O22−, that have a state of −1 each)– The oxygens in the following molecules have an oxidation state of −2: H2O, CH3OH, ClO3.– The following oxygens have an oxidation state of −1: H2O2; R−O−O−H; R−O−O·• Peroxides are two oxygens connected by a single bond. The remaining bond on each oxygen can attach to other atoms.– Oxygen in its molecular form, namely O2, has a value of 0 by the first rule, which takes precedence over this rule.Oxidation States#4) The oxidation state of hydrogen is +1 when bonded to non-metals (electronegativity greater than H) and −1 when associated with metals (electronegativity less than H).The following have a +1 state: CH4, H2SO4, NH3, HClThe following have a −1 state: LiH, NiH2Hydrogen is almost always in the +1 state. The metal hydrides (where the hydrogen is present in the H−form) are very rare and very reactive. You see them in batteries (nickel metal hydride, NiMH).Oxidation States#5) The oxidation state of fluorine is −1 in all compounds. Other halogens generally have an oxidation state of −1 except for oxyanions and fluoro-halogens.The oxygen rule preempts the halogen rule.The halogen atoms in the following substances has a −1 oxidation state each: CH3Cl, CBr4, HCl, PCl3, CaF2.The halogens in the following substances do not have a value of −1. These are examples of the oxyanions: ClO4−, ClO−, BrO3−, IO4−, etc.The value of the halogens in their elemental form (F2, Cl2, Br2, I2) is 0 based on the first rule.Oxidation States#6) The sum of the oxidation states of a molecule must sum to the net charge on the molecule. In neutral molecules, the oxidation states must sum to zero.In ionic molecules, the oxidation states must sum to the charge on the molecule.Note: that oxidation states are not actual charges, so they are a book-keeping item like formal charges in Lewis diagrams.Oxidation State Summary1) Element rule: – All atoms in their elemental form have an oxidation state of 0.2) Monoatomic ion rule:– Monoatomic ions have an oxidation state equal to their charge3) Oxygen rule: – Oxygen generally has a −2 state in compounds except peroxides (which are −1).4) Hydrogen rule: – Hydrogen is +1 when bonded to non-metals and −1 when associated with metals.5) Halogen rule: – Fluorine is always −1 in compounds. Other halogens are generally −1 (but oxyanions are an exception).6) The oxidation states of the atoms in a molecule must sum to the charge on the molecule.Examples:What is the oxidation state of:The sulfur in H2SO4The phosphorus in PH3The carbon in CH4A chlorine atom in CCl4The bromine in BrOThe nitrogen in NOThe nitrogen in NH3The chlorine in FClA hydrogen atom in NiH2The iodine in IO4The phosphorus in H3PO4+2 for the hydrogens, −8 for the oxygens, so S = +6+3 for the hydrogens, so P = −3+4 for the hydrogens, so C = −4−1 (each) by halogen rule−2 for oxygen, must sum to −1, so Br = +1−2 for oxygen, so N = +2+3 for the hydrogens, so N = −3Fluorine always