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Chapter 4 solution basics and electrochemistry. Concentration . Percent composition by mass (Mass of wanted/ mass of solution (whole))x 100% Molarity M = mol of solute/ L of solution m olality m= mol of solute/ kg of solvent Mole Fraction X = mole of desired / mole total . Redox.
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Concentration • Percent composition by mass • (Mass of wanted/ mass of solution (whole))x 100% • Molarity • M = mol of solute/ L of solution • molality • m= mol of solute/ kg of solvent • Mole Fraction • X = mole of desired / mole total
Redox • Oxidation Number and normal half reactions see handout • Remember • OILRIG or LEO the lion goes GER • Red Cat • An Ox • Fat Cat
Voltaic cells are spontaneous Separated into two cells Is a battery Anode is negative Cathode is positive AnOx Red Cat Fat Cat Electrolytic cells are forced to work by using a power source Single container NEEDS a battery Anode is positive Cathode is negative AnOx Red Cat Fat Cat EPA (electrolytic positive anode)
Voltaic cells • E cell = E oxi + E red (all at standard condiditons) • What ever half reaction has the highest (most positive) E value on the reduction potential table is the one that is reduced, the other gets flipped to become oxidized.
When cell is not at standard conditions, use Nernst Equation E = Eo – RT ln Q nF This formula is not given to you…. Or You can explain what is happening with LeChatlier’s principle
Acid Redox • Write separate equations for the oxidation and reduction half-reactions. • For each half-reaction, • Balance all the elements except hydrogen and oxygen • Balance oxygen using H2O • Balance hydrogen using H+ • Balance the charge using electrons • If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions. • Add the half-reaction, and cancel identical species • Check that the elements and charges are balanced
Base Redox • Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ were present • To both sides of the equation obtained above, add a number of OH- ions that is equal to the number of H+ ions. (eliminate H+ by forming H2O) • Form H2O on the side containing both H+ and OH- ions, and eliminate the number of H2O molecules that appear on both sides of the equation. • Check that elements and charges are balanced.