OXIDATIONREDUCTION REACTIONS involve electron transfer

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OXIDATION-REDUCTION REACTIONS involve electron
transfer

• Terms to Know
• LEO the Lion goes GER!
• Losing Electrons is Oxidation, Gaining Electrons
  is Reduction
• OIL RIG oxidation is loss, reduction is gain
  (of electrons)
• Oxidation the loss of electrons, increase in
  charge
• Reduction the gain of electrons, reduction of
  charge
• Oxidation number the assigned charge on an atom
  
• Oxidizing agent (OA) the species that is
  reduced and thus causes oxidation
• Reducing agent (RA) the species that is
  oxidized and thus causes reduction

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• Rules for Assigning Oxidation Statesyou know
  most of this already!
• The oxidation state of an atom in an element is
  ZERO including allotropes i.e. N2, P4, S8.
• 2. The oxidation state of a monatomic ion is the
  same as its charge.
• 3. In its compounds, fluorine is always assigned
  an oxidation state of -1.
• 4. Oxygen is usually assigned an oxidation state
  of -2 in its covalent compounds, such as CO,
  CO2, SO2, and SO3. Exceptions to this rule
  include peroxides (compounds containing the O22-
  group), where each oxygen is assigned an
  oxidation state of -1, as in hydrogen peroxide
  (H2O2), and OF2 in which oxygen is assigned a 2
  oxidation state.
• 5. In its covalent compounds with nonmetals,
  hydrogen is assigned an oxidation state of 1.
  Metal hydrides are an exception H is at the end
  of the chemical formula since it has an
  oxidation state of 1-.
• 6. The sum of the oxidation states must be zero
  for an electrically neutral compound. For a
  polyatomic ion, the sum of the oxidation states
  must equal the charge of the ion.

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Exercise 16

• Assign oxidation states to all atoms in the
  following.
• a. CO2
• b. SF6
• c. NO3-

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Exercise 17

• When powdered aluminum metal is mixed with
  pulverized iodine crystals and a drop of water is
  added to help the reaction get started, the
  resulting reaction produces a great deal of
  energy. The mixture bursts into flames, and a
  purple smoke of I2 vapor is produced from the
  excess iodine. The equation for the reaction is
• 2A1(s) 3I2(s) ? 2A1I3(s)
•  
• For this reaction, identify the atoms that are
  oxidized and reduced, and specify the oxidizing
  and reducing agents.

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• Ex. 18 Metallurgy, the process of producing a
  metal from its ore, always involves
  oxidation-reduction reactions. In the metallurgy
  of galena (PbS), the principal lead-containing
  ore, the first step is the conversion of lead
  sulfide to its oxide (a process called roasting)
• 2PbS(s) 3O2(g) ? 2PbO(s) 2SO2(g)
•  
•  
• The oxide is then treated with carbon monoxide to
  produce the free metal
•  
• PbO(s) CO(g) ? Pb(s) CO2(g)
•  
• For each reaction, identify the atoms that are
  oxidized and reduced, and specify the oxidizing
  and reducing agents.

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• Balancing Redox Reactions by Half Reaction Method
• 1) Divide the equation into oxidation and
  reduction half reactions. OILRIG
• 2) Balance all elements besides hydrogen and
  oxygen.
• 3) Balance Os by adding H2Os to the appropriate
  side of each equation.
• 4) Balance Hs by adding H
• 5) Balance the charge by adding electrons.
  OILRIG again
• 6) Multiply the half reactions to make electrons
  equal for both half-reactions.
• 7) Cancel out any common terms and recombine the
  two half reactions.
• 8) IF BASIC, neutralize any H by adding the
  SAME NUMBER of OH- to EACH side of the balanced
  equation. This creates some waters that will
  cancel!
• 9) CHECK!!

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• Sample Problem Assign oxidation states to all
  atoms in the following equation, identify the
  oxidation and reduction half reactions, and the
  OA and RA and balance (in acidic).
• MnO4-(aq) Fe2(aq) ? Mn2(aq) Fe3(aq)

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• Sample Problem Balance the following equation
  using the half-reaction method.
• (acidic) MnO4-(aq) I-(aq) ? Mn2(aq)
  I2(aq)

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• (basic) Ag(s) CN- O2 ? Ag(CN)2-(aq)

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• Exercise 19
• Potassium dichromate (K2Cr2O7) is a bright orange
  compound that can be reduced to a blue-violet
  solution of Cr3 ions. Under certain conditions,
  K2Cr2O7 reacts with ethyl alcohol (C2H5OH) as
  follows
•  
• H(aq) Cr2O72-(aq) C2H5OH(l) ? Cr3(aq)
  CO2(g) H2O(l)
•  
• Balance this equation using the half-reaction
  method. (in acidic solution)

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Trivia Question

• According to College Board, how many AP classes
  are offered?
• 34!!!!!!!
• Art History, Biology, Calculus (AB BC),
  Chemistry, Chinese Language and Culture, Computer
  Science (A AB), Macroeconomics, Microeconomics,
  English Language, English Literature,
  Environmental Science, European History, French
  Language, French Literature, German Language,
  Comparative Government Politics, U.S.
  Government Politics, Human Geography, Italian
  Language and Culture, Japanese Language and
  Culture, Latin Literature, Latin Vergil, Music
  Theory, Physics (B C), Psychology, Spanish
  Language, Spanish Literature, Statistics, Studio
  Art, U.S. History, World History

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Two types of electrochemical cells

• Galvanic (voltaic) cells which are spontaneous
  chemical reactions (battery)
• Electrolytic cells which are non-spontaneous
  and require external e- source
• BOTH of these fit into the category entitled
  Electrochemical cells

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Parts of Galvanic Cell

• Anode--the electrode where oxidation occurs.
  After a period of time, the anode may appear to
  become smaller as it falls into solution. (An
  Ox)
• Cathode-- the anode where reduction occurs.
  After a period of time it may appear larger, due
  to ions from solution plating onto it. (Red Cat)
• Electrons flow from anode to cathode

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Other Parts of Galvanic Cell

• inert electrodesused when a gas is involved OR
  ion to ion involved such as Fe3 being reduced to
  Fe2 rather than Fe0. Made of Pt or graphite.
• Salt bridge -- a device used to maintain
  electrical neutrality in a galvanic cell. This
  may be filled with agar which contains a neutral
  salt or it may be replaced with a porous cup.
• Standard cell notation (line notation) -
  anode/solution// cathode solution/ cathode Ex.
  Zn/Zn2 (1.0 M) // Cu2 (1.0M) / Cu
• Voltmeter - measures the cell potential (emf) .
  Usually is measured in volts.

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Standard Reduction Potentials

• Reading the reduction potential chart
• ? elements that have the most positive reduction
  potentials are easily reduced (in general,
  non-metals)
• ? elements that have the least positive
  reduction potentials are easily oxidized (in
  general, metals)

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Calculation E of a cell

• Decide which element is oxidized or reduced using
  the table of reduction potentials. Remember
  THE MORE POSITIVE REDUCTION POTENITAL GETS TO BE
  REDUCED.
• Write both equations AS IS from the chart with
  their voltages.
• Reverse the equation that will be oxidized and
  change the sign of the voltage this is now
  E?oxidation
• Balance the two half reactions do not multiply
  voltage values
• Add the two half reactions and the voltages
  together.
• E?cell E?oxidation E?reduction ? means
  standard conditions 1atm, 1M, 25?C

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• a. Consider a galvanic cell based on the reaction
• Al3(aq) Mg(s) ? Al(s) Mg2(aq)
• Give the balanced cell reaction and calculate E
  for the cell.

A 0.71 V
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• A galvanic cell is based on the reaction
• MnO4-(aq) H(aq) ClO3-(aq) ? ClO4-(aq)
  Mn2(aq) H2O(l)
•  
• Give the balanced cell reaction and calculate E
  for the cell.

0.32 V
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• Standard cell notation (line notation) Ion
  sandwich in alphabetical order
• Anode metal/anode ion//cathode ion//Cathode metal
• For Reaction M N ? N M
• Anode Cathode
  (alphabetical order!)
• M(electrode)M (solution)N(solution)N(electro
  de)
• - indicates phase boundary
• - indicates salt bridge
•  
• EX Zn / Zn2 (1.0M)// Cu2 (1.0M) / Cu

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• Sample Problem Calculate the cell voltage for
  the following reaction. Draw a diagram of the
  galvanic cell for the reaction and label
  completely.
• Fe3(aq) Cu(s) ? Cu2(aq) Fe2(aq)

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• Exercise 2
• Calculate the cell voltage for the galvanic cell
  that would utilize silver metal and involve iron
  (II) ion and iron (III) ion. Draw a diagram of
  the galvanic cell for the reaction and label
  completely.

Ecell 0.03 V
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CELL POTENTIAL, ELECTRICAL WORK FREE ENERGY

• The work that can be accomplished when electrons
  are transferred through a wire depends on the
  push or emf which is defined in terms of a
  potential difference in volts between two
  points in the circuit
• Ø      ? - w q
• q Moles of electrons x charge of one mole of
  electrons (Faradays constant)
• When a cell produces a current, the cell
  potential is positive and the current can be used
  to do work THEREFORE ? and work have opposite
  signs!

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?Go -nFEo

• G Gibbs free energy (must be for spontaneous
  process)
• n number of moles of electrons, from balanced
  redox equation
• F 1 Faraday or 96,485 J/V ? mol (sometimes
  rounded off to 96,500)
•  
• So it follows that
• -Eo implies nonspontaneous.
• Eo implies spontaneous (would be a good
  battery!)

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• Using the table of standard reduction potentials,
  calculate ?G for the reaction
• Cu2(aq) Fe(s) ? Cu(s) Fe2(aq)
•  
• Is this reaction spontaneous?

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• Using the table of standard reduction potentials,
  predict whether 1 M HNO3 will dissolve gold metal
  to form a 1 M Au3 solution.

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DEPENDENCE OF CELL POTENTIAL ON CONCENTRATION

• Ecell reaction assumes that all concentrations
  are 1.0 M.
• Voltaic cells at NONstandard conditions
  LeChatliers principle can be applied.
• An increase in the concentration of a reactant
  will favor the forward reaction and the cell
  potential will increase.
• An increase in the concentration of a product
  will favor the reverse reaction and the cell
  potential will decrease.

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• For the cell reaction
• 2Al(s) 3Mn2(aq) ? 2Al3(aq) 3Mn(s)
• Ecell 0.48 V
•  
• predict whether Ecell is larger or smaller than
  Ecell for the following cases.
•  
• a. Al3 2.0 M, Mn2 1.0 M
• b. Al3 1.0 M, Mn2 3.0 M

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• When cell is not at standard conditions (not 1.0
  M and not 25ºC), use Nernst Equation
•  
• RT
• E Eo - ------- ln Q
• nF
•  
• R Gas constant 8.315 J/K? mol
• F Faraday constant (96,485)
• Q reaction quotient productscoefficient/reac
  tantscoefficient
• E Energy produced by reaction
• T Temperature in Kelvins (add 273!)
• n of electrons exchanged in BALANCED redox
  equation

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Nernst Equation at 25º C

• E cell Eºcell 0.0591 log (Q)
• n

Remember Q is reaction quotient If 2A 3B ? D
4C and all are in solution
Q DC4 A2B3
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Nernst Equation

• Determine Eocell and Ecell based on the
  following half reactions
• VO2 2H e- ? VO2 H2O E 1.00
• Zn2 2e- ? Zn E -0.76V
•  
• Where
• T 25C
• VO2 2.0 M
• H 0.50 M
• VO2 1.0 x 10-2 M
• Zn2 1.0 x 10-1 M

Eocell 1.76 V Ecell 1.89 V
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Cell potential at Equilibrium

• When the cell is at equilibrium, there is no
  longer a push of electrons
• the Ecell is 0
• Now the Nernst equation is
• 0 Eºcell 0.0591 log K
• n

K is the equilibrium constant same formula as Q
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Calculating Equilibrium Constant K Ex. 8

• For the oxidation-reduction reaction
•  
• S4O62-(aq) Cr2(aq) ? Cr3(aq) S2O32-(aq)
•  
• The appropriate half-reactions are
•  S4O62- 2e- ? 2S2O32- E 0.17V
  Cr3 e- ? Cr2 E -0.50 V
  
•  
• Balance the redox reaction, and calculate E and
  the equilibrium constant K (at 25C).

E 0.67 V
K 1022.6 4.0 x 1022
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CONCENTRATION CELLS

• We can construct a cell where both compartments
  contain the same components BUT at different
  concentrations.
• Silver will be deposited on the right electrode,
  thus lowering the concentration of Ag in the
  right compartment. In the left compartment the
  silver electrode dissolves producing Ag ions
  to raise the concentration of Ag in solution.

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Exercise 6

• Determine the direction of electron flow and
  designate the anode and cathode for the cell
  represented here

Fe2 will be formed on left side anode
(oxidized) Fe (s) will be formed on right side
cathode (reduced)
FATCAT e flow from left to right
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Electrolytic cells NON spontaneous cells

• Used to separate ores or plate out metals.
• Important differences between a voltaic/galvanic
  cell and an electrolytic cell
• 1) Voltaic (galvanic) cells are spontaneous and
  electrolytic cells are forced to occur by using
  an electron pump or battery or any DC source.
• 2) A voltaic cell is separated into two half
  cells to generate electricity an electrolytic
  cell occurs in a single container.
• 3) A voltaic or galvanic cell IS a battery, an
  electrolytic cell NEEDS a battery
• 4) AN OX and RED CAT still apply BUT the polarity
  of the electrodes is reversed. The cathode is
  Negative and the anode is Positive. Electrons
  still flow FATCAT.
• Usually use inert electrodes

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Electrolytic Stoichiometry

• Used to determine how much chemical change occurs
  with the flow of a given current over a certain
  amount of time
• Example copper plating Cu2(aq) 2e- ? Cu
  (s)
• How much copper is plated when a current of 10.0
  amps (1 amp 1 coul/sec) is passed through
  copper solution for 30.0 mins?
• Steps needed to solve
• Current and time ?quantity of charge ? moles of
  electrons ? moles of copper ? g of copper

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• Step 1 calculate the amount of charge in
  Coulombs
• 10.0 amps 10.0 C/s x number of sec coulombs

18,000 C
Step 2 convert coulombs of charge to moles of
electrons using Faradays constant (96,485 C/mol
e-)
0.187 mol e-
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• Step 3 Convert mol e- to mol copper

0.187 mol e- x 1 mol Cu 0.0935 mol Cu
2 mol e-
Step 4 convert mol Cu to g Cu
0.0935 mol Cu x 63.546 g Cu 5.94 g Cu
1 mol Cu
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From your free response ?s

• In an electrolytic cell, Cu is produced by the
  electrolysis of CuSO4. Calculate the maximum
  mass of Cu that can be deposited by a direct
  current of 100. amperes passed through 5.00 L of
  2.00 M CuSO4 for a period of 1.00 hour.
• 119 g Cu

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• How long must a current of 5.00 A be applied to a
  solution of Ag to produce 10.5 g silver metal?

1,880 secs or 31.3 min
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Trivia Question

• What was the claim to fame of the Honda Accord in
  2008 (besides being the prestigious ride of your
  chemistry teacher)?
• It was the most stolen car in the U.S.

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Trivia Question Part 2

• How many wins does Nicks fantasy football team
  have?
• ZERO!
• http//www.youtube.com/watch?vTX3UqY8KZpU

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Reduction Potentials
The more positive the reduction potential, the
stronger the oxidizing agent. Reduction
potentials are like an activity of metals chart
in reverse. The more active metals (Group 1 and
2), the lower the reduction potential. The less
active, precious metals have larger E
values. Also, the greater the reduction
potential, the easier it will be to plate the
metal using electrolysis the lower voltage
needed during electrolysis.
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• An acidic solution contains the ions Ce4 , VO2
  , and Fe3 . Using the E values, give the order
  of oxidizing ability of these species and predict
  which one will be reduced at the cathode of an
  electrolytic cell at the lowest voltage.

Ce4 has greatest standard reduction potential,
followed by VO2 , and Fe3 . Therefore, Ce4
will be reduced at the lowest voltage.