An informative exploration by JC and Petey B.

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An informative exploration by JC and Petey B.
Unit 5 Electrochemistry
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Oxidation Numbers
All oxidation and reduction reactions involve the
transfer of electrons between substances.
Ag accepts electrons from Cu and is reduced to
Ag Ag is the oxidizing agent.
2 Ag(aq) Cu(s) ? 2 Ag(s) Cu2(aq)
Cu donates electrons to Ag and is oxidized to
Cu2 Cu is the reducing agent.
Losing electrons means oxidation. Gaining
electrons means reduction.
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What are oxidation numbers?

• The oxidation number of an atom in a molecule is
  defined as the electric charge an atom has.
• Example Al3
• Al3 has an oxidation number of 3.
• But how do we do this with more complicated
  molecules?!

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Tips for determining oxidation numbers

• Each atom in a pure element has an oxidation
  number of zero.
• For ions consisting of a single atom the
  oxidation number is equal to the charge of the
  ion.
• Fluorine is always -1 in compounds with other
  elements.
• Cl, Br, and I are always -1 in compounds except
  when combined with O or F.
• The oxidation number of H is 1 and of O is -2.
• The algebraic sum of the oxidation number in a
  neutral compound must be zero in an ion, the sum
  must be equal to the overall ion charge.

Example Cr2O72-
First, recognize that the net charge must be
-2. Then, assign an oxidation number of -2 to the
Os. (-2)7 (x)2 -2 Therefore, x 6.
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Balancing Redox Reactions An example on with
acid.

• C2H5OH(aq) Cr2O72-(aq) ? CH3CO2H(aq) Cr3(aq)
• First, identify what is being oxidized and
  reduced
• Cr (6?3) Cr2O72- is being reduced.
• C (-2 ?0) C2H5OH is being oxidized.
• Find the two half reactions
• C2H5OH ? CH3CO2H
• Cr2O72- ? Cr3
• Then, balance the half reactions for mass
• C2H5OH H2O ? CH3CO2H 4 H
• 14 H Cr2O72- ? 2 Cr3 7 H2O
• Now, balance the half reactions for charge
• C2H5OH H2O ? CH3CO2H 4 H 4 e-
• 6 e- 14 H Cr2O72- ? 2 Cr3 7 H2O

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• We then take the half reactions and multiply them
  by the appropriate factors to make the number of
  electrons on each side equal
• 3 C2H5OH H2O ? CH3CO2H 4 H 4 e-
• 2 6 e- 14 H Cr2O72- ? 2 Cr3 7 H2O
• Add the two balanced half reactions
• 3 C2H5OH 3 H2O 12 e- 28 H 2 Cr2O72- ?
• 3 CH3CO2H 12 H 12 e- 4 Cr3
  14 H2O
• Eliminate commons reactants and products
• 3 C2H5OH(aq) 16 H(aq) 2 Cr2O72-(aq) ?
  CH3CO2H(aq) 4 Cr3(aq) 11 H2O(l)

That was a pretty basic example. Heres an
example thats even more basic!
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The Basic Concept

• Given SnO22-(aq) ? SnO32-(aq)
• First, note the change in oxidation number of Sn,
  from 2 to 4.
• Separate into half reactions (this one is already
  done).
• Balance for mass
• Since the left side is deficient in oxyigen, add
  the oxygen-rich OH-.
• For every two OH-s we use, we need one H2O on
  the opposite side.
• So 2 OH- SnO22- ? SnO32- H2O
• Next, balance for charge
• 2 OH- SnO22- ? SnO32- H2O 2 e-

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Electrochemical Cells

• Electrochemical cells are very closely related to
  oxidation-reduction reactions because the
  transfer of electrons results in a potential
  difference.
• The rest of this slide is empty.
• Move on to the next one.
• Or else.

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How it works

• Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
• Electrons flow through the wire from the Zn
  electrode (anode) to the Cu electrode (cathode).
• A salt bridge provides a connection between the
  half-cells for ion flow thus, SO42- ions flow
  from the copper to the zinc compartment.
• Electrons always flow from the anode to the
  cathode.
• (mnemonic alphabetical order)
• Oxidation takes place at the anode, reduction at
  the cathode.
• AnOx, RedCat
• Ions flow through the salt bridge in the opposite
  direction of the electrons.

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An Electro-Demo
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All About Potential

• The standard potential Eo is a quantitative
  measure of the tendency of the reactants in their
  standard states to proceed to products in their
  standard states.
• Free energy is associated with the same
  characteristics, and is defined as
• ?Gorxn -nFEo
• where nnumber of moles of electrons transferred
  in a balanced redox reaction and f is the faraday
  constant, 9.65x104

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Calculating Cell Potential
Given the cell illustrated has a potential of
EO0.51 V at 25 oC. The net ionic equation
is Zn(s) Ni2(aq, 1M) ? Zn2(aq, 1M)
Ni(s) What is the value of Eo for the half-cell
Ni2(aq) 2e- ? Ni(s) ?
Solution For the anode, Zn, we know the
potential is 0.76 V from the table of standard
reduction potentials. Note, we had to change the
sign from the table because our reaction is
Zn(s) ? Zn2(aq) 2e-. Since the net reaction is
the sum of the half reactions EonetEoZnEoNi
Therefore, EoNi 0.51-0.76 -0.25V
Ta da!
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A note about using the table of standard
reduction potentials

• All potentials listed are for reduction
  reactions the sign must be switched for
  oxidation.
• All half reactions are reversible.
• The more positive the value of the reduction
  potential, the reaction as written is more likely
  to occur as a reduction. Given two half
  reactions, the one with the more positive Eo is
  the one that will occur as on oxidation.
• Changing the stoichiometric coefficients for a
  half-reaction does not change the value of Eo.

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Non-standard conditions

• EEo (RT/nF)ln(Q)
• Q Reaction quotient
• Qproducts/reactants
• F Faraday constant
• 9.65x104 joules/(voltsmole)
• R Gas constant
• 8.315 joules/(Kmole)

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Mass ?? Current

• Current I (amperes, A)
• charge (coulombs, C) / time (seconds, s)
• Example
• A current of 1.50 A is passed through a solution
  containing silver ions for 15.0 minutes. The
  voltage is such that silver is deposited at the
  cathode. What mass of silver is deposited?
• Ag(aq) e- ? Ag(s
• Calculate the charge passed in 15.0 minutes
• QIt(1.5A)(15.0 min)(60.0 sec/min)1350 C
• Next, calculate the number of moles of electrons
• (1350 C) ((1 mol e-)/(9.65x104 C))0.0140 mols e-
• Finally, calculate mass of silver deposited
• (0.0140 mols e-) ((1 mol Ag)/(1 mol e-)) ((107.9
  g Ag)/(1 mol Ag))
• 1.51 g Ag

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Credits

• Special thanks to
• The video camera

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• The book
• The still camera

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• The video capture box
• Delicious water (H2O(l))

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And of course

• The Skipster himself

Our grades just dropped. A lot. Oh well.
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The End

• The End
• But to be continued???
• No, probably not