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An informative exploration by JC and Petey B.

Unit 5: Electrochemistry. An informative exploration by JC and Petey B. 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.

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An informative exploration by JC and Petey B.

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  1. Unit 5: Electrochemistry An informative exploration by JC and Petey B.

  2. 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.

  3. 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?!

  4. 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 O’s. (-2)*7 + (x)*2 = -2 Therefore, x = +6.

  5. 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

  6. 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. Here’s an example that’s even more basic!

  7. 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-

  8. 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.

  9. 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.

  10. An Electro-Demo

  11. 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 n=number of moles of electrons transferred in a balanced redox reaction and f is the faraday constant, 9.65x104

  12. Calculating Cell Potential Given: the cell illustrated has a potential of: EO=+0.51 V at 25 oC. The net ionic equation is: Zn(s) + Ni2+(aq, 1M) Zn2+(aq, 1M) + Ni(s) What is the value of Eofor 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: Eonet=EoZn+EoNi Therefore, EoNi = 0.51-0.76= -0.25V Ta da!

  13. 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.

  14. Non-standard conditions • E=Eo – (RT/nF)ln(Q) • Q= Reaction quotient • Q=[products]/[reactants] • F= Faraday constant • 9.65x104 joules/(volts*mole) • R= Gas constant • 8.315 joules/(K*mole)

  15. 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: • Q=I*t=(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

  16. Credits Special thanks to: • The video camera

  17. The book • The still camera

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

  19. And of course… • The Skipster himself Our grades just dropped. A lot. Oh well.

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

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