An informative exploration by JC and Petey B.

Oxidation Numbers All oxidation and reduction reactions involve the transfer of electrons between substances. 2 Ag + (aq) + Cu (s)  2 Ag (s) + Cu 2+ (aq) Ag + accepts electrons from Cu and is reduced to Ag; Ag + is the oxidizing agent. Cu donates electrons to Ag + and is oxidized to Cu 2+ ; Cu is the reducing agent. Losing electrons means oxidation. Gaining electrons means reduction.

What are oxidation numbers? The oxidation number of an atom in a molecule is defined as the electric charge an atom has. Example: Al 3+ –Al 3+ has an oxidation number of +3. But how do we do this with more complicated molecules?!

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: Cr 2 O 7 2- 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.

Balancing Redox Reactions: An example on with acid. C 2 H 5 OH (aq) + Cr 2 O 7 2- (aq)  CH 3 CO 2 H (aq) + Cr 3+ (aq) First, identify what is being oxidized and reduced: –Cr (+6  +3); Cr 2 O 7 2- is being reduced. –C (-2  0); C 2 H 5 OH is being oxidized. Find the two half reactions: –C 2 H 5 OH  CH 3 CO 2 H –Cr 2 O 7 2-  Cr 3+ Then, balance the half reactions for mass: –C 2 H 5 OH + H 2 O  CH 3 CO 2 H + 4 H + –14 H + + Cr 2 O 7 2-  2 Cr H 2 O Now, balance the half reactions for charge: –C 2 H 5 OH + H 2 O  CH 3 CO 2 H + 4 H e - –6 e H + + Cr 2 O 7 2-  2 Cr H 2 O

We then take the half reactions and multiply them by the appropriate factors to make the number of electrons on each side equal: –3 [C 2 H 5 OH + H 2 O  CH 3 CO 2 H + 4 H e - ] –2 [6 e H + + Cr 2 O 7 2-  2 Cr H 2 O] Add the two balanced half reactions: –3 C 2 H 5 OH + 3 H 2 O + 12 e H Cr 2 O 7 2-  – 3 CH 3 CO 2 H + 12 H e Cr H 2 O Eliminate commons reactants and products: –3 C 2 H 5 OH (aq) + 16 H + (aq) + 2 Cr 2 O 7 2- (aq)  CH 3 CO 2 H (aq) + 4 Cr 3+ (aq) + 11 H 2 O (l) That was a pretty basic example. Here’s an example that’s even more basic!

The “Basic” Concept Given: SnO 2 2- (aq)  SnO 3 2- (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 H 2 O on the opposite side. –So: 2 OH - + SnO 2 2-  SnO H 2 O Next, balance for charge: –2 OH - + SnO 2 2-  SnO H 2 O + 2 e -

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. –M–M–M–Move on to the next one. Or else.

How it works Zn (s) + Cu 2+ (aq)  Zn 2+ (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, SO 4 2- 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.

An Electro-Demo

All About Potential The standard potential E o 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: –∆G o rxn = -nFE o where n=number of moles of electrons transferred in a balanced redox reaction and f is the faraday constant, 9.65x10 4

Calculating Cell Potential Given: the cell illustrated has a potential of: E O =+0.51 V at 25 o C. The net ionic equation is: Zn (s) + Ni 2+ (aq, 1M)  Zn 2+ (aq, 1M) + Ni (s) What is the value of E o for the half-cell : Ni 2+ (aq) + 2e -  Ni (s) ? Solution: For the anode, Zn, we know the potential is V from the table of standard reduction potentials. Note, we had to change the sign from the table because our reaction is: Zn (s)  Zn 2+ (aq) + 2e -. Since the net reaction is the sum of the half reactions: E o net =E o Zn +E o Ni Therefore, E o Ni = = -0.25V Ta da!

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 E o is the one that will occur as on oxidation. Changing the stoichiometric coefficients for a half- reaction does not change the value of E o.

Non-standard conditions E=E o – (RT/nF)ln(Q) –Q= Reaction quotient Q=[products]/[reactants] –F= Faraday constant 9.65x10 4 joules/(volts*mole) –R= Gas constant joules/(K*mole)

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 - )/(9.65x10 4 C))= mols e - –Finally, calculate mass of silver deposited: ( mols e - ) ((1 mol Ag)/(1 mol e - )) ((107.9 g Ag)/(1 mol Ag))= 1.51 g Ag

Credits Special thanks to: The video camera

The book The still camera

The video capture box Delicious water (H 2 O (l) )

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

The End The End –B–B–B–But to be continued??? No, probably not