Adam Rosenbloom and Olga Lozovskaya  This one’s for you, Mr. Hinton Chuga Chuga Chuga Chuga Choo Choo!

Oxidation Numbers A number assigned to each element in a compound in order to keep track of the electrons during a reaction The oxidation number of pure elements’ atoms is zero (Br 2 ) For monatomic ions, the oxidation number equals the ion’s charge (K+) Fluorine is always -1 in compounds with other elements. Cl, Br, and I are -1 unless they are with oxygen or fluorine. H is almost always +1 and O is almost always -2. The sum of the oxidation numbers in a neutral compound must be zero; in a polyatomic ion the sum must equal the overall charge of the ion.

Redox Reactions In an oxidation-reduction reaction, there is a transfer of electrons from one species to another. The reactant that is oxidized loses an electron, and its oxidation number increases. The reactant that is reduced gains an electron, and its oxidation number decreases. Example: A + B  A n+ + B n- A is being oxidized and since it donates an electron to B, it is also the reducing agent. B is being reduced and since it accepts an electron from A, it is the oxidizing agent.

Balancing Redox Reactions (Acidic Solution) Ag + (aq) +HCHO (aq)  Ag (s) +HCO 2 H (aq) Step 1: Divide into two half reactions Ag + (aq)  Ag (s) HCHO (aq)  HCO 2 H (aq) Step 2: Balance elements other than H and O. Step 3: Balance O by adding H 2 O. HCHO (aq) + H 2 O  HCO 2 H (aq) Step 4: Balance H by adding H +. HCHO (aq) + H 2 O  HCO 2 H (aq) + 2H + Step 5: Balance charge by adding electrons: Ag + (aq) + e-  Ag (s) HCHO (aq) + H 2 O  HCO 2 H (aq) + 2H + + 2e- Step 6: Make electrons cancel so you can add the half-reactions. 2(Ag + (aq) + e-  Ag (s) ) Step 7: Combine the two half-reactions: 2Ag + + HCHO (aq) + H 2 O  HCO 2 H (aq) + 2H + + Ag (s)

Balancing Redox Reactions (Basic Solution) This is the same process as with an acidic solution, except that Steps 3 and 4 are different: Step 1: Divide into two half reactions Step 2: Balance elements other than H and O. Step 3: Balance O by adding OH -. Step 4: Balance H by adding H 2 O. Step 5: Balance charge by adding electrons. Step 6: Make electrons cancel so you can add the half-reactions. Step 7: Combine the two half-reactions.

Electrochemical Cells A current is generated by immersing metals with different reduction potentials into solutions of their ions connected by a salt bridge. An electrochemical cell produces electric current due to the transfer of electrons from the anode to the cathode. At the negative electron, the cathode, reduction takes place and electrons are deposited. The cathode gains mass. At the positive electron, the anode, oxidation takes places and electrons are given off. The anode loses mass.

Example of a Wet Cell The reduction potential of Zinc (II) is –0.763 V. The reduction potential of Ni is –0.25 V. Since the Ni has the higher reduction potential, it experiences reduction. The half-reaction for zinc is then flipped for it to be oxidized. To get E 0 (the standard electric potential), subtract the substance oxidized from the one that is reduced. ( = V) TO BE CONTINUED…

E 0 continued… The equation for the relationship between E 0 and Gibbs Free Energy is: ΔG 0 rxn = -nFE 0  n is the number of moles of electrons transferred  F is the is the charge of one mole of electrons, which is the Faraday Constant ( x 10 4 Joules / V / mol)\ Also, don’t forget that even if you need to multiply a half-reaction by some factor to balance the number of electrons transferred, DO NOT (upon penalty of death) multiply the E 0 too.

Using the Faraday Constant for Mass- Current Relationships I have no idea what’s going on right now. A current of 2.3 amps is passed through a solution containing potassium ions for 17 minutes. The voltage is such that potassium is deposited at the cathode. What mass, in grams, of potassium is deposited at the cathode? Step 1: Calculate the charge (number of Coulombs) passed in 17 minutes. Charge= current * time Charge=(2.3A)(17 min)(60 sec/min)=2346 C Step 2: Calculate the number of moles of electrons C (1 mol e- / 9.65 *10 4 ) =.0243 mol e- Step 3: Calculate the number of moles of potassium and then the mass of potassium deposited. (.0243 mol e- )(1 mol K/ 1 mol e-)(39.1 g/1 mol K) =.95 g of Potassium

Cells at Non-Standard Conditions Nernst Equation- This equation can help you correct E 0 to be a proper value under non- standard conditions. E = E 0 – (RT / nF) ln (Q), where Q is the reaction quotient and R is J / K / mol When E = 0, that means the reaction is at equilibrium, so E 0 = RT / nF ln (K eq ) I bet you forgot about me… Thank you Marc Potempa and Matt Meshulam for this lovely information.

Good-bye! Thanks, electrochemistry!