5.111 Lecture Summary #26

26.1 5.111 Lecture Summary #26 Oxidation/Reduction Lecture 3 and Nernst Equation Topics: Adding and Subtracting Half-Cell Reactions, Chapter 12 From Friday’s handout Standard States and Cell Potentials ∆G°cell = -nℑ∆E°cell ∆E°cell = cell potential (cell voltage) in which products and reactants are in their standard states Units for ∆E is volts. Example - Calculate ∆E°cell for Zn (s) | Zn2+ (aq) || Cu2+ (aq) |Cu (s) anode Zn (s) ⇒ Zn2+ (aq) + 2e- ⇒ Zn (s) (oxidation) cathode Cu2+ (aq) + 2e- ⇒ Cu (s) ∆E°(cell) = standard reduction potential for the couple at cathode (reduction) minus standard reduction potential for the couple at anode ∆E°(cell) = E°(cathode) - E° (anode) Look up Standard REDUCTION Potentials (E°) in back of book (measured against S.H.E) Zn2+ (aq) + 2e- ⇒ Zn (s) Cu2+ (aq) + 2e- ⇒ Cu (s) E° = -0.7628 volts E° = +0.3402 volts ∆E°(cell) = E°(cathode) - E° (anode) E° (Cu2+/Cu(s)) - E°(Zn2+/Zn(s)) = 0.3402 - (-0.7628 ) = 1.103 volts Is the flow of electrons spontaneous? ∆G°cell = -nℑ∆E°cell So, if ∆E°cell is positive, ∆G°cell will be negative. Is a reaction spontaneous when ∆G° is negative? 26.2 Galvanic Cell is an electrochemical cell in which a ____________________ chemical reaction is used to generate an electric current. Electrolytic Cell uses electrical energy provided by an external circuit to carry out ________________________ reactions. Summary Whether the cell operates spontaneously can be determined by ∆Ecell. ((+) = spontaneous) ∆Ecell can be calculated from the Standard Reduction Potentials (E°) of half-cell reactions. Meaning of standard reduction potential E° A large positive E° means the element or compound is easy to reduce ex. F2 (g) + 2e- ⇒ 2F- E°=+2.87 volts (easy to add electrons to F2) positive E°, negative ∆G°, favorable Is F2 a good oxidizing agent? A large positive E° means the oxidized species of the couple is very oxidizing. A large negative E° means the element or compound is hard to reduce ex. Li+1 + e- ⇒ Li (s) E°=-3.045 volts (hard to add electrons to Li+1) negative E°, positive ∆G°, not favorable Is Li+1 a good oxidizing agent? A large negative E° means the reduced species of the couple is very reducing. Example: What is ∆E° for the cell reaction: 2Fe3+ (aq) + 2I- (aq) ⇒ 2Fe2+ (aq) + I2 (s)? balanced reaction at the cathode: balanced reaction at the anode: 26.3 Standard Reduction Potentials are: E°(Fe3+/Fe2+) = +0.770 V E° (I2/I-) = +0.535 V ∆E°(cell) = E°(cathode) - E° (anode) = Is the reaction spontaneous? Which is the better oxidizing agent: Fe3+, I2? Which is the better reducing agent: I-, Fe2+? Question: Vitamin B12 has a large negative reduction potential, so how is it reduced in the body? Vitamin B12 needs to be reduced to be active. Proper functioning of an enzyme that requires vitamin B12 and folic acid is thought to be necessary for preventing heart disease and birth defects. Where do you get vitamin B12 and folic acid in your diet? and how is the vitamin B12 reduced? Today’s material Adding and Subtracting Half-Cell Reactions to Calculate ∆E° for a New Half-Cell Reaction What if you need to know E° for the half-cell reaction: Cu2+ (aq) + e- ⇒ Cu+ (aq), but it is not available in the Table in the book? However, values of E° for other reactions involving Cu are available. One can add or subtract half-cell reactions with known E° to form the new half-cell reaction: Half-cell reactions: Cu2+ (aq) + 2e- ⇒ Cu (s) Cu (s) ⇒ Cu+ (aq) + eCu2+ (aq) + e- ⇒ Cu+ (aq) Standard Reduction Potentials are: E° (Cu2+/Cu(s)) = +0.340 V E° (Cu+/Cu(s)) = +0.522 V 26.4 and calculate the E° for that new half-cell reaction as follows: ∆G°new = ∆G°reduction - ∆G°oxidation or -n3ℑE°3 (new) = -n1ℑE°1(reduction) + n2ℑE°2(oxidation) E°3 = n1E°1 (reduction) - n2E°2 (oxidation) n3 E°3 = ( 2 )(0.340V) - (1)(0.522V) (1) = 0.158 V = E° (Cu2+/Cu+) NERNST EQUATION An exhausted battery is a sign that the cell reaction has reached equilibrium. At equilibrium, a cell generates zero potential difference across its electrodes. To understand this, we need to know how the cell potential changes with cell composition. What do we know already about equilibrium and the components of a reaction? We know that ∆G changes as the concentrations of the components change until equilibrium is reached, then ∆G = 0. ∆G = ∆G° + RT ln Q What do we know about the relationship between ∆G° and ∆E°? ∆G° = -nℑ∆E° Combining: -nℑ∆E = -nℑ∆E° + RT ln Q Dividing by -nℑ: ∆E = ∆E° - RT ln Q nℑ (NERNST EQ.) Example: Calculate ∆E (the cell potential, potential difference, emf) at 25 °C of a cell in which the concentration of Zn2+ ions is 0.10 M and Cu2+ is 0.0010 M. Cu2+ (aq) + Zn (s) ⇒ Zn2+ (aq) + Cu (s) 26.5 Step 1: Calculate ∆E °(Cell) from the E° for the half-reactions. Standard Reduction Potentials Cu2+ (aq) + 2e- ⇒ Cu (s) E° (Cu2+/Cu(s)) = +0.340 V 2+ Zn (s) ⇒ Zn (aq) + 2e E° (Zn2+/Zn(s)) = -0.7628 V ∆E°(cell) = E°(cathode) - E° (anode) = Step 2: Calculate Q for Cu2+ (aq) + Zn (s) ⇒ Zn2+ (aq) + Cu (s) Q= Step 3: Find n Step 4: Use the Nernst Eq. ∆E = ∆E° - RT ln Q nℑ ∆E = 1.103 V - (8.315 J K-1mol-1)(298 K) ln (1.0 x 102) (2)(96485 Cmol-1) ∆E = 1.103 V - 0.0592 = +1.044 V 1 JC-1 = 1 V NOTE: UNITS and CONSTANTS At 25.00°C , RT ℑ = If you use log instead of ln (8.3145 J K-1mol-1)(298.15 K) (96485 Cmol-1) ∆E = ∆E° - RT log Q nℑ ∆E = ∆E° - 0.025693 V ln Q n OR = 0.025693 V Use (2.303)(0.025693) = 0.0592 V ∆E = ∆E° - 0.0592 V log Q n 26.6 What about at EQUILIBRIUM? Q=? ∆G = ? ∆G = ∆G° + RT ln Q ∆G° = - RT ln K ∆G° = -nℑ∆E° Combining: -RT ln K = -nℑ∆E° OR: ln K = nℑ∆E° RT Can calculate K from standard potentials! Now the answer to the biochemical question How is vitamin B12 reduced in the body? Vitamin B12 is reduced by a protein called flavodoxin. E° for vitamin B12 is -0.526 V E° for flavodoxin is -0.230 V Is the reduction of vitamin B12 by flavodoxin spontaneous? ∆E°(cell) = E°(reduction) - E° (oxidation) = E° (vitamin B12) - E° (flavodoxin) = -0.526 V -(-0.230 V) = -0.296 V ∆G° = -nℑ∆E° = -(1)(96485 Cmol-1)(-0.296 V) = +28.6 kJ/mol Vitamin B12 is a better reducing agent than flavodoxin. Vitamin B12 should reduce flavodoxin not the other way around. So why don't we all have heart disease and megaloblastic anemia? Answer: S-adenosylmethionine provides the energy to drive the reaction. The ∆G° for the cleavage of S-adenosylmethionine is -37.6 kJ/mol