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An Introduction to Electrochemistry Redox Reaction

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An Introduction to Electrochemistry Redox Reaction An Introduction to Electrochemistry Redox Reaction ox1 + red2 <=> red1 + ox2 1 Outline 1. Review Concepts: Redox Reaction, Cell, Potential Balance an Equation Nernst Equation 2. Redox Titration 3. Potentiometry 4. Voltammetry and Electrochemical Sensor 12.1 Balancing Oxidation- Reduction Reactions • Mass Balance: the amount of each element present at the beginning of the reaction must be present at the end. • Charge Balance: electrons are not lost in a chemical reaction. Half Reactions • Half-reactions are a convenient way of separating oxidation and reduction reactions. 2 Half Reactions • The half-reactions for Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe3+(aq) are Sn2+(aq) Sn4+(aq) +2e- 2Fe3+(aq) + 2e- 2Fe2+(aq) • Oxidation: electrons are products. • Reduction: electrons are reagents. Balancing Equations by the Method of Half Reactions • Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple). 3 How? 1. Write down the two half reactions. 2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H+. d. Finish by balancing charge by adding electrons. 3. Multiply each half reaction to make the number of electrons equal. 4. Add the reactions and simplify. 5. Check! Balancing Equations by the Method of Half Reactions + - 2- 2+ 16H (aq) + 2MnO4 (aq) + 5C2O4 (aq) 2Mn (aq) + 8H2O(l) + 10CO2(g) 4 Exercise: Try to Balance: Sn+2 + Fe+3 <=> Sn+4 + Fe+2 +2 - +3 +2 Fe + MnO4 <=> Fe + Mn Sn+2 + Fe+3 <=> Sn+4 + Fe+2 5 +2 - +3 +2 Fe + MnO4 <=> Fe + Mn Important Redox Titrants and the Reactions Oxidizing Reagents (Oxidants) (1) Potassium Permanganate + 2+ MnO4 + 8H + 5e Mn + 4H2O + MnO 4 + 4H + 3e MnO2 (s) + 2H2O 2 MnO4 + e MnO4 6 Important Redox Titrants and the Reactions Oxidizing Reagents (Oxidants) (2) Potassium Dichromate 2 + 3+ Cr2O7 +14H + 6e 2Cr + 7H2O 2 4 + + 3+ 2+ Cr2O7 + 3U + 2H 2Cr + 3UO2 + H2O Important Redox Titrants and the Reactions Oxidizing Reagents (Oxidants) (2) Potassium Iodate 1 IO + 6H + + 5e I + 3H O 3 2 2 2 7 Important Redox Titrants and the Reactions Reducing Reagent ( Reductants ) (2) Potassium Iodide 1 I I + e 2 2 Important Redox Titrants and the Reactions Reducing Reagent ( Reductants ) (2) Sodium Thiosulfate 2- 2- 2S2O3 S4O6 +2e 8 12.2 Nernst Equation • Voltaic cells consist of – Anode: Zn(s) Zn2+(aq) + 2e2 – Cathode: Cu2+(aq) + 2e- Cu(s) – Salt bridge (used to complete the electrical circuit): cations move from anode to cathode, anions move from cathode to anode. • The two solid metals are the electrodes (cathode and anode). 9 The salt bridge allows charge transfer through the solution and prevents mixing. Fig. 12.1. Voltaic cell. 10 Galvanic Cells anode cathode oxidation reduction spontaneous redox reaction 19.2 11 • This rearranges to give the Nernst equation: RT E = E° lnQ nF • The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K: 0.0592 E = E° logQ n • (Note that change from natural logarithm to base-10 log.) • Remember that n is number of moles of electrons. Nernst Equation aA + bB + ne- <=> cC + dD 0.0592 [C]c[D]d E = Eo - ---------- log ---------------- at 25oC n [A]a[B]b 0.0592 [Re d] E = E 0 log n [Ox] 12 Standard Electrode Potentials Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Reduction Reaction - + 2e + 2H (1 M) H2 (1 atm) E0 = 0 V Standard hydrogen electrode (SHE) 19.3 Standard Reduction (Half-Cell) Potentials •The SHE is the cathode. It consists of a Pt electrode in a + tube placed in 1 M H solution. H2 is bubbled through the tube. • For the SHE, we assign + - 2H (aq, 1M) + 2e H2(g, 1 atm) • E°red of zero. • The potential of a cell can be calculated from standard reduction potentials: E°cell = E°red (cathode) E°red (anode) 13 Standard Reduction (Half-Cell) Potentials 2+ - • Consider Zn(s) Zn (aq) + 2e . We measure Ecell relative to the SHE (cathode): E°cell = E°red(cathode) - E°Ox(anode) 0.76 V = 0 V - E°red(anode). • Therefore, E°red(anode) = -0.76 V. • Standard reduction potentials must be written as reduction reactions: 2+ - Zn (aq) + 2e Zn(s), E°red = -0.76 V. 14 12.3 Cell Diagram anode//cathode Zn/ZnSO4 (0.1 M)//CuSO4 (0.2 M)/Cu If a cell, Ecell = Ecathode - Eanode if Ecell => positive, spontaneous as written 15 o - The more positive the E , the better oxidizing agent is the oxidized form (e.g., MnO4 ). The more negative the Eo, the better reducing agent is the reduced form (e.g., Zn). ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) 16 12.4 Redox Titration Curve 12.4 Redox Titration Curve 17 12.4 Redox Titration Curve Derivation of a titration curve Fe+2 + Ce+4 <=> Ce+3 + Fe+3 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M +4 Ce in a medium that is 1.0 M in H2SO4. EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 E = Eo - (0.0592/n) log([red]/[ox]) At 0.00 mL of Ce+4 added, initial point no Ce+4 present; minimal, unknown [Fe+3]; thus, insufficient information to calculate E 18 Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with +4 0.1000 M Ce in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe V M (15.00 mL)(0.1000 M) [Fe+3] = ---------------Ce Ce = ---------------------------- VFe + VCe (50.00 + 15.00)mL = 2.308 x 10-2 M EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe [Fe+3] = 2.308 x 10-2 M VFe MFe - VCe MCe [Fe+2] = ------------------------ VFe + VCe 19 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe [Fe+3] = 2.308 x 10-2 M (50.00 mL)(0.0500 M) - (15.00 mL)(0.1000 M) [Fe+2] = -------------------------------------------------------- (50.00 + 15.00)mL = 1.54 x 10-2 M EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe [Fe+3] = 2.308 x 10-2 M[Fe+2] = 1.54 x 10-2 M for Fe+2 -> Fe+3 Eo = 0.69 v 0.0592 [Fe+2] o Ecell = Ecell - ----------- log ----------- n [Fe+3] 20 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe [Fe+3] = 2.308 x 10-2 M[Fe+2] = 1.54 x 10-2 M for Fe+2 -> Fe+3 Eo = 0.69 v 0.0592 1.54 x 10-2 Ecell = 0.68 v - ---------- log ----------------- v = 0.69 v 1 2.31 x 10-2 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 At 25.00 mL of Ce+4 added, VFeMFe = VCeMCe, equivalence point o o nFe EFe + nCe ECe 1(1.44) + 1(0.68) Ecell = ----------------------- = ------------------------ v nFe + nCe 1 + 1 = 1.06 v 21 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 26.00 mL of Ce added, VCeMCe > VFeMFe V M (50.00 mL)(0.0500 M) [Ce+3] = ----------------Fe Fe = ------------------------------- VFe + VCe (50.00 + 26.00)mL = 3.29 x 10-2 M 22 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 26.00 mL of Ce added, VCeMCe > VFeMFe [Ce+3] = 3.29 x 10-2 M VCe MCe - VFe MFe [Ce+4] = --------------------------- VFe + VCe EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 26.00 mL of Ce added, VCeMCe > VFeMFe [Ce+3] = 3.29 x 10-2 M (26.00 mL)(0.1000 M) - (50.00 mL)(0.0500 M) [Ce+4] = ------------------------------------------------------- (50.00 + 26.00)mL = 1.32 x 10-3 M 23 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.
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