Redox Reaction - the basics ox + red <=> red + ox Ch.14-16 1 2 1 2 Oxidizing Reducing Electrochemistry Agent Agent Redox reactions: involve transfer of electrons from one species to another. Oxidizing agent (oxidant): takes electrons Reducing agent (reductant): gives electrons Redox Reaction - the basics Balance Redox Reactions (Half Reactions) Reduced Oxidized 1. Write down the (two half) reactions. ox1 + red2 <=> red1 + ox2 2. Balance the (half) reactions (Mass and Charge): a. Start with elements other than H and O. Oxidizing Reducing Agent Agent b. Balance O by adding water. c. balance H by adding H+. Redox reactions: involve transfer of electrons from one d. Balancing charge by adding electrons. species to another. (3. Multiply each half reaction to make the number of Oxidizing agent (oxidant): takes electrons electrons equal. Reducing agent (reductant): gives electrons 4. Add the reactions and simplify.) Fe3+ + V2+ → Fe2+ + V3 + Example: Balance the two half reactions and redox Important Redox Titrants and the Reactions reaction equation of the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple). Oxidizing Reagents (Oxidants) - 2- 2+ MnO4 (qa ) + C2O4 (qa ) → Mn (qa ) + CO2(g) (1)Potassium Permanganate +qa -qa 2-qa 16H ( ) + 2MnO4 ( ) + 5C2O4 ( ) → − + − 2+ 2+ MnO 4 +8H +5 e → Mn + 4 H2 O 2Mn (qa ) + 8H2O(l) + 10CO2( g) MnO − +4H+ + 3 e − → MnO( s )+ 2 H O Example: Balance 4 2 2 Sn2+ + Fe3+ <=> Sn4+ + Fe2+ − − 2− MnO4 + e→ MnO4 2+ - 3+ 2+ Fe + MnO4 <=> Fe + Mn 1 Important Redox Titrants and the Reactions Important Redox Titrants and the Reactions Oxidizing Reagents (Oxidants) Oxidizing Reagents (Oxidants) (2) Potassium Dichromate (3) Potassium Iodate 2− + − 3+ Cr2 O 7 +14 H + 6 e → 2 Cr + 7 H2 O 1 IO− +6 H+ + 5 e − → I + 3 H O 3 2 2 2 2− 4+ + 3+ 2+ CrO2 7 + 3U + 2H → 2Cr + 3UO2 + HO2 Important Redox Titrants and the Reactions Important Redox Titrants and the Reactions Reducing Reagent ( Reductants ) Reducing Reagent ( Reductants ) (1) Potassium Iodide (2) Sodium Thiosulfate 2- 2- ¯ 1 2S2O3 S4O6 +2e II− → + e− 2 2 Galvanic Cells - Components Galvanic Cells - Line Notation A galvanic (voltaic) cell uses a spontaneous chemical reaction to generate electricity. – Electrodes (cathode and anode). – Salt bridge: cations move from anode to cathode, anions move from cathode to anode. Line notation Cd(s) | Cd(NO3)2 (aq) || AgNO3 (aq) | Ag(s) Phase boundary Salt bridge Phase boundary 2 Standard Electrode Potentials Standard Reduction (Half-Cell) Potentials Standard hydrogen electrode (S.H.E.) • The S.H.E. is the cathode. It consists of a Pt electrode in a Standard reduction potential (Eo) is the voltage associated tube placed in 1 M H+ solution. H is bubbled through the with a reduction reaction at an electrode when all solutes are 2 1 M and all gases are at 1 atm. Cathode tube. • For the S.H.E., we assign + - 2H (qa , 1M) + 2e → H2 ( g, 1 atm) Reduction Reaction • E°red of zero. • The potential of a cell can be calculated from standard 2H+ (1 M ) +2e- H (1 atm) 2 reduction potentials: Eo = 0 V E°cell = E°red(cathode)− E°red (anode) Standard Reduction (Half-Cell) Potentials Standard Potentials We use hydrogen (S.H.E.) • Consider Zn(s) → Zn2+(qa ) + 2e-. We measure E + − ° cell H(aq)+ e↔ 1/2H2 (g ) E ≡ 0.000 relative to the S.H.E. (cathode): We can measure Eº for other half-reactions, relative to E°cell = E°red(cathode) - E°red(anode) the hydrogen reaction, e.g. for silver: 0.76 V = 0 V - E°red(anode). Ag+(aq)+ e− ↔ Ag(s) E° = 0.799 • Therefore, E°red(anode) = -0.76 V. • Standard reduction potentials must be written as Standard reduction potentials are listed in Ap. H in º reduction reactions: your book. E for the H2 reaction is for the reaction º 2+ - at 25 C Zn (qa ) + 2e → Zn(s), E°red = -0.76 V. Nernst Equation for a Half-Reaction • The Nernst Equation relates the potential of the half reaction to reagent concentrations • For the half-reactions: b ° RT [B] - ° EE= − ln aA+ n e ↔ b B E nF []A a - ° c d aA+ b B + n e ↔ cC + dD E ° RT [CD] [ ] EE= − ln nF [][]ABa b RT EE= ° − lnQ nF # of moles of electrons 3 Nernst Equation for a Half-Reaction Nernst Equation for a Complete Reaction At 298K (25oC) ox1 + red2 <=> red1 + ox2 0.05916 V EE=° − logQ n EE=+ − E − Example (Nernst) Example (Net Reaction) • Write the Nernst equation for the reduction of • Find the voltage for the Ag-Cd cell and state if the reaction is spontaneous if the right cell contained 0.50 M AgNO3(aq) and if phosphoric acid to solid white phosphorous: the left contained 0.010 M Cd(NO3)2(aq) + - • 1) + − ° H3 PO4 + 5H + 5e↔ 1/4P(4 s )+ 4H2 O E° = -.402 2Ag+ 2e ↔ 2Ag(s ) E+ = 0.799 V + − ° 0.05916 1 Cd+ 2e ↔ Cd(s ) E− = −0.402 V E = −0.402 − log 5 5 []H PO H+ 3 4 [] • 2) 0.05916 1 E+ = 0.799 − log 2 = 0.781 V • Note that multiplying the reaction by any factor 2 []0.50 0.05916 1 does not affect Eº or the calculated E: • 3) E = −0.402 − log = −0.461V − 2 []0.010 2H PO + 10H+ + 10e- ↔1/2P (s )+ 8H O E° = -.402 3 4 4 2 • 4) EEE= +− − = 0.781− (− 0.461)= + 1.242 V 0.05916 1 • 5) E = −0.402 − log + − 2 + 10 2Ag+ 2e↔ 2Ag(s ) 10 []H PO H + 2+ 3 4 [] + − Cd()s 2Ag Cd 2Ag()s Cd+ 2e↔ Cd(s ) + ↔ + Determination of the Equivalence Point Redox Titration Curve Aox + Bred <=> Ared + Box EXAMPLE: Derive the titration curve for 50.00 mL At equivalence point, Ecell=0: of 0.0500 M Fe2+ with 0.00, 0.0592 []A 0.0592 []B 15.00, 25.00, 26.00 mL ° red ° red 4+ EA − log = EB − log 0.1000 M Ce in a medium nA []Aox nB []Box that is 1.0 M in HClO4. Potential of saturated ABBA= and = at the equivalence point red ox red ox calomel electrode is 0.241 o o V. nAA E+ nBBE Valid for simple E = Redox expressions nAB+ n 4 EXAMPLE: Derive the titration curve for 50.00 mL of EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 2+ 0.0500 M Fe with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M M Ce4+ in a medium that is 1.0 M in HClO .Potential of 4+ 4 Ce in a medium that is 1.0 M in HClO4. Potential of saturated calomel electrode is 0.241 V. saturated calomel electrode is 0.241 V. Titration reaction: Fe2+ + Ce4+ <=> Ce3+ + Fe3+ Fe2+ + Ce4+ <=> Ce3+ + Fe3+ 4+ Fe3+ + e- <=> Fe2+ Eo = 0.767 V At 15.00 mL of Ce added, VFeMFe > VCeMCe Ce4++ e- <=> Ce3+ Eo = 1.70 V “Buffer region” At 0.00 mL of Ce4+ added, initial point no Ce4+ present; VM minimal, unknown [Fe3+]; thus, insufficient information to []Fe3+ = Ce Ce VV+ calculate E Fe Ce (15.00mL )(0.1000M ) 0.0592 []Fe2+ = =2.308 × 10−2 M EE= ° − log − 0.241 (50.00+ 15.00)mL n []Fe3+ EXAMPLE: Derive the titration curve for 50.00 mL of EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 4+ 4+ Ce in a medium that is 1.0 M in HClO4. Potential of Ce in a medium that is 1.0 M in HClO4. Potential of saturated calomel electrode is 0.241 V. saturated calomel electrode is 0.241 V. Fe2+ + Ce4+ <=> Ce3+ + Fe3+ Fe2+ + Ce4+ <=> Ce3+ + Fe3+ At 15.00 mL of Ce4+ added, V M > V M 4+ Fe Fe Ce Ce At 15.00 mL of Ce added, VFeMFe > VCeMCe “Buffer region” “Buffer region” [Fe3+ ]= 2.308 × 10−2 M Half-reactions: Fe3+ + e- <=> Fe2+ Eo = 0.767 V [Fe3+ ]= 2.308 × 10−2 M [Fe2+ ]= 1.538 × 10−2 M 2+ VMVM− []Fe = Fe Fe Ce Ce 0.0592 []Fe2+ VVFe+ Ce EE= ° − log − 0.241 n []Fe3+ (50.00mL )(0.0500M )− (15.00mL )(0.1000M ) −2 = =1.54 × 10 M 0.0592 1.54× 10−2 (50.00+ 15.00)mL =0.767 − log −0.241 = 0.533V 1 2.038×10−2 EXAMPLE: Derive the titration curve for 50.00 mL of EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 4+ 4+ Ce in a medium that is 1.0 M in HClO4. Potential of Ce in a medium that is 1.0 M in HClO4. Potential of saturated calomel electrode is 0.241 V. saturated calomel electrode is 0.241 V. Fe2+ + Ce4+ <=> Ce3+ + Fe3+ Fe2+ + Ce4+ <=> Ce3+ + Fe3+ 4+ 4+ At 25.00 mL of Ce added, VFeMFe = VCeMCe, At 26.00 mL of Ce added, VFeMFe < VCeMCe, Equivalence point After equivalence point Half-reactions: Fe3+ + e- <=> Fe2+ Eo = 0.767 V 3+ VMFeF e Ce4++ e- <=> Ce3+ Eo = 1.70 V []Ce = VVFe+ Ce n E+ n E E = Fe Fe CeC e − 0.241 (50.00mL )(0.0500M ) n+ n = =3.29 × 10−2 M Fe Ce (50.00+ 26.00)mL 0.767+ 1.70 = −0.241 = 1.23 − 0.241 = 0.99 V 1+1 5 EXAMPLE: Derive the titration curve for 50.00 mL of EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 4+ 4+ Ce in a medium that is 1.0 M in HClO4.