Electrochemistry and Electrode Potentials

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Presentation transcript:

Electrochemistry and Electrode Potentials Chapter 13 &14 Electrochemistry and Electrode Potentials

Oxidation: a loss of electrons to an oxidizing agent Reduction: a gain of electrons from a reducing agent

16A Principles Reduction-oxidation reaction (redox reaction) Ox1 + Red2  Red1 + Ox2 An oxidizing substance: Ma+ + ne-  M(a-n)+ An reducing substance: Ma+  M(a-n)+ + ne-

16B Electrochemical Cells (1) Galvanic (Voltaic) cell: a chemical reaction spontaneously occurs to produce electrical energy. Ex: lead storage battery (2) Electrolytic cell: electrical energy is used to force a nonspontaneous chemical reaction to occur. Ex: electrolysis of water

the anode: oxidation occurs the cathode: reduction occurs Salt bridge: allows charge transfer through the solutions but prevents mixing of the solutions. Fig 16-2 Cu: reducing agent Ag+: oxidizing agent

(於1953, the 17th IUPAC meeting決定半反應以「還原反應」來表示 ) Electrode potential: the tendency of the ions to give off or take on electrons. Normal Hydrogen Electrode (NHE) or Standard Hydrogen Electrode (SHE) 2H+ + 2e- = H2 or H+ + e- = 1/2H2 Eo (標準電位) = 0.000 V Table 16.1 (於1953, the 17th IUPAC meeting決定半反應以「還原反應」來表示 )

Potential are dependent on [con]. & temp. Standard reduction potential: activity=1 The more positive the electrode potential, the greater the tendency of the oxidized form to be reduced. The more negative the electrode, the greater the tendency of the reduced form to be oxidized.

Related to free energy

The Nernst Equation

Ex: Predict whether 1M HNO3 will dissolve gold metal to form 1M Au3+?

Cell representation anodesolutioncathode

The Nernst Equation & [C] effect Activities should be used in the Nernst equation. We will use concentrations here because titrations deal with large potential changes, and the errors are small by doing so. Table 16-1, 每個離子之活性(activity)皆為1, 叫「標準還原電位」

Dependence of the cell potential on [C] E is the reduction potential at the specified concentrations n: the number of electrons involved in the half-reaction R: gas constant (8.3143 V coul deg-1mol-1) T: absolute temperature F: Faraday constant (96,487 coul eq-1) at 25°C  2.3026RT/F=0.05916

ex: [C] & Ecell standard conditions: [C]=1M what if [C]≠1M? [Al3+]=2.0M, [Mn2+]=1.0M Ecell<0.48V [Al3+]=1.0M, [Mn2+]=3.0M Ecell>0.48V

After兩個半反應reached eq., the cell voltage necessarily becomes zero and the reaction is complete.

Ex: One beaker contains a solution of 0.020 M KMnO4, 0.005 M MnSO4, and 0.500 M H2SO4; and a second beaker contains 0.150 M FeSO4 and 0.0015 M Fe2 (SO4)3. The 2 beakers are connected by a salt bridge and Pt electrodes are placed one in each. The electrodes are connected via a wire with a voltmeter in between. What would be the potential of each half-cell (a) before reaction and (b) after reaction? What would be the measured cell voltage (c) at the start of the reaction and (d) after the reaction reaches eq.? Assume H2SO4 to be completely ionized and equal volumes in each beaker.

5Fe+2 + MnO4- + 8H+ = 5Fe+3 + Mn+2 + 4H2O  Pt/Fe+2(0.15 M), Fe+3(0.003 M)//MnO4-(0.02 M), Mn+2(0.005 M), H+(1.00 M)/Pt (a) EFe = EoFe(III)/Fe(II) – (0.059/1) log [Fe+2]/[Fe+3] = 0.771 – 0.059 log (0.150)/(0.0015 × 2) = 0.671 V EMn = EoMnO4-/Mn+2 – (0.059/5)log [Mn+2]/[MnO4-][H+]8   = 1.51 – 0.059/5 log (0.005)/(0.02)(1.00) 8 = 1.52 V   (b) At eq., EFe = EMn, 可以含鐵之半反應來看, 先找出平衡時兩個鐵離子的濃度,得 EFe = 0.771 – 0.059 log (0.05)/(0.103) = 0.790 V (c) Ecell = EMn - EFe = 1.52 – 0.671 = 0.849 V (d) At eq., EFe = EMn, 所以Ecell = 0 V

16C-7 Limitation to use E0 The sources of differences: For Fe3+ + e-  Fe2+ (1) Use [conc] vs ax (activities) gFe(II)/ gFe(III) = 0.4/0.18 at m = 0.1M (2) Other equilibria: complexes Fe(III) with Cl-, SO4-2 are more stable than those of Fe(II).

16C-7 Formal Potential Ex: Ce4+ + e-  Ce3+ E°=1.6V with H+A- E°≠1.61V 則酸之陰離子會與之起某種程度的螯合,而影響他們的活性  再是標準還原電位Eo改叫形式電位Eo` Formal potential: (E°’) The standard potential of a redox couple with the oxidized and reduced forms at 1M concentrations and with the solution conditions specified. Ex: Ce4+/Ce3+ in 1M HCl E°’=1.28V

Dependence of potential on pH Many redox reactions involved protons, and their potentials are influenced greatly by pH.

Dependence of potential on complexation: Complexing one ion reduces its effective concentration, which changes the potential. In effect, we’ve stabilized the Fe+3 by complexing it, make it more difficult to reduce.

Ex: Systems involving ppt Calculate Ksp for AgCl at 25℃ & E =0.58V

Sol:

Ex 17-4: Calculate the cell potential for Ag∣AgCl(sat’d), HCl(0 Ex 17-4: Calculate the cell potential for Ag∣AgCl(sat’d), HCl(0.0200 M)∣H2(0.800 atm), Pt Sol: 2H+ + 2e-  H2(g) E0H+/H2 = 0.000 V AgCl(s) + e-  Ag(s) E0AgCl/Ag = 0.222 V = -0.0977 V Eleft = 0.222 – 0.0592 log[Cl-] = 0.222 – 0.0592 log 0.0200 = 0.3226 V Ecell = Eright – Eleft = -0.0977 – 0.3226 = -0.420 V 2H+ + 2Ag(s)  H2(g) + 2AgCl(s)

17B Calculating Redox Equilibrium Constants Ex 17-6: Calculate the equilibrium constant for the reaction 2Fe3+ + 3I-  2Fe2+ + I3- Sol: 2Fe3+ + 2e-  2Fe2+ E0 = 0.771 V I3- + 2e-  3I- E0 = 0.536 V

Example 17-7 Calculate the equilibrium constant for the reaction 2MnO4- + 3Mn2+ + 2H2O  5MnO2(s) + 4H+ Sol: 2MnO4- + 8H+ +6e-  2MnO2(s) + 4H2O E0 = +1.695 V 3MnO2(s) + 12H+ + 6e-  3Mn2+ + 6H2O E0 = +1.23 V EMnO4-/MnO2 = EMnO2/Mn2+

17C Constructing Redox Titration Curves Example 17-8 Obtain an expression for the equivalence-point potential in the titration of 0.0500 M U4+ with 0.1000 M Ce4+. Assume that both solutions are 1.0 M in H2SO4. U4+ + 2Ce4+ + 2H2O  UO22+ + 2Ce3+ + 4H+ Sol: UO22+ + 4H+ + 2e- → U4+ + 2H2O E0 = 0.334 V Ce4+ + e-  Ce3+ E0' = 1.44 V

Table 17-1 Electrode Potential versus SHE in Titrations with 0.100 M Ce4+ Potential, V vs. SHE Reagent Volume, mL 50.00 mL of 0.0500 M Fe2+ 50.00 mL of 0.02500 M U4+ 5.00 0.64 0.316 15.00 0.69 0.339 20.00 0.72 0.352 24.00 0.76 0.375 24.90 0.82 0.405 25.00 1.06 ←Equivalence → Point 0.703 25.10 1.30 26.00 1.36 30.00 1.40 Note: H2SO4 concentration is such that [H+] = 1.0 M throughout.

17D Oxidation/Reduction Indicators Self-indication: If the titrant is highly colored, this color may be used to detect the end point. Ex : MnO4-  Mn2+ purple faint pink

Starch indicator: This indicator is used for titrations involving iodine Starch + I2  dark-blue color complex

Redox Indicators: Oxind + ne-  Redind These are highly colored dyes that are weak reducing or oxidizing agents that can be oxidized or reduced Oxind + ne-  Redind

A potential equal to 2×(0.059/n)V is required for a sharp color change The redox indicator reaction must be rapid and reversible. Table 17.2 Ex: (1) Ferroin: [tris(1,10-phenanthroline)ion(II) sulfate] for titrations with cerium(IV) (2) Starch/Iodine soln.

18B Reducing Agents Thiosulfate: stable to air oxidation Iron(II): E0 = 0.771V for titration of cerium(IV), chromium(VI), vanadium(V) indicator: ferroin or diphenylamine sulfonate.

18C Oxidizing Agents Potassium permanganate (KMnO4) E0=1.51 In neutral solution: MnO4-MnO2 In acid solution: MnO4-Mn2+ Autocatalytic decomposition: Standardization: Na2C2O4 5H2C2O4+2MnO4-+6H+ 10CO2+2Mn2++8H2O

Cerium (IV): Ce4+ / H2SO4: E0 = 1.44V; Ce4+ / HClO4: E0 = 1.70V the rate of oxidation of chloride ion is slow is stable in H2SO4 (NH4)2Ce(NO3)6 can be obtained as a primary standard. indicator: Ferroin

Potassium dichromate: K2Cr2O7 a slightly weaker oxidizing agent than KMnO4 primary standard Cr2O72-  Cr3+ E0 = 1.33~1.00V in 1M HCl