Transcript Slide 1
Lecture 3.1 Denoting Electrochemical Cells So, instead of drawing: Pt Pt 1 M KCL KCl (satd.) Glass Frit 1 M HCl Hg2Cl2 10-3 M Fe3+ Hg 10-3 M Fe2+ Pt SCE How about a short-hand method? Pt/Fe3+(10-3M), Fe2+(10-3M), KCl (1M)//KCl(satd.)/Hg2Cl2/Hg/Pt “/” = phase boundary “//” = ionically porous material (liquid junction) + 0.241 V vs. NHE Hg2Cl2 + 2e- SCE 2Hgo + 2Cl- Another Example: Pt/Fe(CN)63-(10-3M), Fe(CN)64-(10-4M), KCl (1M)//KCl(satd.)/Hg2Cl2/Hg/Pt or Pt/Fe(CN)63-(10-3M), Fe(CN)64-(10-4M), KCl(0.1 M)//SCE We find the Ecell is .010V different. Why? Lecture 3.2 Look closer: Cl- K+ K+ ClCl- 0.1M or 1.0M K+ K+ Cl- K+ K+Cl- K+ ClCl- Cl- Cl- K+ Cl- K+ K+ 1M There exists a driving force, DE, due to dC/dx of the ions, both K+ and ClThe ions want to “equilibrate” but would take a long time to do so. The potential is known as: - membrane potential - diffusion potential - Liquid Junction Potential This potential ( E = i Rion motion membrane) will alter E measurement. Lecture 3.3 Types of L. J. s All result in LJP due to different chemical potentials. 1. Same ion / different [ ion ] c+x-llc+x2. Different ions / same [ ion ] c+x-llc+y3. Different ions / different [ ion ] c+x-llc+yWhat else? 4. Different solvent! Different ions!........ …………..Ayy!!!!! Ecell = ENernst + Ej Make corrections? Measure it? Okay but minimize it! Use ions with similar ionic mobilities and [high]. See page 72!! R + d d s M+ Xu1 u2 u1 ≈ u2 (KCl) Lecture 3.4 Let’s move on to more details of what happens when a Polarized Electrode is placed in a solution with electrolyte and various redox species. Pt – WE, SCE – Ref, Pt – Aux H2O, 1.0 M HCl, 1 mM Fe3+, 1 mM Cr3+ What does voltammogram look like? 1. What reactions can occur at WE? - What are E 0 of all species? H2O, Cr3+, H+, Cl-, Fe3+, Pt What is E 0 vs. SCE? - What are electroactive species? - What is electrode material? Fe3+/2+ E 0 = + 0.77 V vs. NHE 0 Cr3+/2+ E = - 0.41 V vs. NHE SCE 0 + 0.242 V - 0.242 V NHE 0 Hg2Cl2 + 2e- 2Hg + 2ClSat’d KCl So always subtract + 0.242 from NHE to get SCE value. E 0 Lecture 3.5 3 Fe / Fe 2 vs. SCE = +0.53 V E 0 Cr 3 / Cr 2 vs. SCE = -0.65 V fast +eH+ -e- ½ H2 Record Voltammogram: Pt e- + Fe3+ ( Fe2+ 1000x il of Fe3+/ Fe2+) No Cr3+ Cr2+! ilim ilim ic 0 E vs. SCE ia H2O 2H2O ½ O2 + 2H+ + 2eO2 + 4H+ + 4e- Hmmm… Try Hg drop electrode: e- + Cr3+ Cr2+ ic 0 ia Hg E vs. SCE Hg2+ + 2eSo, both k and G limitations! Lecture 3.6 How do e-s get into molecules? Well, molecules have orbitals (MO) where either bonding, non-bonding, or anti-bonding electrons can reside. We will remove/add e- to these orbitals where non-bonding e- can reside. “effective p-stat” Potentiostat and Energy of Molecular orbitals eenergy (Eelectrode more - ) So, recall DG = -nFEelectrode. We can make energy of electrode what we like (within reason). So, once M is within ~ 10 Å, e- transfers. Unoccupied MO (LUMO) Occupied MO (HOMO) Lecture 3.7 or Electrode Solution (molecule) Electrode Solution (molecule) But, in order for E – T to occur, must also allow D – L to charge. What does i – E curve (voltammogram) look like for just HCl? (Pt WE) H+ + e- 0 ½ H2 E vs. SCE No E –T occurs! But, some i flows. qm + + + + + WHY? - + + + + qs - + + + + + - + + + Lecture 3.8 qm = -qs , Always!!! Basic Capacitor: NOT C C D.C. Voltage Source (fixed) q Coulom b farad Eapp Volt Depending on the metal, there will be some potential or some point on the potential axis where no charge will reside on the metal. *THIS IS NOT NECESSARILY 0 V vs. REF!!!* P Z C qm ~ 0 o of e i r n o t h a r g e bulk +, - conc sol’n bulk +, - conc C+, At PZC x C+, - At Eapp x What is physical structure of the interface ( Double-Layer) ? + solvent dipole - Solvated or ligated cation of electrolyte (non-specifically absorbed) M+ Z+ Non-solvated cation (electrolyte, etc.) (typically specifically adsorbed) Inner Helmholtz Plane (IHP) qi Lecture 3.9 Outer Helmholtz Plane (OHP) qi + qo = qsoln qo E-T occurs at OHP for freely diffusing materials. EMetal (~10-200Å depending on [SE]) Energy ESoln E x distance (Å) IHP OHP ~3-5Å DE 0.1V 6 2 x 10 V / cm 8 Dx 5x 10 cm Lecture 3.10 Response from D-L with E step: Eapp i(t ) Rs i(t) 0.37 Eapp Rs Eapp Rs et / RsCDL = RsCDL Time to decrease by 0 t 0 0.63 of Eapp Rs What if we actually scan E? Rs CDL v i Scan E at Eapp dE dt Ei Rs ~cst i at cst v i(t) Ei 0 dE (V/s) dt t t Lecture 3.11 i(t) = vCd l If RsCdl is small and v is cst Ei R vCdl s t / Rs C dl e i(t)~vCdl So what if we scan E to Efinal and reverse? i(t) Eapp Ef vCdl 0 Ei -vCdl tr 0 tr t t or vs. Eapp vCdl i(E) 0 Eapp