Electrochemistry Flashcards

Question

How do you determine whether a species os unstable with respect to disproportionation?

Answer 1

A species that lies above the gradient between two points shows that is unstable with respect to disproportionation

Answer 2

If a the middle species in a set of three is in a 'valley' on the Frost diagram

Answer 3

Write down all the species in the latimer diagram Write down the oxidation states of each Write down the electrode potentials for each step Multiply each electrode potential by the nukber of electrons Add the NE values that would be required to get the species to the elemental form

Answer 4

Lowest point on the graph

Answer 5

E = E^o - RT/nF lnQ Q = reaction quotient - supposed to be acitivity of products/reactats but we simplify to concentration The activity of a solid is 1

Answer 6

For m-proton and n-electron processes dE/dpH = -m/n \* (RT ln10) / F

Answer 7

If the line of the latimer diagram is steeper at one pH this means the reduction potential is larger and therefroe the reaction is more feasible at this pH

Answer 8

plots of electrode potentials vs pH They map out the possible stable phases of an electrochemical system and compare this to the redox reactivity

Answer 9

Water is only stable within a restricted potential window At suffieciently negative pH water is susceptible to reduction: 2H+ + 2e^- H₂ E^o = 0V Moving to higher pHs the electrodempotential changes with gradient -59 mV At sufficiently high potantials water is susceptible to oxidation: O₂ (g) + 4H⁺ + 4e^- 2H₂0

Answer 10

The energy difference between the starting materials and the transition state If the products are at a lower energy than the reactants the reaction is usually exothermic - this gives negative delta G value

Answer 11

Theoretically if an applied voltage is greater than Ered this should drive oxidation In reality the applied voltage must be greater than the overpotential for oxidation to occur

Answer 12

Theoretically if an applied voltage is more negative than Ered this should drive reduction In reality it must be more negative than the overpotential

Answer 13

A kinetic barrier to the reaction manifests itself as an Ecell that is smaller than the thermodynamic value This qualifies as an overpotential Ecell = E_cathode - E_anode - E_{overpotential} If there is an overpotential the energy released is lower than the theortical maximum

Answer 14

The requirement for greater input of energy

Answer 15

- 0.3V is the potential for NAD⁺/NADH at pH7 0. 8V is the potential for O₂/H₂O at pH7 Photosynthesis has an overpotential of approx. 0.3V E_{photosynthesis} = -0.3 - (0.8 + 0.3) = -1.4V DG_{photosynthesis} = -nFE = -4 x 96485 x -1.4 = 540 kJ mol^-1

Answer 16

Total charge (C) = amount of e^- (mol) x F (Cmol^-1) For the zinc copper cell2 electrons flow for every mol of Cu consumed (2 x 0.4 = 0.8 mol of electrons) So 0.8 x 96485 = 77188 C This gives no information about how quickly the equilibrium is reached

Answer 17

monitoring the rate and energetics of an electrochemical reaction All basic dynamic electrochemistry experiments measure electrical current (rate of flow of e-) while controlling the electrode potential Cthermodynamic driving force

Answer 18

The 3 electrode cell The potential of the working electrode is measured relative to that of the reference electrode while current flows between the working electrode and the counter electrode

Answer 19

Negative current = electrons moving out of the working elevctrode surface Positive current = electrons moving into the working electrode surface

Answer 20

Current arising from a redox process occurring at a working electrode Positive faradaic current indicates an oxidative process Negative faradaic current indicates a reduction reaction is occurring Current due to a non redox potential is described as non faradaic

Answer 21

Standard electrode potentials are measured relative to the standard hydrgen electrode at pH0, 1 bar H₂ and Pt electrpocatalyst This is not always practical Reference electrodes are used instead which have highly stable redox potentials vs the hydrigen electrode reference electrode potential is measured V vs SHE so the potential of a cell measured is also defined as V vs SHE

Answer 22

If current flows through the reference electrode this will perturb the ratio of oxidesed:reduced species This would change the redox potential

Answer 23

The solvent system Must have suffiecient conductivity so that the solution resistance does not convolute experiment results In aq solution NaCl is commonly used In organic solvents tetrabutylammonium hexafluorophosphate is a common salt to use.

Answer 24

Measure current as a function of time while constant voltage is applied applied voltage is E_app (V vs Ref) Time of hold - t_hold

Answer 25

Voltage is modulated as a function of time while current is monitored - plot current vs voltage

Answer 26

Linear voltage sweep is applied with either a positive (anodic) or negative (cathodic) gradient Control the potential of the working electode (gradient of plot against time is the scan rate) Increasing the potential increases the driving force to remove electrons Measure current vs voltage

Answer 27

A forward and backwards sweep is applied with both a positive and negative gradient Repeated voltage cycles are called scans or sweeps or CVs Control potential of the working electrode with E_initial, E_end and E_turn (tutning point voltage) Measure current

Answer 28

With non-faradaic current (absence of redox active species) a baseline current response is recorded This can be thought of as eqivalent to the solvent ad cuvetter absorbance in UV vis This current is referred to as non-faradaic or capacitive current It arises because of double layer rearrangements at th wokring elecrode surface-solution interface

Answer 29

Steadily decreasing the voltage of the woring electrode charges the surface of the electrode with cations because of the electrostatic attraction and anions move away resulting in negative current. The rate of movement of ions is directly proportional to the rate of change of potential so a constant -ve gradient os measured Reversing the direction of the voltage sweep reverses the direction of ion movement resulting in an equal current of opposite sign as anions move towards the electrode surace and cations move away

Answer 30

Scan rate directly controls rate of movement of ions There is a linear relationship between capacitive current and scan rate

Answer 31

Film electrochemistry is when the analyste is adsorbed onto the working electrode Solution electrochemistry is when the analyte is dissolved in the electrochemical cell solution

Answer 32

The co is reduced from Co²⁺ to Co⁺ The eletrode transfers electrons to the Co catalyst The co catalust transfers its electrons to CO₂ and H⁺ This converts them to CH₄ and H₂O This is thought to be a spontaneous potential scale

Answer 33

For the reduction of CO₂ the measured current is -10mA 8 electrons are used per CH₄ produced - 10 mA = 10 x 10^-3 Cs-1 - ve current means electrons flowing out of the electrode We can find the maximum ate of production of CH₄

Answer 34

We assumed 100% faradaic efficiency assume every 8 electrons produces CH₄ with no side reactions or inefficiencies

Answer 35

- 10 mA current flows for 60 seconds 0. 01 Cs^-1 x 60 s = 0.6 C - total charge flowing out of the electrode 0. 6 C / 96485 C (mol e^-)^-1 x 8 electrons (CH₄)^-1 = 7.7 x 10^-7 mol CH₄

Answer 36

k_cat = i_cat / n x F x m_cat k_cat = rate of catalysis i_cat = current of catalysis m_cat = molar amount of catalyst

Answer 37

Yes catalytic redox chemistry can be intergrated as a function of potential using cyclic voltammetry The voltage of the working electrode will change It is important to consider both faradaic and non-faradaic current

Answer 38

Three windows: 1 - zero faradaic current - potential of the working electrode so positivr that none of the Co²⁺ is reduced and catalysis does not proceed 2 - constant faradaic current - potential of the working electrode os so negative that all the cobalt of converted to Co⁺ so the rate of catalysis is independent of potential 3 - dramatic change in current - equilibrium ratio of M_CO²⁺ : M_CO⁺ is highly dependent on potential

Answer 39

Real data is the sum of faradaic and non-faradaic current

Answer 40

Look at the m (no. protons) and n (no. electrons) values For Co catalyst in reduction of CO₂ m= 8 and n=8 When m = n E drops by 59 mV per pH unit If the E^o value is -0.25 V vs SHE at pH0 then at pH 1 E^o = -0.25 - 0.0059 = -0.31 V

Answer 41

When kcat increases icat increases The plot expands

Answer 42

If the solution includes concentrated acid the ionic strenght will be high enough

Answer 43

Voltage / scan rate

Answer 44

electrode reversibly pushes/pulls electrons into a species adsorbed onto the surface Copper electron transfer proteons can be adsorbed onto graphite or gold electrodes

Answer 45

E_ox = E_red These peak potentials equal the potential of the redox couple being investigated The peak area is equal to the charge passed and provides a measure of the muber of moles of redox active material

Answer 46

Linear relationship as the same amount of redox active species is on the surfact of the electrode in all experiments

Answer 47

Totsal current response = sum of faradaic and non faradaic contributions

Answer 48

E_ox and E_red shift with scan rate If Eox and Ered are plotted against scan rate a trumpet plot of formed This is because the faradaic current is smaller and therefore non-faradaic contributions to the total curretn are more significant The shift reflects kinetics of the electron transfer - rate info contained

Answer 49

Determining HOMO LUMO gap Evaluating homogenous electrocatalytic processes Observing electron transfer through molecules

Answer 50

Same as for film electrochemistry but analyte is in solution Auxilary electrode - completes circuit Reference electrode - defines x axis on voltammogram Working electrode - disc of platinum / gold / glassy carbon All electrodes must be submerged in the solution Electrochemistry is only happening at the surface of the electrode The bulk solution remains unaffected

Answer 51

At a certain negative anmd positive potential the solvent will be reduced and oxidised respectively Between these two potentials is called the solvent window The analyte must undergo redox chemistry within this window

Answer 52

Duck shaped The peak anodic potential is E_pa The peak cathodic potential is E_pc The average of these points is E_1/2 The max current upon oxidation (difference between peak and baseline) is i_pa The max current upon reduction is i_pc

Answer 53

In the beginning there is no current response - the voltage is not high enough toreove and electron The current peaks as the electrons are removed from the anion The current decreases - fewer molecules to oxide in the diffusion sphere Current drops to non zero baseline because molecules are diffusimg into the diffusion sphere and instantly oxidised Current peaks in opposite ciection as electrons are added to the newly oxidised analyte Returns to zero baseline

Answer 54

The solution is under diffusion control The faradaic signal is dominated by loinear diffuciom effects Diffusion is only significant in the direction normal to the electrode surface - reaction and transports are uniform across the surface current is proportional to concentration of analyte and scales with squ root of scan rate

Answer 55

Faster scan rate: shirter diffusion distance greater concentration gradient local diffusion rates increase higher current response

Answer 56

high conc of Fc low conc of Fc+ Fc conc decreases and Fc+ incerases to a max at the duck's bill From here Fc+ decreases and Fc incerases

Answer 57

Electron withdrawing and donating groups EDG - easier to oxidise, harer to reduce, stabalise cation EWG - Harder to oxidise, easier to reduce, destabalise cation

Answer 58

Chemical reversibility Electrochemical reversibility

Answer 59

The Nernst equation tells us that for a 1 electron transfer process DE = | E_pa - E_pc | = 59 mV This should not change with scan rate This shift in peak potential is an incrase in DE which implies slow electron transfer amd therefore poor electrochemical reversibility When peak potentials shift at higher scan rates it shows it is less reversible at higher scan rates

Answer 60

Reversibility decreases if the product formed after the first redox reaction undergoes a chemical reaction In a fully reversible process | i_pc/i_pa| = 1 In a non reversible process the return wave could be diminished or shifted

Answer 61

E = electron transfer process C = chemical change A --(-e^-)--\> A⁺ ---(C)---\> B EC mechanisms A ---(C)---\> B ---(E)---\> B⁺

Answer 62

electron transfer process is followed by a chemical process where the new species is redox inert Leads to peak in forward scan and none in return scan Sometimes chemical step is slow enough that at fast scan rates the electrochemical measurement occurs faster than the chemical reaction so the CV looks reversible

Answer 63

electron transfer step occurs after there has been a chemical reaction Return wave is much larger than forward scan

Answer 64

Some molecules can undergo multiple redox processes e.g Naphthalenediimide - two 1-electron reductions Shows up as two combined ducks

Answer 65

A discrete species in which two or more of the same redox active fragments exist in different oxidation states Three clases Class I: valence trapped - no electronic comunication - no e- transfer Class II: moderately coupled - some electronic comunication - barrier to electron transfer Class III: fully delocalised - complete electronic communication - electrons spread over both centres

Answer 66

First the molecule is neutral First 1 electron transfer occurs At this pooint a mixed valence species is generated - one centre is oxidised and the other is not After the second oxidation there is no longer a mixed values species

Answer 67

Comparing DE_1/2 values - can evaluate the stability of MV state Can infer the extent of electron transfer

Answer 68

In equilibrium with the neutral starting material and over-oxidised species Represented as comproprtionation constant - Kc The higher the value of kc the more stable the MV ion is with repect to conversion to the neutral or doubly oxidised species Greater stability is inferred to be due to greater electonic coupling

Answer 69

Variation of the Nernst equation: Kc = e^(DE_1/2 n₁n₂F/RT)