chem. 1b – 11/3 lecture

Chem. 1B – 11/3 Lecture

Announcements I• Exam 2 - Results

– Average about 65%– Broader Distribution than

Exam 1– Time was more of an issue– Several questions with low %

correct were easy conceptual questions

– Questions with lowest % correct:

Score Range

#

90 - 104 7

80s 22

70s 20

60s 38

50s 28

<50 21Version 24% 27% 31% 37% 40% 41%

A 23 21 12 9 19 17

B 21 19 10 11 18 22

Announcements II

• Lab– Starting Experiment 9 (Wed./Thurs.)– Next Week: Quiz 8 (Experiment 9 + 10 Questions

+ Electrochem Questions)• Mastering – Some problems recently?• Today’s Lecture

– Electrochemistry (Ch. 18)• Some review + basic concepts today• Definitions• Standard Half-Cells and Cells• Standard Reduction Potential

Chapter 18 Electrochemistry

• Electrochemical Reactions– Balancing Redox Reactions:

• 6 step method:1) Assign oxidation states2) Separate overall reaction into oxidation and reduction

reactions3) Balance each half reaction with respect to mass in order a)

mass all elements other than H, O, b) O by adding H2O, c) by adding H+, d) Add OH- to both side if in alkaline sol’n

4) Balance each half reaction for charge by adding electrons5) Use common multiplier to get equal numbers of electrons for

each half-reaction6) Add each half reaction together to get net reaction without

electrons as reactants or productsNote: steps 5 and 6 are skipped if stopping at half reactions


• Electrochemical Reactions– Balancing Redox Reactions:

• Examples (unbalanced):AgNO3(aq) + Zn(s) ↔ Ag(s) + Zn(NO3)2(aq)HClO(aq) + Fe2+(aq) ↔ Cl2(g) + Fe3+(aq)MnO4

- (aq) + C2O42-(aq) ↔ Mn2+(aq) + CO2(g)


• Electrochemical Reactions – Different Forms– “Beaker” Reactions

• Products form along with heat (assuming H < 0)• Little control of reaction• Products co-mingled (from reduction and oxidation)• Example: nail “rusts” (oxidation of Fe, reduction of O2)

– Voltaic (Galvanic) Cells• Oxidation and reduction reactions may be divided into different

parts (half-cells sometimes physically separated through two reaction cells)

• Two electrodes are also needed• Reaction can be “harnessed” through voltage/power production• Examples: batteries, pH measuring electrodes


• Electrochemical Reactions – Different Forms– Electrolytic Cell

• In this type of cell, external electrical energy is used to force unfavorable reactions (e.g. 2H2O(l) ↔ 2H2(g) + O2(g)) to occur

• Also requires two electrodes – but some differences from electrodes of voltaic cells

• Examples: Production of Cl2 gas from NaCl(aq), production of H2 gas from water (above), instruments that measure degree of oxidation/reduction at specific voltages (analogous to spectrometers)


• Voltaic Cells - Description of how example cell works– Reaction on anode =

oxidation– Anode = Zn electrode (as the

Eº for Zn2+ is less than for that for Ag+)

– So, reaction on cathode must be reduction and involve Ag

– Oxidation produces e-, so anode has (–) charge (galvanic cells only); current runs from cathode to anode

– Salt bridge allows replenishment of ions as cations migrate to cathode and anions toward anodes

Salt Bridge

voltmeter

Zn(s)

ZnSO4(aq)

Ag(s)

AgNO3(aq)

GALVANIC CELL

Zn(s) → Zn2+ + 2e-

Ag+ + e- → Ag(s)

– +


• Basic Electrical Quantities– Current: the flow of electrons (although defined

where a positive current has electrons moving backwards)

– Current units: Amperes (A) with 1 A = 1 C/s and 1 C = 1 Coulomb where 1 electron (elementary charge) has a value of 1.60 x 10-19 C

– Potential or Voltage: The potential energy associated with the movement of charge (e.g. to electrode of opposite sign)

– Potential units: Volts (V) = 1 J/C


• Basic Electrical Quantities – From Voltaic Cells– Current: related to the flow of electrons– Potential: related to the reaction occurring

(more energetic means higher potential)– The ability of a metal (or other elements) to

reduce can be measured under standard conditions

– Example: Zn(s) + 2Ag+(aq) ↔ Zn2+(aq) + 2Ag(s)If [Ag+] and [Zn2+] = 1 M, Ecellº = 1.56 V

Chapter 18 ElectrochemistryVoltaic Cells

• Cell notation– Example Cell:Zn(s)|Zn2+(aq)||Ag+ (aq)|Ag(s)

Salt Bridge

voltmeter

Zn(s)

ZnSO4(aq)

Ag(s)

AgNO3(aq)

GALVANIC CELL

left side for anode (right side for cathode)

“|” means phase boundary

“||” means salt bridge


• Example Questions– Given the following cell, answer the following

question:MnO2(s)|Mn2+(aq)||Cr3+(aq)|Cr(s)– What compound is used for the anode?– What compound is used for the cathode?– Write out both half-cell reactions and a net

reaction

Chapter 18 Electrochemistry• Given the following

cell, write the cell notation:

Salt Bridge

voltmeter – reads +0.43 V

Pt(s)

FeSO4 (aq), Fe2(SO4)3(aq)

Ag(s)

NaCl(aq)

GALVANIC CELL

AgCl(s)

+–

Note: In this case the Pt(s) is an “inert” electrode (provides electrons but doesn’t react

Chapter 18 ElectrochemistryStandard Reduction Potential

• A cell used to determine the standard reduction potential consists of two half cells

• One half-cell, the anode, is the standard hydrogen electrode (2H+

(aq) + 2e- ↔ H2(g))• Eanodeº = 0 (defined)• Other is the test cell (compound

being reduced when half-cell is coupled to standard hydrogen electrode (oxidation electrode)

• Both cells under standard conditions (1 M, 1 atm)

• Ecellº = Ecathodeº• The SHE is not actually used much

any more (just a reference for relative potential)

Ag(s)

AgNO3(aq)

Pt(s)

H+(aq)H2(g)

Chapter 18 ElectrochemistryStandard Reduction Potential

• Meaning of Values– Half-cells that exhibit positive values

have electrodes with compounds that easily reduce (e.g. Ag+(aq), MnO4

-, PbO2(s))

– Half-cells that exhibit negative values have electrodes that easily oxidize (e.g. alkali metals)

– What if we have two half-cells (neither SHE), can we find Ecellº?

Example: Zn(s)|Zn2+(aq)||Ag+ (aq)|Ag(s) Ecellº = ?

Eº = 0

Ag+ reductionEº = +0.80 V

Zn2+ reductionEº = -0.76 V


• Example Question– An Ag/AgCl electrode is a common reference

electrode. What is the standard potential of a cell made up of a Cu2+ solution being reduced to Cu(s) and AgCl(s) being reduced to Ag(s)?

E°(Cu2+ + 2e- ↔ Cu(s)) = 0.34 VE°(AgCl(s) + e- ↔ Ag(s) + Cl- (aq)) = 0.22 VWhat is the balanced reaction and what species

must be present at 1 M?

chem. 1b – 11/3 lecture

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