Download Ch.14-16 Electrochemistry Redox Reaction and more Study notes Electrochemistry in PDF only on Docsity!
Ch.14-
Electrochemistry
Redox reactions: involve transfer of electrons from one
species to another.
Oxidizing agent (oxidant): takes electrons
Reducing agent (reductant): gives electrons
Oxidizing Agent
Reducing Agent
ox 1 + red 2 <=> red 1 + ox 2
Redox Reaction - the basics
Redox Reaction - the basics
Redox reactions: involve transfer of electrons from one
species to another.
Oxidizing agent (oxidant): takes electrons
Reducing agent (reductant): gives electrons
3 2 2 3 Fe V Fe V
Oxidizing Agent
Reducing Agent
Reduced Oxidized
ox 1 + red 2 <=> red 1 + ox 2
1. Write down the (two half) reactions.
2. Balance the (half) reactions (Mass and Charge):
a. Start with elements other than H and O.
b. Balance O by adding water.
c. balance H by adding H+.
d. Balancing charge by adding electrons.
(3. Multiply each half reaction to make the number of
electrons equal.
4. Add the reactions and simplify.)
Balance Redox Reactions (Half Reactions)
Example: Balance the two half reactions and redox
reaction equation of the titration of an acidic solution of
Na 2 C 2 O 4 (sodium oxalate, colorless) with KMnO 4 (deep
purple).
MnO 4 - ( aq ) + C 2 O 4 2-( aq ) → Mn2+^ ( aq ) + CO 2 ( g )
16H+( aq ) + 2MnO 4 - ( aq ) + 5C 2 O 4 2-( aq ) →
2Mn2+( aq ) + 8H 2 O( l ) + 10CO 2 ( g )
Example: Balance
Sn2+^ + Fe3+^ <=> Sn4+^ + Fe2+
Fe2+^ + MnO 4 -^ <=> Fe3+^ + Mn2+
Important Redox Titrants and the Reactions
Oxidizing Reagents (Oxidants)
(1) Potassium Permanganate
MnO 4 −+ 8 H ++ 5 e −→ Mn^2 ++ 4 H 2 O
MnO 4 −+ 4 H ++ 3 e −→ MnO 2 ( s )+ 2 H 2 O
− − −
2
MnO 4 e MnO 4
Important Redox Titrants and the Reactions
Oxidizing Reagents (Oxidants)
(2) Potassium Dichromate
Cr 2 O 72 −+ 3 U^4 ++ 2 H+→ 2 Cr^3 ++ 3 UO 22 ++H 2 O
Cr O H e Cr^3 H 2 O
2
2 7 +^14 +^6 →^2 +^7
− + − +
Oxidizing Reagents (Oxidants)
(3) Potassium Iodate
IO (^) 3 H e I 2 3 H 2 O 2
1
−
Important Redox Titrants and the Reactions
Important Redox Titrants and the Reactions
Reducing Reagent ( Reductants )
(1) Potassium Iodide
I → I 2 + e
Important Redox Titrants and the Reactions
Reducing Reagent ( Reductants )
(2) Sodium Thiosulfate
2S 2 O 3 2-^ S 4 O 6 2-^ +2e¯
Galvanic Cells - Components
- Electrodes (cathode and anode).
- Salt bridge: cations move from anode to cathode, anions move from cathode to anode.
A galvanic (voltaic) cell uses a spontaneous chemical reaction to generate electricity.
Galvanic Cells - Line Notation
Cd(s) | Cd(NO 3 ) 2 (aq) || AgNO 3 (aq) | Ag(s)
Phase boundary Salt bridge Phase boundary
Line notation
Nernst Equation for a Half-Reaction
At 298K (25oC)
Q n
E E log
- 05916 V = ° −
Nernst Equation for a Complete Reaction
E = E + − E −
ox 1 + red 2 <=> red 1 + ox 2
Example (Nernst)
- Write the Nernst equation for the reduction of phosphoric acid to solid white phosphorous:
- Note that multiplying the reaction by any factor does not affect E º^ or the calculated E :
H 3 PO 4 + 5H+^ +5e-↔1/4P 4 ( s )+ 4 H 2 O E °= -.
[ ][ ]
5
H 3 PO 4 H
log
E =− 0. 402 − +
2H 3 PO 4 + 10H+^ +10e-↔1/2P 4 ( s )+ 8 H 2 O E °= -.
[ ] [ ] 2 10
H 3 PO 4 H
log
E =− 0. 402 − +
Example (Net Reaction)
- Find the voltage for the Ag-Cd cell and state if the reaction is spontaneous if the right cell contained 0.50 M AgNO 3 ( aq ) and if the left contained 0.010 M Cd(NO 3 ) 2 ( aq ) - 1)
2 Ag +^ + 2 e−↔ 2 Ag( s ) E +°= 0. 799 V
Cd +^ + 2 e−↔Cd( s ) E °−=− 0. 402 V
[ ]
0. 781 V
log
E += 0. 799 − 2 =
[ ]
0. 461 V
log
E −=− 0. 402 − =−
E = E + − E −= 0. 781 −(− 0. 461 )=+ 1. 242 V
2 Ag ++ 2 e−↔ 2 Ag( s )
Cd +^ + 2 e−↔Cd( s ) Cd(^ )^2 Ag Cd^2 Ag()
s + +↔^2 ++ s
Determination of the Equivalence Point
A B
o B B
o A A n n
nE nE E
=
and at theequivalence point
[ ]
[ ]
log
[ ]
[ ]
log
red ox red ox
ox
red B
B ox
red A
A
A B B A
B
B
n
E
A
A
n
E
At equivalence point, Ecell=0:
Aox + Bred <=> Ared + Box
Valid for simple Redox expressions
Redox Titration Curve
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 Ce4+^ in a medium
that is 1.0 M in HClO 4.
Potential of saturated
calomel electrode is 0.
V.
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
Ce4+^ in a medium that is 1.0 M in HClO 4. Potential of
saturated calomel electrode is 0.241 V.
Titration reaction: Fe2+^ + Ce4+^ <=> Ce3+^ + Fe3+
Fe3+^ + e-^ <=> Fe2+^ Eo^ = 0.767 V
Ce4++ e-^ <=> Ce3+^ Eo^ = 1.70 V
At 0.00 mL of Ce4+^ added, initial point no Ce4+^ present;
minimal, unknown [Fe3+]; thus, insufficient information to
calculate E
[ ]
[ ]
log
3
2
Fe
Fe
n
E E
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.
M Ce4+^ in a medium that is 1.0 M in HClO 4 .Potential of
saturated calomel electrode is 0.241 V.
Fe2+^ + Ce4+^ <=> Ce3+^ + Fe3+
At 15.00 mL of Ce4+^ added, VFeMFe > VCeMCe
“Buffer region”
M
mL
mL M
V V
Fe VM Fe Ce
Ce Ce
2
3
[ ]
−
= ×
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
Ce4+^ in a medium that is 1.0 M in HClO 4. Potential of
saturated calomel electrode is 0.241 V.
Fe2+^ + Ce4+^ <=> Ce3+^ + Fe3+
At 15.00 mL of Ce4+^ added, VFeMFe > VCeMCe
“Buffer region”
[ Fe^3 +]^ = 2. 308 × 10 −^2 M
M
mL
mL M mL M
V V
V M VM
Fe Fe Ce
Fe Fe Ce Ce
2
2
[ ]
−
= ×
[ Fe^3 +^ ]= 2. 308 × 10 −^2 M [ Fe^2 +^ ]= 1. 538 × 10 −^2 M
V
Fe
Fe
n
E E
log^1.^5410
0. 767 0.^0592
[ ]
[ ]
log
2
2
3
2
×
= − ×
−
−
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
Ce4+^ in a medium that is 1.0 M in HClO 4. Potential of
saturated calomel electrode is 0.241 V.
Fe2+^ + Ce4+^ <=> Ce3+^ + Fe3+
At 15.00 mL of Ce4+^ added, VFeMFe > VCeMCe
“Buffer region”
Half-reactions: Fe3+^ + e-^ <=> Fe2+^ Eo^ = 0.767 V
0. 241 1. 23 0. 241 0. 99 V
Fe Ce
Fe Fe Ce Ce
n n
n E n E
E
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
Ce4+^ in a medium that is 1.0 M in HClO 4. Potential of
saturated calomel electrode is 0.241 V.
Fe2+^ + Ce4+^ <=> Ce3+^ + Fe3+
At 25.00 mL of Ce4+^ added, VFeMFe = VCeMCe,
Equivalence point
Half-reactions: Fe3+^ + e-^ <=> Fe2+^ Eo^ = 0.767 V
Ce4++ e-^ <=> Ce3+^ Eo^ = 1.70 V
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
Ce4+^ in a medium that is 1.0 M in HClO 4. Potential of
saturated calomel electrode is 0.241 V.
Fe2+^ + Ce4+^ <=> Ce3+^ + Fe3+
At 26.00 mL of Ce4+^ added, VFeMFe < VCeMCe,
After equivalence point
M
mL
mL M
V V
VM
Ce
Fe Ce
Fe Fe
2
3
[ ]
−
= ×
Example: Find the equilibrium constant
This can be divided into two half reactions:
Cu( s )+ 2Fe^3 +^ ↔Cu^2 ++ 2 Fe^2 +
2Fe 3 +^ + 2 e−↔ 2 Fe^2 +( s ) E °+= 0. 771 V
E °^ = E +°− E °−= 0. 771 − 0. 339 = 0. 432 V
( 0. 05916 ) 14
(2)(0.432)
= 10 0.^05916 = 10 = 4 × 10
(^) nE °
K
Cu 2 +^ +2e-↔Cu( s ) Eo-= 0. 339 V
Cells as Chemical Probes
Two Equilibria: (1) Equilibrium between two half-cells (2) Equilibrium within each half-cell
•The half-reaction that you write must involve species that appear in two oxidation states in the cell. •The reaction is in equilibrium in the right cell is not the net cell reaction:
AgCl( s ) ↔Ag+^ ( aq )+Cl−( aq )
AgCl ( s )+ e−^ ↔ Ag ( s )+ Cl -^ ( aq , 0. 10 M ) E +°= 0. 222 V 2 H +^ ( aq ,? M )+2e-^ ↔ H 2 ( g , 1. 00 bar ) Eo-= 0 V
Survival Tips
(1) Write half-reactions and their standard potentials (2)Write Nernst equation for the net reaction and put in all the known quantities. (3) Solve for the unknown concentration and use that concentration in the chemical equilibrium equation to solve the problems.
Ex. p.286-
Electrochemistry, chemical equilibrium, solubility, complex formation, and acid-base chemistry
Important Biochemical Reactions
- Formal potential , Eº’ , meant to define potentials under conditions of biochemistry
Potentiometry
- Potentiometry : the use of electrodes to measure voltages
from chemical reactions.
- Electroactive species : can donate or accept electrons at
an electrode; can be measured as the part of a galvanic
cell (analyte)
- Reference electrode : we then connect the analyte half-
reaction to a second cell with a fixed composition (known
potential), the 2nd half-cell is called reference electrode.
- Indicator electrode : responds to analyte - Metal electrodes: inert metals, e.g., Pt, Au
- Ion-selective electrodes: respond to specific analytes
S.H.E.
(again)
The standard reduction potential, Eo, for each half- cell is measured when different half-cells are connected to S.H.E. S.H.E. || Ag+^ (aq. =1) | Ag(s) Standard means that the activities of all species are unity. Not practical for regular use due to the hydrogen gas
AgCl() e Ag() Cl 0. 222 V
Fe e Fe 0. 771 V
3 - 2
− °
°
s s E
E
[ ] [ ]
3
2
- 222 - .059log Cl
Fe
- 059 logFe
=
= − +
E
E
E = E +− E -
Reference Electrode Indicator Electrode
Reference Electrodes
Detect Fe2+^ /Fe3+^ in solution: a Pt wire (indicator electrode) in the half-cell and connect this half cell to a 2nd half-cell at a constant potential.
AgCl() e Ag() Cl 0. 222 V
Fe e Fe 0. 771 V
3 - 2
− °
°
s s E
E
The entire left half-cell containing appropriate solutions and a salt bridge.
Reference Electrodes
Silver-Silver Chloride Electrode
- The difference in E is due to activity coefficients.
- A double junction electrode: avoid to mix Cl-^ with the analyte
w/saturatedKCl 0. 197 V
AgCl() e Ag() Cl 0. 222 V
Ag|AgClElectrode:
= +
E
s s E
Hg | Hg 2 Cl 2 Electrode:
Saturated Calomel Electrode (S.C.E.)
w/saturatedKCl 0. 241 V
1/2Hg 2 Cl 2 () e- Hg() Cl 0. 268 V = +
E
s l E
Voltage Conversions between Different Reference Scales
- 0.351V+ 0. 241 V= -0.110V
An indicator electrode has a potential of -0.351 V with respect to an S.C.E., what’s its potential with respect to an S.H.E.?
Indicator Electrodes - Metal Electrodes
- Metal electrodes: develop potential in response to a redox
reaction on their surface
- Pt is mostly inert, not participating in reactions
- It simply allows electron transfer to/from solution
- Platinum is the most common metal indicator electrode
Gold is also an inert metal indicator electrode
- Carbon electrodes are often used because many redox
reactions are very fast on a carbon surface
The Glass Membrane of a pH Electrode
•Cross section of the glass membrane of a pH electrode. •H+^ can diffuse into the membrane to replace the metal ions through binding to oxygen in glass (ion- exchange equilibrium).
[ ]
constant (0.0592)pH at 25 C
constant (0.0592)logH
Responseoftheelectrode:
O Out
Out E
E
β
β = −
β: ~1.00, electromotive
efficiency measured during calibration.
Errors in pH Measurement
1. Calibration standards (±0.01 pH)
2. Junction potential (~0.01 pH)
3. Junction potential drift (recalibrate every 2 hrs)
4. Sodium error (when [H+] is low and [Na+] is high)
5. Acid error (strong acid, the glass surface is saturated
with H+)
6. Equilibration time (~30s with adequate stirring)
7. Hydration of glass (A dry electrode requires several
hours of soaking)
8. Temperature (calibrate at same T as measurement)
9. Cleaning (remove hydrophobic liquid)
Selectivity Coefficient
coefficient: the
relative response of
the electrode to
different species
less interference there
is due to ion X
K, Rb
K,Cs
5 K,Na
= ×
−
k
k
k
responseto A
responsetoX
k A, X =
)log [ ] at 25 C
n
constant (
Responseofion-selectiveelectrode:
= ± + ^ O
A (^) x A,X x
E β a (k a)
Specifications for Electrochemical
Techniques
Advantages
Linear response to analyte over wide dynamic range
Nondestructive
Short response times
Unaffected by color/turbidity (limited matrix effects)
Cheap
Disadvantages
Sensitivity (High detection limits)
Not universal
Voltammetry
- A collection of techniques in which the relation between current and voltage is observed during electrochemical process.
- Can be used to (1) Study electroactivity of ions and molecules at the electrode/solution interface (2) Probe coupled chemical reactions and measure electron transfer rates (3) Examine electrode surfaces
- An electrochemical cell consists of a working (analyzing) electrode, an auxiliary (counter) electrode, and a reference electrode. The control device is a potentiostat.
Calibration Curve
[A]
ip
Anodic stripping voltammetry (ASV): analytes are reduced and deposited into (onto) an electrode. They are reoxidized during the stripping step. e.g. Cd2+(aq) + 2e = Cd(Hg) Deposition Step Cd(Hg) – 2e = Cd2+^ (aq) Stripping
Cathodic stripping voltammetry (CSV) : typically anions are oxidized and deposited onto an electrode with subsequent stripping via a negative potential scan.
e.g. 2I-^ + 2Hg -2e = Hg 2 I 2 at a Hg electrode Deposition step Hg 2 I 2 + 2e = 2 Hg + 2I-^ cathodic stripping, reduction Trace analysis (enhanced sensitivity) can be realized since sample analytes are preconcentrated from a large-volume dilute solution into (onto) a small-volume electrode under forced convection.
Stripping Analysis
