Redox Reaction - the basics
Ch.14-16
Electrochemistry
ox1 + red2 <=> red1 + ox2
Oxidizing
Agent
Reducing
Agent
Redox reactions: involve transfer of electrons from one
species to another.
Oxidizing agent (oxidant): takes electrons
Reducing agent (reductant): gives electrons
Redox Reaction - the basics
Reduced
Oxidized
ox1 + red2 <=> red1 + ox2
Oxidizing
Agent
Balance Redox Reactions (Half Reactions)
Reducing
Agent
Redox reactions: involve transfer of electrons from one
species to another.
Oxidizing agent (oxidant): takes electrons
Reducing agent (reductant): gives electrons
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.)
Fe3+ + V 2+ → Fe 2+ + V 3+
Example: Balance the two half reactions and redox
reaction equation of the titration of an acidic solution of
Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep
purple).
MnO4-(aq) + C2O42-(aq) → Mn2+ (aq) + CO2(g)
16H+(aq)
+ 2MnO4
+
→
2Mn2+(aq) + 8H2O(l) + 10CO2(g)
-(aq)
5C2O42-(aq)
Example: Balance
Sn2+ + Fe3+ <=> Sn4+ + Fe2+
Fe2+ + MnO4- <=> Fe3+ + Mn2+
Important Redox Titrants and the Reactions
Oxidizing Reagents (Oxidants)
(1) Potassium Permanganate
−
MnO 4 + 8 H + + 5e − → Mn 2 + + 4 H 2 O
−
MnO4 + 4 H + + 3e − → MnO2 ( s ) + 2 H 2O
MnO + e → MnO
−
4
2−
−
4
1
Important Redox Titrants and the Reactions
Important Redox Titrants and the Reactions
Oxidizing Reagents (Oxidants)
Oxidizing Reagents (Oxidants)
(2) Potassium Dichromate
(3) Potassium Iodate
Cr2O7
2−
Cr2 O 7
2−
+ 14 H + + 6e − → 2Cr 3+ + 7 H 2O
+ 3U 4+ + 2H + → 2Cr 3+ + 3UO 2
−
IO3 + 6 H + + 5e − →
2+
+ H 2O
Important Redox Titrants and the Reactions
Reducing Reagent ( Reductants )
1
I 2 + 3 H 2O
2
Important Redox Titrants and the Reactions
Reducing Reagent ( Reductants )
(1) Potassium Iodide
(2) Sodium Thiosulfate
2S2O32-
1
I → I 2 + e−
2
−
Galvanic Cells - Components
S4O62- +2e¯
Galvanic Cells - Line Notation
A galvanic (voltaic) cell uses a
spontaneous chemical reaction
to generate electricity.
– Electrodes (cathode
and anode).
– Salt bridge: cations
move from anode to
cathode, anions move
from cathode to
anode.
Line notation
Cd(s) | Cd(NO3)2 (aq) || AgNO3 (aq) | Ag(s)
Phase boundary
Salt bridge
Phase boundary
2
Standard Electrode Potentials
Standard Reduction (Half-Cell) Potentials
Standard hydrogen electrode (S.H.E.)
Standard reduction potential (Eo) is the voltage associated
with a reduction reaction at an electrode when all solutes are
1 M and all gases are at 1 atm. Cathode
Reduction Reaction
2H+ (1
M ) + 2eEo =
H2 (1 atm)
0V
• The S.H.E. is the cathode. It consists of a Pt electrode in a
tube placed in 1 M H+ solution. H2 is bubbled through the
tube.
• For the S.H.E., we assign
2H+ (aq, 1M) + 2e- → H2 (g, 1 atm)
• E°red of zero.
• The potential of a cell can be calculated from standard
reduction potentials:
E °cell = E °red (cathode ) − E °red (anode )
Standard Potentials
Standard Reduction (Half-Cell) Potentials
We use hydrogen (S.H.E.)
• Consider Zn(s) → Zn2+(aq) + 2e-. We measure Ecell
relative to the S.H.E. (cathode):
E°cell = E°red(cathode) - E°red(anode)
0.76 V = 0 V - E°red(anode).
• Therefore, E°red(anode) = -0.76 V.
• Standard reduction potentials must be written as
reduction reactions:
Zn2+(aq) + 2e- → Zn(s), E°red = -0.76 V.
H +(aq) + e − ↔ 1/2H 2 ( g )
E ° ≡ 0.000
We can measure Eº for other half-reactions, relative to
the hydrogen reaction, e.g. for silver:
Ag +(aq) + e − ↔ Ag(s)
E ° = 0.799
Standard reduction potentials are listed in Ap. H in
your book. Eº for the H2 reaction is for the reaction
at 25º C
Nernst Equation for a Half-Reaction
• The Nernst Equation relates the potential of the half
reaction to reagent concentrations
• For the half-reactions:
aA + ne ↔ bB
-
E
E = E° −
°
aA + bB + ne - ↔ cC + dD
E = E° −
E°
RT
ln Q
nF
E = E° −
RT [B]
ln
nF [A]a
b
RT [C] [D]
ln
nF [A ]a [B]b
c
d
# of moles of
electrons
3
Nernst Equation for a Half-Reaction
Nernst Equation for a Complete Reaction
At 298K (25oC)
ox1 + red2 <=> red1 + ox2
E = E° −
0.05916V
log Q
n
E = E+ − E−
Example (Net Reaction)
Example (Nernst)
• Write the Nernst equation for the reduction of
phosphoric acid to solid white phosphorous:
H 3PO 4 + 5H + + 5e - ↔ 1/4P4 ( s) + 4H 2O
0.05916
1
E = −0.402 −
log
5
[H 3PO 4 ] H +
E ° = -.402
Cd + + 2e − ↔ Cd( s)
[ ]
5
• Note that multiplying the reaction by any factor
does not affect Eº or the calculated E:
2H3 PO 4 + 10H + + 10e- ↔ 1/2P4 ( s) + 8H 2 O
0.05916
1
E = −0.402 −
log
10
[H 3PO 4 ]2 H +
E ° = -.402
[ ]
10
Determination of the Equivalence Point
Aox + Bred <=> Ared + Box
At equivalence point, Ecell=0:
°
EA −
0.0592
[A ]
0.0592
[B ]
°
log red = EB −
log red
nA
[ Aox ]
nB
[ Box ]
Ared = Box and Bred = Aox at the equivalence point
E=
o
n A E A + nB E B
n A + nB
o
• Find the voltage for the Ag-Cd cell and state if the reaction is
spontaneous if the right cell contained 0.50 M AgNO3(aq) and if
the left contained 0.010 M Cd(NO3)2(aq)
• 1) 2Ag + + 2e − ↔ 2Ag( s)
E+° = 0.799 V
Valid for simple
Redox expressions
E−° = −0.402 V
• 2) E+ = 0.799 − 0.05916 log 1
= 0.781 V
2
[0.50]2
• 3) E = −0.402 − 0.05916 log 1
= −0.461V
−
[0.010]
2
• 4)
E = E+ − E− = 0.781 − ( −0.461) = +1.242 V
• 5)
+
−
2Ag + 2e ↔ 2Ag( s)
Cd( s) + 2Ag + ↔ Cd 2+ + 2Ag( s)
Cd + + 2e − ↔ Cd ( s )
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 HClO4.
Potential
of
saturated
calomel electrode is 0.241
V.
4
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 HClO4. Potential of
saturated calomel electrode is 0.241 V.
Titration reaction: Fe2+ + Ce4+ <=> Ce3+ + Fe3+
Fe3+ +
e- <=>
Fe2+
Ce4++ e- <=> Ce3+
Eo =
0.767 V
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
E = E° −
0.0592
[ Fe 2+ ]
log
− 0.241
n
[ Fe 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 HClO4. 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
V M −V M
[ Fe ] = Fe Fe Ce Ce
VFe + VCe
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 HClO4.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+ ] =
=
VCe M Ce
VFe + VCe
(15.00mL)(0.1000M )
= 2.308 ×10 −2 M
(50.00 + 15.00)mL
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 HClO4. 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
[ Fe3+ ] = 2.308 × 10 −2 M
2+
=
(50.00mL)(0.0500M ) − (15.00mL)(0.1000M )
= 1.54 ×10 −2 M
(50.00 + 15.00)mL
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 HClO4. 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
4+
3+
Ce + e <=> Ce
Eo = 1.70 V
E=
=
nFe EFe + nCe ECe
− 0.241
nFe + nCe
0.767 + 1.70
− 0.241 = 1.23 − 0.241 = 0.99 V
1+1
E = E° −
[ Fe 2+ ] = 1.538 ×10 −2 M
0.0592
[ Fe 2 + ]
log
− 0.241
n
[ Fe3+ ]
= 0.767 −
0.0592
1.54 ×10 −2
log
− 0.241 = 0.533V
1
2.038 ×10 −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 HClO4. 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
[Ce3+ ] =
=
VFe M Fe
VFe + VCe
(50.00mL)(0.0500 M )
= 3.29 ×10 − 2 M
(50.00 + 26.00) mL
5
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 HClO4. Potential of
saturated calomel electrode is 0.241 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 HClO4. Potential of
saturated calomel electrode is 0.241 V.
Fe2+ + Ce4+ <=> Ce3+ + Fe3+
Fe2+ + Ce4+ <=> Ce3+ + Fe3+
At 26.00 mL of Ce4+ added, VFeMFe < VCeMCe,
After equivalence point
At 26.00 mL of Ce4+ added, VFeMFe > VCeMCe
After equivalence point
Fe3+ + e- <=> Fe2+
Half-reactions:
Ce4++ e- <=> Ce3+
[Ce3+ ] = 3.29 × 10 −2 M
V M −V M
[Ce 4+ ] = Ce Ce Fe Fe
VFe + VCe
=
(26.00mL)(0.1000 M ) − (50.00mL)(0.0500M )
= 1.32 × 10 −3 M
(50.00 + 26.00)mL
[Ce3+ ] = 3.29 × 10 −2 M
E = E° −
= 1.70 −
Eo = 0.767 V
Eo = 1.70 V
[Ce 4+ ] = 1.32 × 10 −3 M
0.0592
[Ce 3+ ]
log
− 0.241
n
[Ce 4+ ]
0.0592
3.29 ×10 −2
log
− 0.241 = 0.63 V
1
1.32 ×10−3
Galvanic Cells
A galvanic (voltaic) cell uses a spontaneous
chemical reaction to generate electricity.
Theoretical curve for
titration of 100.0 mL
of 0.50 0 M Fe2+ with
0.100 M Ce4+ in 1 M
HClO4. (p.330-331)
– Electrodes (cathode and
anode).
– Salt bridge: cations
move from anode to
cathode, anions move
from cathode to anode.
– produces electricity
when the cell reaction
is not at equilibrium.
0.0592
[ Fe 2+ ]
log
− 0.241
[ Fe 3+ ]
n
0.0592
= 0.767 −
log 1 − 0.241 = 0.526 V
1
E = E° −
Free Energy and Electrochemical Reaction
E and Equilibrium Constants
• The free energy change, ∆G, for a chemical reaction at
constant T, P equals the maximum possible electrical work
that can be done by the reaction on its surroundings.
• A galvanic cell produces electricity because the cell
reaction is not at equilibrium
• At equilibrium, E for the net reaction can thus be
related to K
aA + ne - ↔ cC
E+°
Work = − ∆G = E × q
∆G = − nFE
• A spontaneous reaction (∆G<0) E>0
• Think: how the equilibrium constant can be related to free
energy?
dD + ne - ↔ dD
E = E+ − E− = E+° −
E = E° −
E-°
.05916
.05916
[C]
[B]
log a − E−° −
log d
n
n
[A]
[D]
c
b
0.05916
0.05916
[C] [D]
log a b = E ° −
log Q
n
n
[A] [B]
E = 0 → E° =
c
0.05916
log K
n
d
at 25°C
K = 10
at 25°C
nE °
0.05916
at 25° C
When the cell is at equilibrium, E=0 and Q=K
6
Cells as Chemical Probes
Example: Find the equilibrium constant
3+
2+
Cu ( s ) + 2Fe ↔ Cu + 2Fe
Two Equilibria:
2+
(1) Equilibrium between two half-cells
(2) Equilibrium within each half-cell
AgCl ( s) + e − ↔ Ag (s ) + Cl - (aq,0.10M )
E+° = 0.222 V
This can be divided into two half reactions:
2Fe 3+ + 2e − ↔ 2Fe 2+ ( s)
2 H + ( aq, ? M ) + 2e - ↔ H 2 ( g ,1.00bar )
E+° = 0.771 V
Cu 2+ + 2e- ↔ Cu (s )
•The half-reaction that you
write must involve species that
appear in two oxidation states
in the cell.
E o- = 0.339 V
E ° = E+° − E−° = 0.771− 0.339 = 0.432 V
K = 10
nE °
0.05916
•The reaction is in equilibrium
in the right cell is not the net
cell reaction:
(2)(0.432)
= 10 ( 0.05916) = 4 ×1014
AgCl( s ) ↔ Ag + (aq) + Cl − ( aq)
Important Biochemical Reactions
Survival Tips
Electrochemistry, chemical equilibrium, solubility,
complex formation, and acid-base chemistry
(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.
E o- = 0 V
• Formal potential,
Eº’, meant to
define potentials
under conditions
of biochemistry
Ex. p.286-287
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 halfreaction 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 halfcell 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
7
Reference Electrodes
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.
Reference Electrode
Fe3+ + e - ↔ Fe 2+
AgCl ( s ) + e ↔ Ag ( s) + Cl
−
[Fe ]
[Fe ]
= 0.222 - .059log[Cl ]
The entire left half-cell containing
appropriate solutions and a salt bridge.
°
-
E = 0.222 V
2+
E+ = 0.771 − .059 log
E-
Indicator Electrode
E+° = 0.771 V
-
3+
-
Fe 3+ + e - ↔ Fe 2 +
E = E+ − E-
E+° = 0.771 V
AgCl( s) + e ↔ Ag( s ) + Cl −
-
Silver-Silver Chloride Electrode
E-° = 0.222 V
Saturated Calomel Electrode (S.C.E.)
Ag | AgCl Electrode :
AgCl( s) + e - ↔ Ag ( s) + Cl−
E-° = +0.222 V
w/ saturated KCl
E = +0.197 V
Hg | Hg2Cl2 Electrode:
• The difference in E is due to activity
coefficients.
• A double junction electrode: avoid to
mix Cl- with the analyte
Voltage Conversions between Different
Reference Scales
1/2Hg 2 Cl 2 ( s) + e - ↔ Hg(l ) + Cl −
E ° = +0.268 V
w/ saturated KCl
E = +0.241 V
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
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.?
– 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
- 0.351V + 0.241V = -0.110V
8
Indicator Electrodes-Metal electrodes
• Metal electrodes: develop potential in response to a redox reaction on
their surface
• Inert metal indicator electrode: allows electron transfer to/from the
solution but not participating in reactions, e.g., Pt (the most common
one), Au.
• Silver Indicator Electrode
Ag + ( s ) + e- ↔ Ag( s )
E+° = 0.799 V
1/2Hg 2Cl 2 ( s ) + e - ↔ Hg (l ) + Cl −
E−° = 0.241 V
1
E = E+ − E− = 0.799 − 0.059 log
Ag +
[ ]
− 0.241
[ ]
E = 0.558 + 0.059 log Ag +
Junction Potential
Example: 100.0 mL solution containing 0.100 M NaCl
was titrated with 0.100 M AgNO3 and monitored with a
S.C.E. What voltage reading would be observed after 65.0
mL?
+
−
Ag + Cl → AgCl( s )
Ve = 100.0 mL
[ ]
E = 0.558 + 0.05916log Ag +
[Cl ] = (0.350)(0.100 M) 100.0
165.0
−
K
1.8 ×10
[Ag ] = [Cl
] = 0.021
+
sp
−
−10
= 0.021 M
= 8.5 ×10 −9
E = 0.558 + 0.05916 log (8.5 ×10-9 ) = 0.081 V
Ion Mobilities and Liquid Junction Potentials
• Junction potential: A voltage difference develops whenever
dissimilar electrolyte solutions are in contact.
• Happens at the salt bridge/solution interface since different ions
have different mobilities in water.
• A major (fundamental) source of error in a potential measurement.
Ion-Selective
Electrodes
(ISE)
Glass Electrode (pH Combination Electrode)
• The most common ISE
• A glass membrane
selectively binds H+
• Two Ag|AgCl reference
electrodes measure the
potential difference across
the glass membrane
• An ion-exchange
equilibrium is on the
surface of the glass
membrane
• Responds
selectively to one
ion using an ionselective membrane
• Do not involve
redox reactions
• The electric
potential across the
membrane depends
on [analyte]
• E is measured by
two reference
electrodes
E = constant +
0.05916
log [C
n
+
]
Out
at 25°C n = charge of analyte
Ag(s) | AgCl(s) | Cl - (aq) || H + (aqout ) | H + (aqin ), Cl − (aq) | AgCl( s ) | Ag( s)
9
Errors in pH Measurement
The Glass Membrane of a pH Electrode
•Cross section of the glass
membrane of a pH electrode.
1.
2.
3.
4.
5.
•H+
can diffuse into the
membrane to replace the
metal ions through binding
to oxygen in glass (ionexchange equilibrium).
6.
7.
Response of the electrode :
E = constant + β (0.0592) log[H ]
E = constant − β (0.0592) pH at 25 C
+
Out
O
Out
8.
9.
β: ~1.00, electromotive
efficiency measured during
calibration.
Specifications for Electrochemical
Techniques
Selectivity Coefficient
k A,X
response to X
=
response to A
k
K + , Na +
= 1× 10
k
K + , Cs +
= 0.44
K + , Rb +
= 2.8
Advantages
• The selectivity
coefficient: the
relative response of
the electrode to
different species
−5
Linear response to analyte over wide dynamic range
Nondestructive
Short response times
Unaffected by color/turbidity (limited matrix effects)
Cheap
• The smaller k is, the
less interference there
is due to ion X
Response of ion - selective electrode :
k
E = constant ± β (
[
0.0592
) log a +
n
A
x
]
(k a )
A,X
x
Calibration standards (±0.01 pH)
Junction potential (~0.01 pH)
Junction potential drift (recalibrate every 2 hrs)
Sodium error (when [H+] is low and [Na+] is high)
Acid error (strong acid, the glass surface is saturated
with H+)
Equilibration time (~30s with adequate stirring)
Hydration of glass (A dry electrode requires several
hours of soaking)
Temperature (calibrate at same T as measurement)
Cleaning (remove hydrophobic liquid)
Disadvantages
Sensitivity (High detection limits)
Not universal
at 25 C
O
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.
ip
•
[A]
Calibration Curve
10
Stripping Analysis
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 = Hg2I2
at a Hg electrode
Deposition step
Hg2I2 + 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.
11