2.7 Redox Reactions

1
Version 3.1
Chemistry NCEA L3 3.7 Redox
Gaze
SJ
2013
2
Achievement Criteria
This achievement standard involves describing
oxidation-reduction processes. 1 Processes
include reactions in electrochemical and
electrolytic cells, This includes the use of
reduction potentials and spontaneity of
reactions. Knowledge of preferential discharge in
electrolytic cells is not required. 2
Calculations are limited to include the use of
oxidation numbers, mole ratios and electrode
potentials. 3 Knowledge of appearance of the
following reactants and their products is
required. Oxidants will be limited to O2, Cl2,
I2, Fe3, dilute acid (with metals), H2O2, MnO4
(reacting in acidic, basic or neutral
conditions), Cu2 , Cr2O72 / H, OCl,
concentrated HNO3, IO3, MnO2. Reductants will
be limited to metals, C, H2, Fe2, Br, I, H2S,
SO2, (SO32 , HSO3-), S2O32, H2O2,
H2C2O4. Appropriate information relating to other
oxidants or reductants will be provided. Standard
reduction potentials will be included where
required.
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Redox Terms

• A redox reaction is where one substance is
  oxidised and the other substance is reduced.

Oxidation Reduction
gtloss of electrons gtloss of hydrogen gtgain of oxygen gtgain of electrons gtgain of hydrogen gtloss of oxygen
Oxidation numbers are used to determine what is
oxidised and what is reduced in a reaction.
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Electron transfer

• An Iron nail left in copper sulfate
• Fe(s) Cu(aq)
  Fe(aq) Cu(s)

Copper is reduced gained electrons
Oxidising agent (oxidant)
2
2
Iron is oxidised lost electrons Reducing
Agent (reductant)
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Oxygen transfer

• Iron Ore smelting
• 2Fe2O3(s) 3C(s)
  4Fe(s) 3CO2(g)

Iron oxide is reduced lost oxygen
Oxidising agent (oxidant)
carbon is oxidised gained oxygen Reducing
Agent (reductant)
Iron Ore
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Hydrogen transfer

• Sulfur production
• 2H2S(g) O2(g)
  2S(s) 2H2O(l)

Hydrogen sulphide is oxidised lost hydrogen
Reducing Agent (reductant)
Oxygen gas is reduced gained hydrogen
Oxidising agent (oxidant)
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Half Equations

• A balanced redox equation is broken into two
  half-equations, to show how electrons are
  transferred.

Fe(s) Cu(aq)
Fe(aq) Cu(s)
2
2
Reduction half equation - oxidant is reduced to
a product Fe
Fe 2e
Oxidation half equation reductant is
oxidised to a product Cu 2e
Cu
2
-
-
2
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Summary of terms
Reductant Oxidant
gtreducing agent gtis oxidised gtloses electrons hydrogen gtgains oxygen gtoxidising agent gtis reduced gtgains electrons hydrogen gtloses oxygen
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Oxidants L3
Name Conditions Half equation / colour change /ON Name
permanganate (manganate(VII)) acidified MnO4- 8H 5e- ? Mn2 4H2O manganese (II) purple (7) colourless (2) ion
permanganate (manganate(VII)) neutral MnO4- 4H 3e- ? MnO2 2H2O manganese purple (7) brown (s) (4) dioxide
permanganate (manganate(VII)) alkaline MnO4- e- ? MnO42- manganate purple (7) green (6)
Dichromate (dichromate(VI)) acidified Cr2O72- 14H 6e- ? 2Cr3 7H2O chromic (III) ion orange (6) green (blue) (3)
Hydrogen peroxide acidified H2O2 2H 2e- ? 2H2O water colourless (-1) colourless (-2)
chlorine Cl2 2e- ? Cl- chloride ion pale yellow/green (0) colourless (-1)
Hypochlorite (chlorate (I)) alkaline ClO- H2O 2e- ? Cl- 2OH- chloride ion colourless (1) colourless (-1)
iodine I2 2e- ? 2I- iodide grey (s) (0) colourless (aq)(-1) ion
triiodide
I3- ? I-
I2 iodide Purple brown
(-1) colourless (aq)(-1)
ion
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Oxidants L3
Name Conditions Half equation / colour change /ON Name
chlorate (chlorate(V)) acidified ClO3- 6H 6e- ? Cl- 3H2O chloride colourless (5) colourless (-1) ion
Nitric acid (nitrate ion) concentrated NO3- 2H e- ? NO2 H2O nitrogen colourless (5) brown (g) (4) dioxide
Iron (III) ion Fe3 e- ? Fe2 Iron (II) ions orange (3) pale green (3)
Copper (II) ion Cu2 e- ? Cu copper (I) ion blue (2) white (1)
oxygen O2 4e- ? 2O2- oxide ion colourless (0) colourless (-2)
Dilute acid (H ions) 2H 2e- ? H2 hydrogen gas Colourless (1) colourless (0)
Manganese dioxide acidified MnO2 2H 2e- ? Mn2 2H2O manganese colourless (4) colourless (2) ion
iodate IO3- 2e- ? I2 iodine Colourless (aq) (5) grey (s)(0)
bromate
BrO3- 2e- ? Br-
bromide
ion Colourless (aq) (5)
colourless (-1)
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Reductants L3
Name Half equation / colour change /ON Name
hydrogen H2(g) ? 2H(aq) 2e- colourless (0) colourless (1) hydrogen ion
zinc Zn(s) ? Zn2 2e- grey (0) colourless (2) zinc (II) ion
Iron (II) ion Fe2(aq) ? Fe3(aq) e- pale green (2) orange (3) Iron (III) ion
oxalate ion (ethanediotae (COO-)2) C2O42-(aq) ? 2CO2(aq or g) 2e- colourless (3) colourless (4) carbon dioxide
oxalic acid (ethanedioic acid) H2C2O4(aq) ? 2CO2(aq or g) 2H 2e- colourless (3) colourless (4) carbon dioxide
sulfite ions SO32-(aq) H2O ? SO42-(aq) 2H 2e- colourless (4) colourless (6) sulphate ion
thiosulfate 2S2O32-(aq) ? S4O62-aq) 2e- colourless (2) colourless solution (2.5) tetrathionate
magnesium Mg(s) ? Mg2(aq) 2e- grey (0) colourless (2) magnesium ion
carbon
C(s) 2H2O ? CO2(g)
4H 4e- carbon
black (0)
colourless (4)
dioxide
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Reductants L3
Name Half equation / colour change /ON Name
carbon monoxide CO(g) H2O ? CO2(g) 4H 2e- colourless (2) colourless (4) carbon dioxide
iron metal Fe(s) ? Fe2 2e- silver (0) pale green (2) iron (II) ion
copper metal Cu(s) ? Cu2(aq) 2e- Red/brown (0) blue (2) copper (II) ion
hydrogen sulphide H2S(g) ? S(s) 2H 2e- colourless (-2) yellow ppt (0) sulphur
hydrogen peroxide H2O2(aq) ? O2( g) 2H 2e- colourless (-1) colourless (0) oxygen gas
iodide ion 2I-(aq) ? I2(s) 2e- colourless (-1) grey (0) iodine
bromide ion 2Br-(aq) ? Br2(aq) 2e- colourless (-1) orange (0) bromine
Sulfur dioxide SO2(g) 2H2O ? SO42-(aq) 4H 2e- colourless (4) colourless (6) sulphate ion
carbon
C(s) H2O ? CO(g)
2H 2e- carbon
black (0)
colourless (2)
monoxide
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Oxidation Numbers

• The Oxidation Number (ON) gives the degree of
  oxidation or reduction of an element.
• They are assigned to a INDIVIDUAL atom using the
  following rules.

Elements 0 e.g. Fe H2 Hydrogen atom (not as element) 1 e.g. HCl H2SO4 Except Hydrides -1 e.g. LiH Oxygen atom (not as element) -2 e.g. MnO4- CO2 Except peroxides -1 H2O2
ON
ON
ON
0
0
1
1
-2
-2
ON
ON
-1
-1
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Oxidation Numbers

• The Oxidation Number (ON) gives the degree of
  oxidation or reduction of an element.
• They are assigned to a INDIVIDUAL atom using the
  following rules.

Elements Each atom 0 e.g. Fe H2 Hydrogen atom (not as element) 1 e.g. HCl H2SO4 Except Hydrides -1 e.g. LiH Oxygen atom (not as element) -2 e.g. MnO4- CO2 Except peroxides -1 H2O2
ON
ON
ON
ON
ON
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Oxidation Numbers
Monatomic ions charge e.g Fe Cl Polyatomic ions Sum of charge e.g. MnO4 Because Total charge -1 And Oxygen -2 7 (4x -2) -1 Molecules Sum of on atoms charge e.g. CO2 Because Total charge 0 And Oxygen -2 4 (2x-2) 0
ON
ON
ON
-
2
-
2
-1
4
-2
-2
7
Mn
O
c
O
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Oxidation Numbers
Monatomic ions charge e.g Fe 3 Cl- Polyatomic ions Sum of charge e.g. MnO4- Because Total charge -1 And Oxygen -2 7 (4x -2) -1 Molecules Sum of zero e.g. CO2 0 Because Total charge 0 And Oxygen -2 4 (2x-2) 0
ON
ON
ON
4
-2
-2
7
Mn
O
C
O
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Oxidation Numbers

• Oxidation is a loss of electrons
• and causes an increase in

Reduction is a gain of electrons and causes an
decrease in
ON
ON
Oxidation of Fe Fe Fe
e- Fe has increased
(2 to 3) caused by a loss of
electrons e-
2
Reduction of MnO4 MnO4 5e-
Mn Mn has decreased (7 to
2) caused by a gain in electrons e-
-
3
2
2
-
3
2
7
2
ON
ON
OXIDATION and REDUCTION always occur together.
The electrons lost by one atom are gained by
another atom. This is called a REDOX
reaction. Join the two half equations together.
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Using ON to identify Redox Reactions
What has been oxidised and what has been
reduced? STEP ONE write the ON for each
atom STEP TWO Identify the atom that has had
its ON increased. It is Oxidised C has increased
ON (0 to 4) so C is Oxidised. STEP THREE
Identify the atom that has decreased ON. It is
reduced. Fe has decreased ON (3 to 0) so Fe is
Reduced.
3
-2
0
0
4
-2
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Balancing half equations
1. Write half equation by identifying reactant and product 2. Balance atoms that are not O or H 3. Balance O by adding H2O and H by adding H 4. Balance charge by adding electrons

Balance the half equation for the oxidation of Fe
to Fe
3
2
Fe Fe Atoms already balanced There are no O or H atoms to balance Fe Fe e 2 3 (-1) 1 electron to balance charge
3
2
3
2
Fe Fe e
2
3
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Balancing half equations
1. Write half equation by identifying reactant and product 2. Balance atoms that are not O or H 3. Balance O by adding H2O and H by adding H 4. Balance charge by adding electrons

-
Balance the half equation for the reduction of
MnO4 to Mn
2
-
-
-
MnO4 Mn Atoms already balanced MnO4 8H Mn 4H2O Balance O by adding 4H2O and H by adding 8 H MnO4 8H Total charge 7 Mn 4H2O Total charge 2 Add 5 electrons (e-)
2


2
2
-
-
MnO4 8H 5e
Mn 4H2O
2

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Balancing half equations
Rules e.g. Cr2O72- ? Cr3 1. Assign
oxidation numbers and identify element oxidised
or reduced.
(6)(-2) (3) Cr2O72- ? Cr3 2.
Balance atom no. for element oxidised or reduced
(other than oxygen and hydrogen)
Cr2O72- ? 2Cr3 3. Balance the Oxygen using H2O
Cr2O72-
? 2Cr3 7H2O 4. Use H (acidic conditions) to
balance the hydrogen 14H
Cr2O72- 6e- ? 2Cr3 7H2O 5. Use OH- (in
alkaline conditions) to cancel any H same
amount on both sides 6. Balance charge by
adding electrons (LHS on oxidants RHS on
reductants) 14H Cr2O72-
6e- ? 2Cr3 7H2O 7. Check balance of
elements and charges.
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Electrochemistry
Electrochemistry is the chemistry of reactions
involving the transfer of electrons, redox
reactions. In year 12 the focus was on
electrolytic cells or electrolysis. This
involves a non-spontaneous reaction in which an
external source of electricity provides electrons
with the energy required to bring about a redox
reaction.
In year 13 the focus is on electrochemical cells
in which spontaneous redox reactions use the
energy released from a chemical reaction to
generate electric current. These are called
Galvanic cells or batteries.
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Galvanic Cells and Salt Bridges
Under normal conditions a redox reaction occurs
when an oxidising agent is in contact with a
reducing agent. If the two half reactions are
physically separated, the transfer of electrons
is forced to take place through an external metal
wire. As the reaction progresses a flow of
electrons occurs. This only happens if there is
a full circuit so that there is no net build-up
of charge. To complete this circuit the separate
solutions are connected using a salt bridge which
allows ions to flow and transfer charge.
Typically the salt bridge is a glass tube filled
with a gel prepared using a strong electrolyte
such as KNO3(aq) (which contains ions that do not
react with the electrodes or species in the
solutions. The anions (NO3-) and cations (K)
can move through the salt bridge so that charge
does not build up in either cell as the redox
reaction proceeds.
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Galvanic cells
The oxidation and reduction reactions that occur
at the electrodes are called half-cell
reactions. Zn electrode (anode, oxidation)
Zn(s) ? Zn2(aq) 2e? Cu electrode
(cathode, reduction) Cu2(aq) 2e? ?
Cu(s)
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Electromotive Force
The reduced and oxidised substances in each cell
form a redox couple. The 2 couples in this cell
(the Daniel cell) are Zn2Zn and Cu2Cu. By
convention, when writing redox couples, the
oxidised form is always written first.   The
fact that electrons flow from one electrode to
the other indicates that there is a voltage
difference between the two electrodes. This
voltage difference is called the electromotive
force or emf of the cell and can be measured by
connecting a voltmeter between the two
electrodes. The emf is therefore measured in
volts and is referred to as the cell voltage or
cell potential.
A high cell potential shows that the cell
reaction has a high tendency to generate a
current of electrons. Obviously the size of this
voltage depends on the particular solutions and
electrodes used, but it also depends on the
concentration of ions and the temperature at
which the cell operates.
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Cell Diagrams
Galvanic cells can be represented using cell
diagrams. This is a type of short hand notation
that follows a standard IUPAC convention. For
the copper/zinc cell the standard cell diagram
is Zn(s) Zn2(aq) Cu2(aq) Cu(s) The
vertical lines represent phase boundaries and
represents the salt bridge. The cathode
(reduction reaction) is always shown on the right
hand side and the anode (oxidation) on the left
in a standard cell diagram. The electrons thus
move from left to right in the standard cell
diagram, representing a spontaneous redox
reaction. The electrodes are always written in
at the beginning and end of a cell diagram. This
occurs both if the metal is involved in the redox
reaction (as in the Daniel cell above where the
electrodes are the Cu and Zn), and also if an
inert electrode is used. In each half cell the
reactant appears first, followed by the product.
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Cell Diagrams
An inert electrode must be used in cells in which
both species in a redox couple are in aqueous
solution (MnO4- and Mn2). The inert electrodes
are commonly either platinum, Pt(s) or graphite,
C(s) electrodes. Since the two species in the
redox couple are in solution, they are separated
by a comma rather than a vertical
line.   eg Cu(s) Cu2(aq) MnO4?(aq),
Mn2(aq) Pt(s)   The cell diagram shows two
half cells linked. Each half cell consists of
the oxidant, the reductant and the electrode
(which may be the oxidant or reductant). The two
half cells above are Cu(s)Cu2(aq) and
MnO4?(aq), Mn2(aq)Pt(s).
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Standard Electrode Potentials
Under standard conditions (when the pressure of
hydrogen gas is 1 atm, and the concentration of
acid is 1 mol L-1) the potential for the
reduction reaction is assigned a value of
zero. 2H(aq) 2e? ? H2(g) Eo
0.00 V   The superscript o denotes standard state
conditions. When the hydrogen electrode acts as a
cathode, H ions are reduced, whereas when it
acts as an anode, H2 gas is oxidised.
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Standard Electrode Potential
The overall cell voltage is the sum of the
electric potential at each electrode. If one of
the electrode potentials is known, and the
overall cell voltage is measured, then the
potential of the other electrode can be
calculated by subtraction. Clearly it is best if
all electrode potentials are measured relative to
a particular electrode. In this way, a scale of
relative values can be established. The standard
hydrogen electrode (SHE) is used as the standard
reference electrode, and it has arbitrarily been
given a value of 0.00 V.
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Standard Electrode Reduction Potential
For any redox couple, the standard electrode
(reduction) potential is the voltage obtained
under standard conditions when that half-cell is
connected to the standard hydrogen electrode.
For example, the electrode potential of a
Zn2Zn electrode can be measured by connecting
it to a hydrogen electrode.
Experimentally, the more positive terminal is
always where reduction is occurring in a
spontaneous reaction. In example (a) reduction
occurs in the hydrogen electrode (positive
electrode) while oxidation occurs in the Zn2Zn
compartment (negative electrode). The cell
diagram for this electrochemical cell is Zn(s)
Zn2(aq) H(aq), H2(g) Pt(s)
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Standard Reduction Potentials
Using the standard reduction potentials for many
half reactions have been measured under standard
conditions (at 25 oC). Standard reduction
potentials are provided in examinations.   The
table can be used to decide the relative strength
of species as oxidants or reductants. The
species on the left in the couple with the most
positive reduction potential, will be the
strongest oxidising agent or oxidant. E.g it is
F2(g) (NOT F2 / F?). This means F2 has the
greatest tendency to gain electrons. As the
electrode potential decreases, the strength as an
oxidant decreases.   Conversely the strongest
reducing agent or reductant would have the least
positive (or most negative) e.g. Li(s). This
means Li has the greatest tendency to lose
electrons.
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Using reduction potentials to determine Eocell
In any electrochemical cell, the standard cell
potential (voltage), E0cell, is the difference
between the reduction potentials of the two redox
couples involved. The couple with the more
positive reduction potential will be the
reduction half-cell (cathode). This means that
the Eocell for any combination of electrodes can
be predicted using the relationship   Eocell
Eo(reduction half-cell) - Eo(oxidation
half-cell)   OR Eocell Eo(cathode) -
Eo(anode) OR Eocell Eo(RHE)
- Eo(LHE) (where RHE is the right hand
electrode and LHE is the left hand electrode in
the standard cell diagram).
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Using reduction potentials to determine Eocell
For example, consider the cell Zn(s)
Zn2(aq) Ag(aq) Ag(s) Reduction reaction
is Ag(aq) e? ? Ag(s)
Eo(Ag/Ag) 0.80V Oxidation reaction is
Zn(s) ? Zn2(aq) 2e? Eo (Zn2/Zn)
-0.76V   Eocell Eo(Ag/Ag) - Eo (Zn2/Zn)
0.80 - (-0.76) V 1.56V
1. Any metal that is more reactive (lower Eo
value) will reduce the cation of a less reactive
metal because the Eocell value for the reaction
will be positive. 2. A positive standard cell
potential suggests that the reaction occurs
spontaneously from left to right, as shown in a
standard cell diagram. In reality the reaction
may not appear to proceed as it may be very slow
due to a high activation energy. Another reason
a reaction may not proceed is if the surface of
the metal is coated with an oxide. Aluminium
oxide on the surface of aluminium can protect the
aluminium from undergoing a spontaneous oxidation
reaction
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Predicting whether a Reaction will occur
It is possible to use Eo values to predict
whether a reaction will occur. This simply
involves identifying which species must be
reduced and which species must be oxidised if the
reaction is to proceed spontaneously. The
appropriate reduction potentials are then
substituted into the equation. Eocell
Eo(cathode) - Eo(anode) where Eo(cathode) is
the reduction potential for the half cell where
reduction occurs and Eo(anode) is the reduction
potential for the half cell where oxidation
occurs. If the Eocell calculated is positive,
then the reaction will occur spontaneously.
Conversely, a negative cell potential means the
reaction will not proceed.
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Predicting whether a Reaction will occur
Corrosion an everyday application of a redox
reaction Corrosion is the term usually applied
to the deterioration of metals by an
electrochemical process. One example is the
rusting of iron in the presence of water and
oxygen. Although the reactions involved in
rusting are quite complex, the main steps are as
follows. Step 1 - Oxidation occurs at a region
of the irons surface, the anode. Fe(s) ?
Fe2(aq) 2e? Eo red -0.44 V Step 2
- Electrons travel to some other region of the
metals surface where a variety of cathode
reactions can occur. Step 3 - In acidic medium,
atmospheric oxygen is quickly reduced to H2O.
The acidity of the solution can, in part, be due
to the presence of dissolved acidic gases such as
CO2 and SO2.
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Predicting whether a Reaction will occur
Example Can a solution of acidified potassium
permanganate oxidise the Fe2 present in a
solution of iron (II) nitrate? (Note in
questions such as this you will have to recognise
that ions such as sodium and potassium are
spectator ions.) The unbalanced equation for the
reaction would be MnO4? Fe2 ?
Mn2 Fe3 Reduction reaction is MnO4?
? Mn2 Eo (MnO4? /Mn2) 1.51 V Oxidation
reaction is Fe2 ? Fe3 Eo (Fe3/Fe2)
0.77 V Eo cell Eo (MnO4? /Mn2) - Eo
(Fe3/Fe2) 1.51 - 0.77 0.74 V   Since
the cell potential is a positive value (gt0.00 V)
the reaction should proceed. If it had been
negative then the reaction would not proceed.
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Discussion format of E Cells

• Movement of electrons
• from anode (oxidation gt LEO) lower E value
• to cathode (reduction gt GER) higher E value
• 2. Half equations at each electrode
• anode X- ? X e-
• cathode X e- ? X-
• 3. Movement of salt bridge solutions
• Anions ? Anode
• Cations ? Cathode
• 4. Overall summary
• Oxidation at the anode ? salt bridge anions
  balance cations produced at the electrodes
• Reduction at the cathode ? salt bridge cations
  balance cations removed at electrode

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E Cells Summary
CCR
AAO
a t i o n
a t h o d e
e d u c t i o n
n i o n
n o d e
x i d a t i o n
LEO
GER