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Electrode Potentials
In a redox reaction, electrons are transferred between the reacting chemicals.
Zinc reacts with aqueous copper (II)
ions in a redox reaction:
Zn (s) + Cu2+ (aq)  Zn2+ + Cu (s)
This is an electrochemical cell. It contains
two half cells, within each of which oxidation
and reduction take place separately. The
circuit is completed using a wire which
transfers electrons between the half cells
and a salt bridge which transfers ions
between the half cells.
At the negative
electrode, electrons
are released and flow
through the wire.
Therefore, oxidation
occurs at the
negative electrode;
the zinc pushes its two
electrons into the
wire.
This can be split into two half equations for oxidation
and reduction:
Zn  Zn2+ + 2e- (oxidation)
Cu 2+ + 2e-  Cu (reduction)
A high resistance voltmeter is used to measure the potential
difference between the two half cells. The high resistance of
the voltmeter ensures that current flow is minimised so that
the potential difference is at its maximum (known as the
electromotive force).
At the positive
electrode, electrons
are accepted and
flow into the
electrode. Therefore,
reduction occurs at
the positive
electrode; the
copper accepts the
two electrons from
the zinc.
Types of Half Cell
1. Metal in contact with a solution of its ions
Metal electrode in contact with a solution containing ions of
the same metal. Examples include copper, tin and zinc;
Cu 2+ + 2e-  Cu; Sn 2+ + 2e-  Sn; Zn 2+ + 2e-  Zn
2. Gas in contact with a solution of its ions
Platinum inert electrode in contact with a solution of the ions of the
gas present. Examples include the standard hydrogen electrode and
chlorine:
2H+ + 2e-  H2; Cl2 + 2e-  2Cl-
3. Ions of different oxidation states in solution
An inert platinum electrode is used to allow electrons to pass into the
half cell. The concentrations of the different ions must be the same,
1.0 moldm-3 . Examples of this include iron (III) and iron (II), copper (II)
and copper (I); non metal examples include iodine and iodide,
bromine and bromide, and manganate (VII) and manganese (II).
Standard Electrode Potentials
A hydrogen half cell is used as
the standard measurement of
electrode potentials.
The hydrogen half cell
comprises:
• 1.0 moldm-3
hydrochloric acid
as the source of
H+ ions
• Supply of
Hydrogen gas
• Inert platinum
electrode on
which the half
reaction takes
place.
2H+
+
2e-
 H2
The conditions must be
controlled:
Temperature: 298 K
Solution concentration: 1.0
moldm-3
Pressure of gas: 100kPa
The standard electrode
potential of a half cell, Eθ,
is the e.m.f. Measured
when the half cell is
connected to a standard
hydrogen half cell under
standard conditions of
298K, 1moldm-3 solutions
and 100kPa gas pressure.
The standard electrode potential of a half cell represents its contribution to the cell
e.m.f. The contribution made by the hydrogen half cell to the cell e.m.f. is defined as 0 V.
The sign of the standard electrode potential of a half cell indicates its polarity compared
to the hydrogen half cell.
Calculating Standard Cell Potentials
The standard cell potential is calculated by:
Eθcell = Eθpositive – (Eθnegative)
Eθcell = 0.80 - (-0.76)
= 1.56 V
Determining the redox equation for the cell reaction
Ag+ + e-  Ag
Eθ = 0.80 V
Fe2+ + 2e-  Fe Eθ = -0.44 V
Silver is most positive – therefore it stays
the same as this is reduction. The iron
equation gets reversed – it donates the
electrons and is therefore oxidation. We
then balance by multiplying the silver
equation by 2 to get;
2Ag+ + Fe  2 Ag + Fe2+
1. Decide which is the most positive half
equation. This remains unchanged – it is
reduction.
2. The most negative half equation is reversed
so that it is clear to see that this is the one
which provides the electrons, and proceeds
to the left – this is oxidation.
3. Balance for electrons and add the two half
equations.
The Electrochemical Series
Note: Equations in the electrochemical series are always given as reduction equations, with the electrons on the left hand side.
Example:
Mg2+ 2e-  Mg (E θ =-2.37)
Cu2+ 2e-  Cu (E θ =0.34)
The position of the
magnesium equilibrium will
lie further to the left because
the value is more negative.
The Mg more readily loses
electrons, and is therefore
the stronger reducing agent.
The more positive the Eθ value,
the further the equilibrium lies to
the right: the more readily the
species gains electrons, and
therefore the easier it itself is
reduced.
The more positive the Eθ value, the more likely the reaction is
to occur with a flow of electrons from left to right.
Strongest
Reducing
Agent
Increasing tendency to
accept electrons,
decreasing tendency to
release electrons.
The more negative the Eθ value,
the further the equilibrium lies to
the left: the more readily the
species loses electrons, and
therefore the easier it itself is
oxidised.
Strongest
Oxidising
Agent
Explaining Reaction Feasibility
An oxidising agent can oxidise a
reducing agent with a more
negative E θ value.
Oxidising Agents
Cr2O72- + 6H+ + 6eAg+ + e-
Reducing Agents
2Cr3+ + 7H2O E θ = +1.33v
Ag
E θ = +0.80v
Example: Can acidified dichromate oxidise Ag?
Answer: Acidified Cr2O72- can oxidise Ag because:
 Acidified chromium is a more powerful oxidising agent that Ag+
 E θ for the reduction of acidified chromium is more positive than the E θ for the reduction of
Ag+
 The half cell reaction proceeds in the direction
Cr2O72- + 6H+ + 6e-  2Cr3+ + 7H2O
A reducing agent can reduce an
oxidising agent with a more
positive E θ value.
Oxidising Agents
Reducing Agents
MnO4- + 8H+ + 5eCr2O72- + 14H+ + 6e-
Mn2+ + 4H2O E θ = +1.52
2Cr3+ + 7H2O E θ = +1.33v
Example: Can Cr3+ reduce acidified MnO4-?
Answer: Cr3+ can reduce acidified manganate because:
 Cr3+ is a more powerful reducing agent than Mn2+
 E θ for the reduction of acidified chromium is more negative than the E θ for the reduction
of MnO4 The half cell reaction proceeds in the direction
2Cr3+ + 7H2O  Cr2O72- + 14H+ + 6e-
Effect of Concentration on Cell Voltage
For changes at the negative electrode:
Zn


For changes at the positive electrode:

Zn2+ + 2eIf we move this
equilibrium to the right,
by increasing the
concentration of Zn or
decreasing the
concentration of Zn2+ , the
negative electrode will
become more negative,
because Zn is more able
to donate electrons.
If the negative electrode
becomes more negative,
the cell voltage increases.
 If we move this equilibrium to the left, by
increasing the concentration of Zn2+ or
decreasing the concentration of Zn , the
negative electrode will become more positive –
because less electrons are produced.
 If the negative electrode becomes less
negative, the cell voltage decreases.
Cu2+ + 2e-
Cu

If we move this
equilibrium to the right,
by reducing the
concentration of Cu or
increasing the
concentration of Cu2+ ,
the positive electrode
will become more
positive, because Cu is
more able to accept
electrons.
If the positive
electrode becomes
more positive, the cell
voltage increases.
 If we move this equilibrium to the left, by
increasing the concentration of Cu or
decreasing the concentration of Cu2+ , the
positive electrode will become less positive –
because Cu2+ is more able to donate electrons.
 If the positive electrode becomes less
positive, the cell voltage decreases.
Fuel Cells
Acid Membrane Hydrogen Fuel Cell
Alkali Membrane Hydrogen Fuel Cell
Negative electrode: 2H+ + 2eH2
Positive Electrode: ½ O2 + 2H+ + 2e2H2O
Overall: H2 + ½ O2 H2O
Positive: ½ O2 + H2O + 2e2OHNegative: 2H2O + 2eH2 + 2OHOverall: H2 + ½ O2 H2O
At the negative electrode, hydrogen gas is
oxidised to form H+ ions and electrons. The
electrons flow in the external circuit. The
hydrogen ions diffuse through a permeable
membrane to the positive electrode where they
react with oxygen forming water – the oxygen is
reduced.
At the negative electrode, hydrogen is oxidised.
At the positive electrode, oxygen is reduced. The
OH- ions produced from the reduction of oxygen
diffuse across a partially permeable membrane
and react with the Hydrogen to form water at the
negative electrode.
Fuel Cells
Hydrogen gas is difficult to store. Possible methods of storage include:
1) as a liquid under pressure or low temperature
2) Adsorbed on the surface of a solid material
3) Absorbed within a solid material
Fuels used in
modern cars
include
hydrogen and
other hydrogen
rich fuels such
as methanol or
ethanol which
are converted
into hydrogen
gas by an on
board reformer.
Advantages of Fuel Cell
Vehicles
1. Less pollution as water is
the only product
2. No carbon dioxide is
produced
3. Greater efficiency as fuel
cell vehicles can be up to
60% efficient compared
to 22% for a diesel
vehicle.
Absorption – H2 molecules
Adsorption – H2 molecules attach dissociate into H atoms which
are incorporated as hydrides
to the surface of a material
within a solid lattice.
Disadvantages of Fuel Cell Vehicles
1. Storing and transporting hydrogen
is difficult
2. The adsorber or absorber in fuel
cells have limited lifetimes and high
production costs
3. Toxic chemicals are used in the
production of fuel cells.
The Hydrogen Economy is the use of hydrogen as a major fuel. It may be a fuel for
the future, but there are a number of factors to consider: hydrogen is difficult to
handle and store; hydrogen has not gained general acceptance as a fuel from the
public; the initial manufacture of hydrogen (by electrolysis form water or other
sources) requires energy; there is currently no infrastructure designed to supply
hydrogen as a fuel.