Title: Lesson 4 Voltaic Cells
Learning Objectives:
– Explain in simple terms how voltaic cells use redox reactions to produce
electricity
– Understand that oxidation occurs at the anode and reduction at the
cathode
– Make a series of voltaic cells in order to better understand the how they
work
Refresh

Consider the following three redox reactions.
Cd(s) + Ni2+(aq) → Cd2+(aq) + Ni(s)
Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s)
Zn(s) + Cd2+(aq) → Zn2+(aq) + Cd(s)
a)
Deduce the order of reactivity of the four metals, cadmium, nickel, silver and zinc
and list in order of decreasing reactivity.
b)
Identify the best oxidizing agent and the best reducing agent.
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Electrochemical Cells

The fact that redox reactions involve transfers of electrons suggests a link
between this type of chemical reactivity and electricity.
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Voltaic Cells

Voltaic cells generate electricity from spontaneous redox reactions…

Consider Zin reducing copper ions…

When the reaction is carried out in a single test tube, the electrons flow
spontaneously from the zinc to the copper ions in the solution, energy is released
in the form of heat (exothermic reaction)

We can organize this reaction so that the energy is released in the form of
electrical energy…

We need to separate the two half reactions…

Into half cells and allowing the electrons to flow between them only through an
external circuit. This is a voltaic or a galvanic cell.
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What is a half cell?
If a rod of metal is dipped into a solution of its own ions, an
equilibrium is set up. For example:
Zn(s)
Zn2+(aq) + 2e-
zinc metal
strip
zinc sulfate
solution
(1 mol dm-3)
This is a half cell and the strip of metal is an electrode. The
position of the equilibrium determines the potential difference
between the metal strip and the solution of metal.
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Electrode potentials

In a zinc half-cell, zinc atoms will form ions by releasing electrons
that will make the surface of the metal negatively charged with
respect to the solution

There will therefore be a charge separation, known as an
electrode potential, between the metal and its ions in solution.
 At the same time, ions in the solutions gain electrons to form Zn atoms, so the
equilibrium exists:
 The position of this equilibrium determines the size of the electrode potential in the halfcell, and depends on the reactivity of the metal.
 Because copper is the less reactive metal, in it’s half-cell the equilibrium position for the
equivalent reaction lies further to the right:
 E.g. It has less tendency to lose electrons compared to zinc. Consequently, there are
fewer electrons to the copper metal strip, so it will develop a higher (or less negative)
electrode potential.
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Cells and electrode potentials
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Two connected half-cells
make a voltaic cell
 The equilibrium for the copper half-cell lies
further to the right than the equilibrium in
the zinc half cell.

If we connect these two half-cells by an external wire, electrons will have a
tendency to flow spontaneously from the zinc half-cell to the copper half-cell
because of their different electrode potentials.

The half-cells connected this way are called electrodes.
ANODE
OXIDATION
-VE CHARGE
CATHODE
REDUCTION
+VE CHARGE
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Key Parts of a Voltaic Cell

Anode




Cathode





REMEMBER
 AnOx


Anode-Oxidation
CaRe

Cathode-Reduction
Connects the metal electrodes in each half cell
Electrons flow from anode to cathode
Salt Bridge




Electrode or ‘half-cell’ where reduction happens
Contains the less reactive metal
The positive electrode: accepts electrons
External circuit


Electrode or ‘half-cell’ where oxidation happens
Contains the more reactive metal
The negative electrode: produces electrons
Contains a neutral salt such as potassium or sodium nitrate as it does not interfere with the reactions at the
electrodes
Made of a tube of jelly or a filter paper soaked in salt solution
Ions diffuse in and out to neutralise build up of charge, maintaining the potential difference.
Voltmeter


Measures the difference in potential between half-cells
Could be replaced with other circuitry to do useful work
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Anions
move in the
salt bridge
from the
cathode to
anode.
Cations
move in the
salt bridge
from the
anode to
cathode.
It opposes
the flow of
electrons in
the external
circuit.
Without a salt bridge, no voltage is generated!
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Combining half cells 1
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Representing half cells: cell diagrams
An electrochemical cell can be represented in a shorthand
way by a cell diagram.
Eө = -0.76 V
Anode
Zn(s) | Zn2+(aq) || Cu(aq) | Cu(s)
Eө = +0.34 V
Cathode
The double vertical lines represents a salt bridge. The single lines
represent a phase change between the solid metal and the aqueous metal
ions. Aqueous solutions are placed next to the salt bridge.
The half cell with the greatest negative potential is on the
left of the salt bridge, so Ecell = Eright cell – Eleft cell. In this
case, Ecell = +0.34 – -0.76 = +1.10 V.
The left cell is being oxidized while the right is being reduced.
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Different half-cells make voltaic cells
with different voltages

Any two half-cells can be connected together to make a voltaic cell.

The direction of electron flow and the voltage generated will be determined by the
difference in the reducing strength of the two metals.

Can be judged by the relative position in the reactivity series.

E.g. If we swap the copper half-cell for a silver half cell…
A larger voltage would be produced because
the difference in electrode potentials will be
greater.
Electrons would flow from the zinc (anode) 
silver (cathode)
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If you now swap zinc for copper…


Electrons would flow from copper
(anode) to silver (cathode)
Copper has a greater reducing power,
it has the lower electrode potential.
From these examples we can summarise:
 Electrons flow from anode to cathode
through the external circuit
 Anions migrate from cathode to anode
through the salt bridge
 Cations migrate from anode to
cathode through the salt bridge
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Another example…

The reaction of Mg with Cu2+ ions:

Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)

This reaction involves two electrons being transferred from the Mg to the Cu:





The Mg reduces the copper ions as it is more reactive
This is an exothermic reaction, and the energy is normally released as heat
A voltaic cell forces each half of the reaction to take place in a separate container, with the
electrons moving through a circuit to get from one side to the next


Mg  Mg2+ + 2eCu2+ + 2e-  Cu
This is an exothermic reaction, where the energy is released as electrical rather than thermal
energy
The reactions in Voltaic cells usually involve only metals but do not have to.
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Voltaic Cells Continued
-
+
Anode:
Cathode:
Where
oxidation
happens
Where
reduction
happens
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Constructing Voltaic Cells

You will need to build and measure the potential of voltaic cells comprising
various combinations of the following:






Cu/Cu2+
Fe/Fe2+
Mg/Mg2+
Sn/Sn2+
Zn/Zn2+
Follow the instructions here
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General Reminders…
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Comparisons of half-cell electrode potentials need a
reference point

Potential difference is known as the electromotive force (EMF)

Electrons tend to flow from half-cells: more negative potential  more positive potential

Potential generated is called the cell potential or electrode potential… Symbol is E.

Magnitude of this voltage depends on the difference in tendency of reduction of the halfcells.

Can’t measure an isolated half cell (no electron flow)

So we measure against a fixed reference point… STANDARD HYDROGEN
ELECTRODE
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Standard Electrode Potential, Eo
Half Cell
• This is the potential of a
standard electrode
relative to the standard
hydrogen electrode.
Standard Electrode Potential,
Eo / V
H+(aq) + e- ⇌ ½ H2(g)
0.00
Li+(aq)
-3.04
+
e-
⇌ Li(s)
Mn2+(aq) + 2e- ⇌ Mn(s)
-1.19
Cu2+(aq) + 2e- ⇌ Cu(s)
+0.34
½ Br2(l) + e- ⇌ Br-(aq)
+1.07
• Always measure the
potential of the reduction

• Measured in Volts, V
• Full table in the data
booklet
Look at the table in the data booklet:
 What trends do you notice?
 How do the values relate to your ideas
of reactivity?
 How do the values compare to the
reactivity series you constructed
earlier?
Comparisons of half-cell electrode potentials need
a reference point
• Potential difference is known as the electromotive force (EMF)
• Electrons tend to flow from half-cells: more negative potential  more positive potential
• Potential generated is called the cell potential or electrode potential… Symbol is E.
• Magnitude of this voltage depends on the difference in tendency of reduction of the half-cells.
• Can’t measure an isolated half cell (no electron flow)
• So we measure against a fixed reference point… STANDARD HYDROGEN ELECTRODE
The standard hydrogen electrode
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The Standard Hydrogen Electrode
 Platinum is used as the conducting
metal because it is fairly inert, will not
ionize and can b a catalyst for
reduction
 Surface of the metal is coated in finely
divided platinum to increase surface
area for absorption of hydrogen gas
 The electrode is bathed alternately in
H2(g) and H+(aq), setting up an
equilibrium
Forward - Reduction
Standard electrode potential defined as 0.00 V…
Backward - Oxidation
So we can measure and compare electrode potentials
or other half-cells
Measuring Standard Electrode Potentials
• Temperature = 298K
• Pressure = 100 kPa or 1 atm
• Substances must be pure
• If half cell does not include a solid metal, platinum is used
Connecting wire
Half-cells measured under these
conditions are known as
standard half-cells
Metal Electrode
Aqueous solution of
metal ions:
[M+] = 1.0 mol dm-3
Combining half cells 2
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Measuring Standard Electrode Potentials
• When the standard hydrogen electrode is connected to another standard hall-cell, the EMF generated is
known as the Standard Electrode Potential.
• Given the symbol Eө
+ve value for Eө
(+0.34V) indicates
that this has greater
tendency to be
reduced than H+
Electrons flow from
hydrogen half-cell
(oxidized) to copper
half cell (reduced)
Anode
Overall reaction equation:
Cathode
More reactive metals tend to lose their
electrons…
• H+ will be
reduced
• Electrons flow
toward Hydrogen
• -ve value for Eө
Overall reaction equation:
Anode
OXIDISED
From the examples we can see: Zinc has a lower Eө than hydrogen so can reduce H+…
From the examples we can see: Copper has a higher Eө than hydrogen so cannot reduce H+…
Cathode
REDUCED
Remember, Standard Electrode Potentials are give for
REDUCTION reaction
• Sometimes known as the Standard Reduction Potentials
• Oxidised species on the left, reduced species on the right.
Note:
• The Eө values do not depend on the total number of electrons, no need to
scale according to stoichiometry
The electrochemical series
The electrochemical series is a list of standard electrode
potentials (Eө). The equilibria are written with the electrons on
the left of the arrow, i.e. as a reduction.
Half cell
Half equation
Eө / V
Mg2+(aq) / Mg(s)
Mg2+(aq) + 2e-
Mg(s)
-2.36
Zn2+(aq) / Zn(s)
Zn2+(aq) + 2e-
Zn(s)
-0.76
2H+(aq) / H2(g)
2H+(aq) + 2e-
H2(g)
0
Cu2+(aq) / Cu(s)
Cu2+(aq) + 2e-
Ag+(aq) / Ag(s)
Ag+(aq) + e-
Cu(s)
Ag(s)
+0.34
+0.80
Electrodes with negative values of Eө are better at releasing
electrons (i.e. better reducing agents) than hydrogen.
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Calculating Ecell
The e.m.f of an electrochemical cell, Ecell, is the difference
between the standard electrode potentials of the two half cells.
Ecell = Eө (positive electrode) – Eө (negative electrode)
The positive electrode is taken to be the least negative half
cell, and the negative electrode is the most negative half cell.
This can be worked out from the electrode potentials values
in the electrochemical series.
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Calculating Ecell: worked example
An electrochemical cell is set up using the two half reactions
below. What potential difference Ecell would this cell generate?
Zn2+(aq) + 2e-
Zn(s)
Eө = -0.76 V
Cu2+(aq) + 2e-
Cu(s)
Eө = +0.34 V
Ecell = Eө (positive electrode) – Eө (negative electrode)
The zinc half cell has the more negative potential and so
forms the negative electrode. Therefore:
Ecell = (+0.34) – (-0.76)
= +1.10 V
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Calculating Ecell: combining half equations
To find the overall reaction occurring in the cell as a whole,
the two half equations are added together:
Cu2+(aq) + 2e-
Cu(s)
Because the zinc half cell forms the negative electrode of the
cell, oxidation occurs at this electrode and the half equation
must be reversed:
Zn(s)
Zn2+(aq) + 2e-
The two half equations are added to give the overall
cell reaction:
Zn(s) + Cu2+(aq)
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Zn2+(aq) + Cu(s)
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Calculating Ecell
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Remember, Standard Electrode Potentials are give for REDUCTION reaction
• Sometimes known as the Standard Reduction Potentials
• Oxidised species on the left, reduced species on the right.
• The more positive the Eө value, the more readily it is reduced
Note:
• The Eө values do not depend on the total number of electrons, no need to scale according to
stoichiometry
• A selection of electrode potential values are given in section 24 of the IB data booklet