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Chemistry 1011
Chemistry 1011
Introductory Chemistry II
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Chemistry 1011 Slot 5
TEXT REFERENCE
Masterton and Hurley Chapter 18
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18.2 Standard Voltages
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Cell Voltage
YOU ARE EXPECTED TO BE ABLE TO:
• The force that pushes the electrons through
the external circuit of a cell is known as the
• Define the standard electrode potential of a half cell
• Order species according to their ease of oxidation or
reduction based on a table of standard reduction potentials
• Calculate the net cell voltage, Eo, of a combination of half
cells from standard electrode potential data
• Determine whether a given redox reaction will be
spontaneous or non-spontaneous
Chemistry 1011 Slot 5
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– Potential difference, or
– Electromotive force (emf), or
– Voltage
• It is measured in volts
• The magnitude of the voltage depends on
– The nature of the redox reaction
– The concentrations of the ions in solution, (or
pressures of any gases)
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The Zinc – Hydrogen Voltaic Cell
Standard Voltage
• In order to compare the voltages of different
cells, or to calculate the expected voltage of
a given cell, measurements are taken under
standard conditions:
– Current flow is almost zero
– All ions and molecules in solution are at a
concentration of 1.0 mol/L
– All gases are at a pressure of 1.0 atm
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The Standard Voltage of the
Zinc – Hydrogen Voltaic Cell
Zn(s) + 2H+(aq) →
1.0 mol/L
Standard Half Reaction Voltages
Zn2+(aq) + H2(g)
1.0 mol/L
1.0 atm
Zn | Zn2+ || H+ | H2 | Pt
• Cell voltage with no current flowing is +0.762V
• This is the standard voltage for this cell
Zn(s) + 2H+(aq, 1.0M) → Zn2+(aq, 1.0M) + H2(g, 1.0atm)
Eo = +0.762V
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Obtaining Values for Standard
Half Reaction Voltages
2H+(aq,1.0M) + 2e− → H2(g,1.0atm)
Eored (H+ → H2) = 0.000V
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• Once one half reaction standard voltage is
established, others can be deduced:
• For: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
the standard cell voltage is +1.101V
Zn(s) → Zn2+(aq,1.0M) + 2e−
Cu2+(aq,1.0M) + 2e− → Cu(s)
Eoox (Zn → Zn2+) = +0.762V
Eored (Cu2+→ Cu) = ??V
Since Eo = Eoox + Eored
+1.101V = +0.762V + Eored
Since Eo = Eoox + Eored
Eoox (Zn → Zn2+) = +0.762V
Eored = +0.339V
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Standard Reduction Potentials
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Standard Reduction Potentials
Oxidizing Agent
Li+(aq) + e−
Na+(aq) + e−
Zn2+(aq) + 2e−
Ni2+(aq) + 2e−
2H+(aq) + 2e−
Cu2+(aq) + 2e−
Ag+(aq) + e−
NO3−(aq) + 4H+(aq) + 3e−
MnO4−(aq) + 8H+(aq) + 5e−
F2(g) + 2e−
• Standard half cell voltages are found in
tables of standard potentials
• These are the values for reduction half
reactions based upon the convention that
2H+(aq,1.0M) + 2e− → H2(g,1.0atm)
Eored (H+ → H2) = 0.000V
• Standard reduction potential = Eored
Chemistry 1011 Slot 5
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Obtaining Values for Standard
Half Reaction Voltages
• Standard half reaction voltages are determined by
arbitrarily assigning the value of zero to the
standard reduction half reaction for hydrogen ions
to give hydrogen gas
Chemistry 1011 Slot 5
• Each half reaction has a standard voltage
• Eoox (standard oxidation voltage)
• Eored (standard reduction voltage)
Eo = Eoox + Eored
o
• Only E can be measured - the standard
voltage of a half reaction cannot be
measured directly
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Reducing Agent
→ Li(s)
→ Na(s)
→ Zn(s)
→ Ni(s)
→ H2(g)
→ Cu(s)
→ Ag(s)
→ NO(g) + 2H2O
→ Mn2+(aq) + 4H2O
→ 2F−(aq)
Chemistry 1011 Slot 5
Eored (V)
-3.040
-2.714
-0.762
-0.236
0.000
+0.339
+0.799
+0.964
+1.512
+2.889
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Standard Reduction Potentials
• Elements above hydrogen in the table of standard
reduction potentials will react with a solution of
hydrogen ions to produce hydrogen gas
M(s) + 2H+(aq) → M2+(aq) + H2(g)
M2+(aq) + 2e− → M(s) Eored = negative
M(s) → M2+(aq) + 2e−
Eoox = positive
0.000V
2H+(aq) + 2e− → H2(g) Eored =
M | M2+ || H+ | H2 | Pt
Eocell = positive
• Elements below hydrogen in the table of standard
reduction potentials will NOT react with a solution
of hydrogen ions to produce hydrogen gas
Chemistry 1011 Slot 5
Standard Voltages for Voltaic Cells
• The table of standard reduction potentials gives
standard voltages for reduction half reactions
• Standard voltages for oxidation half reactions are
obtained by reversing these reactions and changing
the sign of the Eored value
→ Zn(s)
Zn2+(aq) + 2e−
Eored = -0.762
• If:
• Then: Zn(s) → Zn2+(aq) + 2e−
Eoox = +0.762
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Chemistry 1011 Slot 5
Oxidizing Agents
Computing Standard Cell Potential
• An oxidizing agent is a species that can gain
electrons
• The standard voltage of a cell is the sum of the
standard potentials for the two half reactions
• For the cell:
Zn | Zn2+ || Cu2+ | Cu
Eoox = +0.762V
Zn(s) → Zn2+(aq) + 2e−
Eored = +0.339V
Cu2+(aq) + 2e− → Cu(s)
Zn(s) + Cu2+(aq) →
– The strongest oxidizing agents are the species
that gain electrons most readily
– They have the largest positive Eored values
– Oxidizing strength increases moving down the
left column of the table of standard reduction
potentials
– Oxidizing agents in the table of standard
reduction potentials can oxidize any species
above
Zn2+(aq) + Cu(s)
• Eocell = Eoox + Eored = + 0.762 + 0.339 = 1.101V
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Strong Reducing and Oxidizing
Agents
Reducing Agents
• A reducing agent is a species that readily
loses electrons
– The strongest reducing agents are the species that
lose electrons most readily
– They have the largest negative Eored values (The
largest positive Eoox values)
– Reducing strength increases moving up the right
column of the table of standard reduction potentials
– Reducing agents in the table of standard reduction
potentials can reduce any species below
Reducing agent causes another species to be reduced - it is oxidized
Li(s) → Li+(aq) + e−
Eoox = +3.040V
Oxidizing agent causes another species to be oxidized - it is reduced
F2(g) + 2e−
R
→ 2F−(aq)
Eored = +2.889V
Table of Standard Reduction Potentials
R = strongest reducing agent
O = strongest oxidizing agent
O
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Spontaneity of Redox Reactions
Reaction of Copper with Dilute
Hydrochloric Acid??
• In order for a redox reaction to occur
spontaneously, the calculated cell potential
MUST BE POSITIVE
• Questions:
• Possible oxidation half reaction:
Cu(s) → Cu2+(aq) + 2e−
Eoox = -0.339V
+
• Possible reduction half reaction (H and Cl− ions are
present - Cl− ions cannot be reduced):
2H+(aq) + 2e− → H2(g)
Eored = 0.000
– Will copper metal be oxidized to Cu2+ ions by
dilute hydrochloric acid?
– Will copper metal be oxidized to Cu2+ ions by
dilute nitric acid?
Chemistry 1011 Slot 5
• Net possible reaction:
Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
• Net calculated cell voltage
Eocell = Eoox + Eored = - 0.339 + 0.000 = - 0.339 V
• Reaction will not be spontaneous i.e no reaction
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Reaction of Copper with Dilute
Nitric Acid??
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Voltaic Cells with Inert Electrodes
• Possible oxidation half reaction:
Cu(s) → Cu2+(aq) + 2e−
Chemistry 1011 Slot 5
Eoox = -0.339V
• Possible reduction half reactions (H+ and NO3− ions are present):
2H+(aq) + 2e− → H2(g)
Eored = 0.000V
NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O
Eored = +0.964V
• Net spontaneous reaction (Add multiples of the two half
reactions so that same #electrons (6) in each half):
3Cu(s) + 2NO3−(aq) + 8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
• Half cells will frequently be constructed
with inert electrodes (often carbon or
platinum)
• The Hydrogen half cell is one example:
H+ | H2 | Pt
• A cell with two inert electrodes might be:
Pt | Fe2+(aq) | Fe3+(aq) || Cl −(aq) | Cl2(g) | Pt
• Net calculated cell voltage:
Eocell = Eoox + Eored = - 0.339 + 0.964 = + 0.629 V
• Reaction will be spontaneous i.e reaction takes place
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The Leclanché Cell
• The Leclanché cell is the ordinary commercial
flashlight battery
Zn | Zn2+ ||MnO2 | Mn2O3 | C
• Anode half reaction:
Zn(s) → Zn2+(aq) + 2e−
Eoox = +0.762V
• Cathode half reaction (complex):
2MnO2(s) + 2NH4+(aq) + 2e− → Mn2O3(s) + 2NH3(aq) + H2O Eored = +0.7 V
• Net cell voltage Eocell = 1.5V
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