Chemistry 1011
TOPIC
Electrochemistry
TEXT REFERENCE
Masterton and Hurley Chapter 18
Chemistry 1011 Slot 5
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18.1 Voltaic Cells
YOU ARE EXPECTED TO BE ABLE TO:
• Identify oxidation-reduction reactions (Review Chapter 4.3)
• Assign oxidation numbers to elements in molecules and ions (Review
Chapter 4.3)
• Balance equations for oxidation-reduction (Redox) reactions (Review
Chapter 4.3)
• Identify the anode and cathode in an electrochemical cell
• Construct a labelled diagram to show the structure of an
electrochemical voltaic cell and describe its operation
• Represent a voltaic cell using the symbolism M1 | M12+ | | M2 | M22+
Chemistry 1011 Slot 5
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Oxidation-Reduction Reactions
• In simple terms, oxidation is the addition of oxygen
to an element
Mg(s) + O2(g) → 2MgO(s)
• Here the Mg is oxidized
Mg → Mg2+ + 2e−
• In this oxidation process, the magnesium loses
electrons
• In general, in an oxidation-reduction (Redox)
reaction, one species loses electrons and the other
gains electrons
• The species gaining electrons is reduced
• The species losing electrons is oxidized
Chemistry 1011 Slot 5
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Oxidation-Reduction Reactions
• When zinc reacts with hydrochloric acid:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
• The zinc atoms lose electrons – they are oxidized
• The H+ ions gain electrons – they are reduced
OXIDATION:
Zn → Zn2+ + 2e−
REDUCTION:
2H+ + 2e− → H2
• Oxidation and reduction take place together
• There is no net change in the total number of
electrons
Chemistry 1011 Slot 5
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Oxidation Number
• The oxidation number is an arbitrary number assigned to
an element in a compound to indicate the charge that
would be present if it were an ion
• The oxidation number of a pure element is 0
– Oxidation number of chlorine in Cl2 is 0
• Oxidation number of an element in a monatomic ion is
equal to the charge on the ion
– Oxidation number of chlorine in NaCl is –1
• In compounds, the oxidation number of oxygen is
usually –2
• In compounds, the oxidation number of hydrogen is
usually +1
Chemistry 1011 Slot 5
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Oxidation Number
• What is the oxidation number of Mn in
MnO4− ?
– Charge due to 4O will be –8
– Ion has overall charge of –1
– Mn is assigned oxidation number of +7
• What is the oxidation number of N in
NH3, N2, N2O, NO, NO2?
• -3, 0, +1 +2 +4
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Balancing Simple Half Reactions
• For simple ions the balanced half reaction
must be balanced for mass and charge
– The numbers of atoms must be the same on
both sides of the equation
– The net charge must be the same on both sides
of the equation
Zn → Zn2+ + 2e−
Cl2 + 2e− → 2Cl−
Chemistry 1011 Slot 5
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Balancing More Complex Half
Reactions
•
•
•
•
•
•
Identify the element being oxidized or
reduced
Assign oxidation numbers to the element on
both sides of the equation
Identify change in oxidation number
Add electrons to balance change in oxidation
number
Balance charge by adding H+ or OHBalance hydrogen by adding H2O
Chemistry 1011 Slot 5
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Balancing More Complex Half
Reactions
• Balance: MnO4−(aq) →
Mn2+(aq) (acid solution)
– Oxidation number of Mn in MnO4−(aq) is 7+
– Oxidation number of Mn in Mn2+(aq) is 2+
• 5 electrons must be added to the Left Hand Side of the
equation to change oxidation number by 5
MnO4−(aq) + 5e− → Mn2+(aq)
• 8 H+ must be added to the LHS to balance for charge
8H+(aq) + MnO4−(aq) + 5e− → Mn2+(aq)
• 4 H2O must be added to the RHS to balance completely
8H+(aq) + MnO4−(aq) + 5e− → Mn2+(aq) + 4H2O(l)
Chemistry 1011 Slot 5
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Balancing More Complex Half
Reactions
• Balance: SO42−(aq) →
SO2(g) (acid solution)
– Oxidation number of S in SO42−(aq) is ?
– Oxidation number of S in SO2(g) is ?
• ?? electrons must be added to the LHS? or RHS? of the
equation to change oxidation number by ??
SO42 −(aq) + ?? → SO2(g)
• ?? H+ must be added to the LHS? or RHS? to balance for
charge
• ?? H2O must be added to the LHS? or RHS? to balance
completely
Chemistry 1011 Slot 5
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Voltaic Cells
• In a voltaic cell, chemical energy is
converted to electrical energy
• In a voltaic cell, an oxidation-reduction
(redox) reaction takes place
• The voltaic cell is constructed so that the
electrons produced by the oxidation half
reaction must travel round an external
circuit to become available for the reduction
half reaction
Chemistry 1011 Slot 5
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Voltaic Cells
• When a strip of zinc is placed in a solution
of copper ions, a redox reaction takes place:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Half reactions:
Zn → Zn2+ + 2e− oxidation
reduction
Cu2+(aq) + 2e− → Cu(s)
• This reaction is spontaneous
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Designing the Voltaic Cell
• Dip zinc metal into zinc ions (zinc sulfate
solution) in one beaker
• Dip copper metal into copper ions (copper
sulfate solution) in a second beaker
• Link the two solutions with a “salt bridge”
• Link the two metal strips with a wire
• Electrons flow from the zinc to the copper
through the wire
• Ions flow through the “bridge”
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The Zinc – Copper Voltaic Cell
• Cell representation:
Zn | Zn2+ || Cu2+ | Cu
–
–
–
–
anode (oxidation) on left;
cathode (reduction) on right;
|| indicates salt bridge
| indicates phase boundary
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Voltaic Cell Operation
• The zinc strip is the ANODE
– (oxidation occurs - Zn
→
Zn2+ + 2e− )
• The copper strip is the CATHODE
– (reduction occurs - Cu2+(aq) + 2e− → Cu(s) )
• The electrons flow around the external circuit
from the anode to the cathode
• Zinc ions are produced in solution around the zinc
strip
• Copper ions are removed from solution around the
copper strip
• Ions migrate across the salt bridge to maintain
electrical neutrality
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Other Metal | Metal Ion Voltaic Cells
• Any two different metals can be used to
produce an electric current
Cu | Cu 2+ || Ag+ | Ag
Zn | Zn2+ || Fe2+ | Fe
Al | Al3+ || Cu2+ | Cu
Zn
Cu
LEMON
Chemistry 1011 Slot 5
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The Zinc – Hydrogen Voltaic
Cell
• Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
Half reactions:
Zn → Zn2+ + 2e−
2H+(aq) + 2e− → H2(g)
oxidation
reduction
Zn | Zn2+ || H+ | H2 | Pt
• The hydrogen half cell consists of hydrogen gas
bubbled over a platinum electrode, submerged in a
solution of acid
Chemistry 1011 Slot 5
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The Zinc – Hydrogen Voltaic Cell
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Structure and Operation of
Voltaic Cells A Summary
• A voltaic cell consists of two half cells joined by
– an external circuit through which electrons flow, and by
– a salt bridge through which ions move
• Each half cell consists of an electrode dipping into an
aqueous solution
– If a metal participates in the cell reaction it will usually be the
electrode; otherwise an inert platinum electrode is used
• In one half cell oxidation occurs at the anode and reduction
occurs at the cathode
– The overall cell reaction is the sumof the half reactions taking
place at anode and cathode
Chemistry 1011 Slot 5
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