NAME __________________________________ NOTES: UNIT 6 (6): REDOX: VOLTAIC CELLS
I-XII) Intro to Redox to Balancing with Half-Reactions
Voltaic Cells … in which a chemical reaction
is used to produce an electrical current
XIII) Electrochemical Cells
Electrolytic Cell … in which an electrical
current from an outside source is used to
generate a chemical reaction
A) Voltaic (Galvanic Cells)
Let’s begin with a demonstration:
Essentially, this is a single replacement
reaction in which a metal species is
oxidized (becoming an oxidized cation),
while a metal cation gains the lost
electrons and is reduced back to a metal.
e.g. Cu2+(aq) + Zn0(s) → Zn2+(aq) + Cu(s)
The voltaic cell (a.k.a. galvanic cell) separates the oxidation and reduction, and forces the electrons
(the electrical current) to flow through an external conductor (a wire). [Note: electrons flowing through a wire
= an electrical current …. as well as ions flowing through a solution …such as an electrolyte solution.]
B) Definition: A voltaic cell produces electrical energy from spontaneous redox reactions taking place
within the cell. It involves the conversion of the free energy of a spontaneous chemical reaction to
the kinetic energy of electrons moving through a wire.
A voltaic cell generally consists of two different electrodes (terminals) which are often metallic, but
sometimes Cgraphite. The electrodes (terminals) are connected by a wire. The two half-cells are
further connected by a salt bridge, or individual half-cells separated by a porous membrane.
http://en.wikipedia.org/wiki/Galvanic_cell
http://scioly.org/wiki/index.php/File:Voltaic_Cell_Salt_Bridge.jpeg
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XIV) Table J & the Reduction Half-Reaction (Potential)
Stronger oxidizing agents
Table J: Table of Reduction Half-Reaction Potentials E0 (V)
+2.87
+1.82
+1.50
+1.44
+1.36
+1.23
+1.07
+0.90
+0.85
+0.80
+0.77
+0.54
+0.52
+0.35
+0.34
+0.16
+0.15
0.00
-0.04
-0.13
-0.14
-0.23
-0.28
-0.40
-0.41
-0.74
-0.76
-1.18
-1.66
-1.70
-2.38
-2.71
-2.76
-2.89
-2.90
-2.92
-3.04
Stronger reducing agents
F2(g) + 2e- → 2F-(aq)
Co3+(aq) + e- → Co2+(aq)
Au3+(aq) + 3 e- → Au(s)
Ce4+(aq) + e- → Ce3+(aq)
Cl2(g) + 2e- → 2Cl-(aq)
O2(g) + 4e- → 2O2Br2(l) + 2e- → 2Br-(aq)
2Hg2+(aq) + 2e- → Hg22+(aq)
Hg2+(aq) + 2e- → Hg(l)
Ag+(aq) + e- → Ag(s)
Fe3+(aq) + e- → Fe2+(aq)
I2(s) + 2e- → 2I-(aq)
Cu+(aq) + e- → Cu(s)
ClO3-(aq) + H2O(l) + 2e- → ClO2-(aq) + 2OH-(aq)
Cu2+(aq) + 2e- → Cu(s)
Cu2+(aq) + e- → Cu+(aq)
Sn4+(aq) + 2e- → Sn2+(aq)
2H+(aq) + 2e- → H2(g)
Fe3+(aq) + 3e- → Fe(s)
Pb2+(aq) + 2e- → Pb(s)
Sn2+(aq) + 2e- → Sn(s)
Ni2+(aq) + 2e- → Ni(s)
Co2+(aq) + 2e- → Co(s)
Cd2+(aq) + 2e- → Cd(s)
Fe2+(aq) + 2e- → Fe(s)
Cr3+(aq) + 3e- → Cr(s)
Zn2+(aq) + 2e- → Zn(s)
Mn2+(aq) + 2e- → Mn(s)
Al3+(aq) + 3e- → Al(s)
Be2+(aq) + 2e- → Be(s)
Mg2+(aq) + 2e- → Mg(s)
Na+(aq) + e- → Na(s)
Ca2+(aq) + 2e- → Ca(s)
Sr2+(aq) + 2e- → Sr(s)
Ba2+(aq) + 2e- → Ba(s)
K+(aq) + e- → K(s)
Li+(aq) + e- → Li(s)
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A) Again: A voltaic cell is a simplified version of what we might refer to as a battery. (Technically the
term “battery” refers to two voltaic cells linked together … but let’s not worry about this)
1) A voltaic cell (galvanic cell, a Daniell cell) *uses a redox chemical reaction to generate
an electrical current.
a) The electrical current originates at the oxidized species,
moves through a machine, (doing work)
and produces a reduction of a chemical species, on the other side
http://www.dummies.com/how-to/content/electrochemical-cells-the-daniell-cell.html
b) Chemists recognize that for every reduction there must be an oxidation.
c) We can, therefore, split a voltaic cell in half … recognizing that it is essentially
a reduction reaction, linked to an oxidation reaction.
d) A table listing possible reduction half-reactions or oxidation reaction sure would
come in handy, to help us create voltaic cells….
Enter … Table J of your reference tables.
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B) Standard Reduction Potentials
1) We use the standard reduction potentials of Table J to describe what can occur at the
ends of a voltaic cell.
a) To help compare potentials chemists have created a set of standard electrode
potentials …
i) These are electromotive force values (emf) relative to a standard electrode
potential, using H2(g) electrode, under the standard conditions of:
25°C and
1 atm with reactions occurring in
molar concentrations of 1 M
ii) Thus, these are concentration dependent and pressure & temperature
dependent
b) By mutual agreement, we have opted to list only reduction potentials … because
the opposite of any reduction ½ reaction, is an * oxidation ½ reaction
Review: Write the oxidation half-reaction given: Zn2+(aq) + 2e- → Zn(s)
* Zn(s) → Zn2+(aq) + 2e-
Review: Given the following reduction half-reaction, write the reverse of it.
I2(s) + 2e- → 2I-(aq)
* 2I-(aq) → I2(s) + 2e-
c) The reduction of a species, generates or consumes a certain number of volts, listed
on table J.
i) a positive voltage = the reduction in question is (comparatively) more likely to
occur than the reduction of H+
ii) a negative value = the reduction in question is less likely to occur relative to
reduction of H+
d) It is not possible to measure the standard reduction potential of a half-reaction
directly … The reduction of H+(aq) to H2(g) was selected as a reference point &
assigned a value of 0.00 V. All other voltages are relative to this standard
potential.
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2) The Standard Hydrogen Electrode (SHE)
Citation: http://www.docbrown.info/page07/equilibria7b.htm
The spontaneous reaction which occurs when a copper electrode in Cu2+(aq) solution is connected to the
SHE (standard H2 electrode) provides chemists with a standard reduction potential for the reduction of
Cu2+ → Cu0, under the standard conditions of 25°C, 1 atm, and 1 M concentrations of H+(aq) & Cu2+(aq)
466
In-Class Practice: Using Table J:
1) Based upon their position on Table J, and / or their values for E0 which of the following is more likely to
occur as a reduction? (Place a check mark next to the ½ reaction of your choice and write down a reason)
Fe2+(aq) + 2e- → Fe(s)
E0 = -0.41
Ag+(aq) + e- → Ag(s)
E0 = +0.80
2) Based upon their position on Table J, and / or their values for E0 which of the following is more likely to
occur as a reduction? (Place a check mark next to the ½ reaction of your choice and write down a reason)
Cu2+(aq) + e- → Cu+(aq)
E0 = +0.16
Sn2+(aq) + 2e- → Sn(s)
E0 = -0.14
a) If we tried to make a voltaic cell by coupling these two reactions, which half-reaction would
be the oxidation half-reaction? Write it as an oxidation half-reaction with a correct value for E0
at standard conditions.
* Sn(s) → Sn2+(aq) + 2e-
3) Given:
E0 = +0.14
Cr3+(aq) + 3e- → Cr(s)
E0 = -0.74
Au3+(aq) + 3 e- → Au(s)
E0 = +1.50
What would be the reduced species and what would be the oxidized species assuming these were
coupled together to make a voltaic cell?
*Cr = oxidized species
Au3+ = reduced species
467
XV) The Voltaic Cell: Construction / Parts / Roles
A) Parts
On/Off Switch
V
Zn0
Cu0
Salt Bridge KNO3(aq)
Zn2+
Cu2+
Zn(NO3)2(aq)
Zn0(s) + Cu2+(aq) 
Cu(NO3)2(aq)
Zn2+(aq)
+ Cu0(s)
HALF CELL(S)
There are 2 different half cells. Each one is made of 1 metal electrode and an ionic solution of the
metal electrode. In one half-cell, the oxidation occurs, in the other, the reduction reaction occurs.*
The voltmeter takes on the role of the machine we wish to operate
ELECTRODE
2 different electrodes
anode
cathode
*is the oxidized species (in most cases) so
it is not reduced (it is the site of reduction)
it is a reactant
…it is an electron collector
it does not gain e-, it conducts them to the
it loses electrons
metal ions in solution.
synonyms:
terminal, metal
it is the negative terminal (electrode)
it is the positive terminal (electrode)
SOLUTIONS
it becomes smaller in mass & the resulting it becomes larger in size as new metal
metallic ions dissolve into the sol’n
atoms are produced
*Ionic (aqueous) solutions for simple voltaic cells
WIRE (external
*electrons travel through the wire.
conductor)
SALT BRIDGE
The operation of the voltaic cell causes a buildup of charge in each half-cell. The salt bridge contains
saltwater and allows dissolved ions to move between the half-cells so that the half-cells remain electrically
neutral.
 On Electrolytes: When electrodes are placed in an electrolyte and a voltage is applied, the
electrolyte will conduct electricity...in the absence of metal wires. Lone electrons normally cannot pass
through the electrolyte; instead, a chemical reaction (reduction) occurs at the cathode consuming electrons
from the anode. Another reaction (oxidation) occurs at the anode, producing electrons that are eventually
transferred to the cathode. As a result, a negative charge cloud develops in the electrolyte around the
cathode, and a positive charge develops around the anode. The ions in the electrolyte neutralize these
charges, enabling the electrons to keep flowing and the reactions to continue. http://en.wikipedia.org/wiki/Electrolyte
468
B) Process for answering questions re: Voltaic Cell:
+ - +-
+ - + - + - + - + -
+
_
+
_
+
_
+
_
Please understand that you should label everything on the diagram and THEN answer the questions.
Step 1
Use Table J to find out which metal ion is more likely to be reduced. LABEL the ion as the reduced
species
Step 2
Assume that the other half cell holds the metal which is oxidized. LABEL it as the oxidized species.
Step 3
Step 4
When the reaction equation is given, go on to Step 4. If not, then write the reaction out and predict the products.
Write the word: DECREASE beneath the reactants and the word INCREASE beneath the products
Step 5
Use the mnemonic: AN OX & RED CA
reduction occurs at the cathode)
to complete the labeling. (Remember: The anode is oxidized and
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C) Focus on the Electrodes (terminals): The electrodes (terminals) in simple voltaic cells are
often metals (Zn0, Cu0, Fe0, Pb0 etc…)
1) Sometimes it is convenient to have an inactive electrode (an electrode that is not
oxidized easily for instance …An inactive electrode is one that
*does NOT become oxidized or react with the surrounding system’s chemicals
a) Pt (platinum) or Cgraphtie are often used when inactive electrodes are required.
e.g.) Pt is used in the Hoffman apparatus (the electrolysis of water)
Cgraphite is used as the cathode in the LeClanche Zn-C battery
www.amazon.com
http://www.mpoweruk.com/cell_construction.htm
b) Sometime a gaseous electrode is created … such as when measuring Standard
Electrode Potentials. In this case, the gaseous electrode is H2(g) … In which a
Pt electrode and H2(g) are used in conjunction with each other
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2) The names given to the 2 electrodes are: *Anode and Cathode
a) Regardless of the electrochemical cell type (voltaic or electrolytic) ….
i) *oxidation occurs at the anode and reduction occurs at the cathode.
Mnemonic: *AN OX and RED CAT
ii) * electrons flow through a wire from anode to cathode (a to c)
iii) In a voltaic cell the anode is assigned a * negative (because e- are moving
away from it) …. and the cathode is assigned a positive sign, because e- are
moving towards it.
B) Essentially these voltaic (galvanic) reactions occur * between the metal atoms of one
species and the cations of a second (different) species
1) Classic examples of these reactions include, but are not limited to:
a) Cu2+ (aq) + Zn0(s) → Zn2+ (aq) + Cu(s)
b) 2 Fe3+ (aq) + 3 Zn0(s) → 3 Zn2+ (aq) + 2 Fe(s)
c) Pb2+ (aq) + Mn0(s) → Mn2+ (aq) + Pb(s)
d) 2 Ag+ (aq) + Ni0(s) → Ni2+ (aq) + 2 Ag(s)
e) 3 Cu2+ (aq) + 2 Al0(s) → 2 Al+3(aq) + 3 Cu0(s)
2) An important inference is that, while electrons are moving towards the cathode, the cathode
acts more as an “electron collector” … and the surrounding cations are the reduced species.
a) The cathode itself (especially when metallic) is not reduced … but rather the site of
reduction. The cations of the cathode half-cell are reduced … and the cathode often
(in simple voltaics) gains in mass ….when the electrodes are active metals. When the
electrodes are inactive metals (Pt) or graphite … mass of the cathode remains the
same.
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3) Symbolism of a voltaic cell: Line Notation:





anode || cathode
A single vertical line (|) indicates a difference in state or phase.
A double vertical line (||) is used to indicate the junction between the half-cells.
A comma is used to separate reactants and produces in the same state or phase, this
will include an inactive electrode
Within a half-cell, the reactants are listed before the products.
The line notation for the anode (oxidation) is written before the line notation for the
cathode (reduction)
a)
Zn(s)| Zn+2(aq) || Pb+2(aq) | Pb(s)
b)
Pt(s)| Sn+4(aq) , Sn+2(aq) || Cl-(aq) |Hg2Cl2(s)|Hg(ℓ)
c)
Pt(s)|ClO3-(aq), H+(aq) || H+(aq), MnO4-(aq), Mn2+(aq) | Pt(s)
describes:
MnO4-(aq) + H+(aq) + ClO3-(aq) → ClO3-(aq) + Mn2+(aq) + H2O(ℓ)
c) Focus upon the function of the Salt Bridge
1) The salt bridge is designed to allow +/- ions to migrate into the half cells, in order to keep
the half-cell ionic solutions, neutral in overall charge.
Anode
Cathode
(oxidized
species)
the cations in the solution surrounding the cathode are the reduced species
a) alternative: Use of a single porous cup
(ceramic cylinder designed to allow for the migration of
ions between the electrolyte solution in the cup and the electrolyte solution in the beaker.
The ions are allowed to pass through the ceramic cup … hence the cup’s porosity
allows the cup to act as a substitute for the salt bridge
animation:
http://www.blackgold.ab.ca/ict/Division4/Science/Div.%204/Voltaic%20Cells/Voltaic.htm
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So, essentially what we have, when speaking of a voltaic cell is something like the following diagram from Nivaldo Tro’s
text p. 867
Animation:
TRY THIS!!!
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf
Study the following diagram for a voltaic cell with the line notation: Fe(s) |Fe3+(aq) || Cu2+(aq)|Cu(s)
What species is reduced?* Cu2+ Which species is oxidized? * Fe0
0
What species is the anode? * Fe
The flow of electrons is from *Fe0
Cu
Fe
0
Which is the cathode? * Cu
to *Cu0 , the electron collector
What is the positive terminal*? *Cu0 What is the negative terminal? *Fe0
What is the oxidizing agent? *Cu2+
What is the reducing agent? * Fe0
The anode will become (smaller/ larger) in size
ans: *smaller
The redox reaction is:
Cu2+
Fe3+
____ + ____ → ____ + ____
The cathode will become (smaller/ larger) in size ans: * larger
473
XVI) FOCUS ON WHY ELECTRONS SPONTANEOUSLY FLOW
FROM ONE ELECTRODE TO ANOTHER
A) In small … The electrons of the anode are * higher (greater) in potential energy than the electrons
of the cathode.
Thus electrons flow spontaneously from a point of high energy to lower energy.
1) The difference in potential energy per electrical charge is called the potential difference
a) it is measured in *volts … the pressure an electron “feels” as it is pushed along
b) 1 Volt (V) is the potential difference required to impart 1 joule (J) of energy to a
charge of 1 coulomb (C) or 1 V = 1 J/C
c) 1 electron has a charge of 1.60 x 10-19 C ….
d) The potential difference between two electrodes of a voltaic cell = *Cell Potential
i) symbol: * Ecell … commonly referred to as the voltage of the cell
e) The cell potential is synonymous with something called the
*electromotive force (emf) …because it is the driving force that pushes electrons
through the wire … it is essentially “causing electron motion”. It too is
synonymous with voltage
f) Thus: cell potential = Ecell = emf = driving force = voltage
Annotated reading: From Dr. Rod Nave http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/electrode.html#c1
3) Summary:

Whenever we assign an electrical potential to a half-reaction, we write the reaction as a
reduction.

The standard electrode (reduction) potential is measured against the hydrogen electrode
…which is assigned an electron volt value of 0.00

The reduction potentials of various species, as seen on the table of reduction potentials may
be – or +

Those with + electron volts are MORE likely to occur as reductions than the H+ reduction.

Those with – electron volt values are less likely to occur as reductions than the H+ reduction.
474
Notice that even though the anode reaction re: the oxidation of I- is written stoichiometrically as 3
times greater than the half-reaction on the table, the E° of the half-reaction did not change in numeric
value (It remained at numerically as 0.54V) …The point is that the numeric value did NOT get
multiplied by 3 …. Why? …. These values are an INTENSIVE PROPERTY!!!! HEY!!!!!
Note as well that were this to be made as a voltaic cell, the total E°cell value of +0.79 V would NOT be
registered on the voltmeter, as SOME energy is lost due to “heat”, as the electrical current passes
through the wire.
This actually brings up something often called “the other diagonal rule”
An upper left to lower right
diagonal between reactants
provides us with a
spontaneous electrochemical
reaction … thus a + sum for
E°cell
The larger the separation
between the reactants, the
larger the sum of E°cell
475
Try This!!!! A voltaic cell is based on the two standard half-reactions:
Al3+(aq) + 3 e- → Al0(s)
Cu2+(aq) + 2 e- → Cu0(s)
Using the table of standard reduction potentials:
a) What is the half-reaction occurring at the anode? *Al0(aq)→ Al3+(aq) + 3 e- +1.66 V
b) Which half-reaction occurs at the cathode? * Cu2+(aq) + 2 e- → Cu0(aq)
+0.34 V
c) What is the standard cell potential? * 2.00 V
d) What is the total number of electrons exchanged when the reaction is balanced? *6ee) What species is the strongest oxidizing agent? *Cu2+(aq)
Try This!!!! A voltaic cell is based on the two standard half-reactions:
Cr3+(aq) + 3 e- → Cr0(s)
Fe3+(aq) + 3 e- → Fe0(s)
Using the table of standard reduction potentials:
a) What is the half-reaction occurring at the anode? * Cr0(s) → Cr3+(aq) + 3 e-
+0.73 V
b) Which half-reaction occurs at the cathode? * Fe3+(aq) + 3 e- → Fe0(s)
-0.04 V
c) What is the standard cell potential? *0.69 or 0.694 V
d) What is the total number of electrons exchanged when the reaction is balanced? *3ee) What species is the strongest oxidizing agent? *Fe3+(aq)
Try This!!!! Based upon the standard reduction potentials, which species is the strongest oxidizing agent?
Ni2+(aq) + 2 e- → Ni0(s)
Ag+(aq) + 1 e- → Ag0(s)
Au3+(aq)+ 3 e- →Au0(s)
- 0.23
+ 0.80 V
+1.50 V
Defend your thinking: Au3+(aq) is the best oxidizing agent of the three. The standard reduction potential with
the greatest positive value is most likely to occur as a reduction. One may infer then that Au3+ is most likely to
be reduced, making it possible for another species to be oxidized … or rather, making Au3+ a good oxidizing
agent …due (probably) in large part to a very strong effective nuclear charge.
476
GUIDED PRACTICE:
Note: the terms *terminal & electrode are synonymous
1) What species is reduced ? _____ Which species is oxidized? _____
What species is the anode ? ____ Which is the cathode ? _____
Cr
Ag
The flow of electrons is from _____ to _____
What is the positive terminal? _____
What is the negative terminal? ____
What is the oxidizing agent ? _____
What is the reducing agent? ____
Cr+3
The anode will become (smaller/larger ) in size
Ag+1
The cathode will become (smaller/larger ) in size
redox rxn that occurs:
2) Which species is reduced ? _______
_______________ → _______________
What species is oxidized? _______
Which species is the anode ? _____ What is the cathode ? _____
Zn
Ag
The flow of electrons is from _____ to _____
What is the positive terminal? ____
What is the negative terminal? ____
What is the oxidizing agent ? _____
What is the reducing agent? ____
The anode will become (smaller/larger ) in size
Ag+1
Zn+2
The cathode will become (smaller/larger ) in size
redox rxn that occurs:
3) Which species is reduced ? _____
What is the anode ? _____
_______________ → _______________
Which species is oxidized? ____
What is the cathode ? _____
The flow of electrons is from _____to _____
Mn
What is the positive terminal? ____
What is the negative terminal? ____
What is the oxidizing agent ? _____
What is the reducing agent? ____
Al
The anode will become (smaller/larger )
Mn+2
redox rxn that occurs:
Al+3
_______________ → _______________
477
PRACTICE For 1-6 one or more of the responses given is (are) correct. Decide which of is correct.
1)
2)
3)
4)
5)
when only I is correct
when only II is correct
when only I and II are correct
when only II and III are correct
when I, II, and III are correct
___ 1. The diagram represents an electrochemical
cell at 298 K . As the cell operates:
___3. The diagram represents an electrochemical
cell at 298 K . As the cell operates:
I) the concentration of Cr+3 ions will decrease
II) the Cr0 electrode will be oxidized
III) the Au0 electrode will become larger
I) Auo acts as the anode
II) the Sn0 will act as the "negative" electrode
III) the concentration of Au3+ will decrease
V
V
Au0
Cr0
Cr+3
Au3+
Au+3
Cr0 + Au+3
Sn0
Au0
Sn2+
Au0 + Cr+3
___2. The diagram represents an electrochemical
cell at 298 K . As the cell operates
I) electrons will flow from Zn0 to Mg0
II) Zn+2 ions will act as an oxidizing agent
III) Mg0 and Zn+2 are the reactants
___4. The diagram represents an electrochemical
cell at 298 K . As the cell operates
I) the iron electrode will decrease in mass
II) the concentration of Fe2+ ion will decrease
III) electrons will flow from Cu0 to Fe0
V
V
0
0
Mg
Zn
Fe0
Mg+2
Mg2+
2+
Zn
Zn+2
Cu0
Fe2+
Fe0 + Cu+2
Cu2+
Fe+2 + Cu0
478
1)
2)
3)
4)
5)
when only I is correct
when only II is correct
when only I and II are correct
when only II and III are correct
when I, II, and III are correct
___5) The diagram represents an electrochemical cell at 298 K. As the cell operates,
I) electrons travel through the salt bridge
II) A redox reaction occurs
III) Au0 is reduced
V
Au0
Fe0
Fe+3
Au+3
___6) The diagram represents an electrochemical cell at 298 K. As the cell operates:
I) electrons travel from Mg0 to Pb0
II) the amount of Mg0 decreases
III) ions migrate through the salt bridge.
V
Pb0
Mg0
Pb+2
Mg0 + Pb+2
Mg+2
Mg+2 + Pb0
Answers:
1) 4
4) 1
2) 4
5) 2
3) 4
6) 5
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7) Which of the following will occur if a strip of Mg is dipped into a solution of Fe(NO3)2, containing
Fe2+ ions?
1) No reaction occurs.
2) The Fe+2 is oxidized to Fe+3.
3) The Mg0 is oxidized.
4) The Mg0 is reduced
8) Given the following voltaic cell, draw the cell and label the cathode, anode, salt bridge, and direction of
flow of the electrons. Al(s)  Al+3(aq)  Pb(s) Pb +2(aq)
9) Draw the voltaic cell based on the following oxidation half-reactions under standard state conditions:
1M, 250C, & 1 atm.
Ag  Ag+ + 1eFe  Fe+3 + 3e-
10) Given the hypothetical elements, W, X, Y, & Z, along with the corresponding ions W+2, X+2, Y+2, & Z+2,
use the information provided below to determine the order in which the ions should be listed in the
Table 7.
List:
+2
+2
Reaction 1: X + Z  X + Z (reacts spontaneously)
Reaction 2: W + X+2 
(does not react)
Reaction 3: Y + Z+2  Y+2 + Z (reacts spontaneously)
11) Using Table J, which of the following will occur when a strip of Al is dipped into a 1 M solution
of Co(NO3)2?
1)
2)
3)
4)
All of these
The Al0 strip will decrease in size.
The Al0 strip will become coated with Co0.
A redox will reaction occur
Selected Answers:
7) 3
10) W2+ ,X2+ ,Z2+ Y2+
11) 1
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NAME ________________________________ BATTERY … NOT THE CRIME, THE GOOD IDEA
DIRECTIONS: The following are fairly old Regents exam questions … but they bring you through the basics.
Select the most correct answer to each of the questions. You will see a good deal of repetition.
1)
2)
For questions 3 – 5 use the diagram below, and your grasp of electrochemistry.
3)
4)
5)
481
6)
For questions 7 – 9 use the diagram below, and your grasp of electrochemistry.
7)
8)
9)
10)
11)
482
For questions 12 – 14 use the diagram below, and your grasp of electrochemistry
12)
13) Identify one metal from Table J that is more easily oxidized than Mg(s)
14)
15)
16)
17)
483
For questions 18 – 19 use the
diagram below, and your grasp
of electrochemistry
18)
Using half reactions, balance the redox equation, using the smallest whole-number coefficients.
Al + Cu2+ → Al3+ + Cu
19)
20)
21)
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Answers:
1) 1 Voltaic cells produce energy by using a spontaneous chemical reaction. Electrolytic cells consume
energy and force a non-spontaneous chemical reaction to occur.
2) 4 Both cations and anions migrate from the salt bridge … in order to maintain electrical neutrality in
each half-cell, as the cell operates.
3) Half-Cell 1 or The Pb/Pb+2 cell or The Pb half-cell.
4) Electrons flow from anode to cathode through the wire… thus, electrons flow from Pb0 to Ag0
5) Use the equation found beneath the picture to help you…. Pb0 → Pb2+ + 2e6) 2
7) Half-cell 2 or the zinc half-cell….
8) Pb+2 +2e- → Pb0
9) Electrons will flow from anode to cathode … thus from zinc to lead
10) 3 essentially, oxidation occurs at the anode, thus you need a balanced half-reaction that represents
oxidation
11) 3 anode and oxidation go together…
12) 12 1 mol of nickel metal is oxidized to nickel (II) ion by losing two moles of electrons, thus 12 mol of
electrons are required for 6 mol.
13) Any metal below Mg0 on Table 7
14) The salt bridge is present to : complete the circuit by allowing ions to migrate between the half-cells
maintain electrical neutrality in the half-cells as the cell operates
15) 1
16) 4 catch onto a trend here????
17) 3
18) coefficients are: 2:3:2:3
19) Aluminum is the anode. The anode is oxidized producing soluble aluminum ions which dissolve into the
surrounding solution.
20) 1
21) 4
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CHALLENGE QUESTIONS
TRY THIS!!! A classic test question: The following two half-reactions occur in a voltaic cell:
Brown and LeMay p. 867
Ni(s) → Ni2+ (aq) + 2 e(electrode = Ni)
Cu2+ (aq) + 2 e- → Cu0 (electrode = Cu)
Which of the following descriptions most accurately describes what is occurring in the half-cell
containing the Cu electrode and Cu2+(aq)?
1)
2)
3)
4)
The electrode is losing mass and the cations from the salt bridge are flowing into the half-cell
The electrode is gaining mass and the cations from the salt bridge are flowing into the half-cell
The electrode is losing mass and anions from the salt bridge are flowing into the half-cell
The electrode is gaining mass and anion from the salt bridge are flowing into the half-cell
ans: *choice 2) You must understand that the solution of the half-cell is something like CuSO(aq) or
Cu(NO3)(aq) in which both Cu2+ cation and 2 (NO3)-(aq) [for instance] exist. As the reaction proceeds,
Cu2+(aq) are reduced … thus becoming Cu0(s) and clinging to the cathode of copper metal. This loss of
+ ions from the solution demands a replacement of said ions … The salt bridge can provide those + ions
designed only to balance out the charge of the aqueous solution … thus allowing the flow of electrons to
continue through the external wire.
TRY THIS!!! Another classic test question:
Brown and LeMay p. 867
Given: Cr2O72-(aq) + 14 H+ (aq) + 6 I-(aq) → 2 Cr3+ (aq) + 3 I2(s) + 7 H2O (ℓ)
The above is a spontaneous reaction. A solution of potassium dichromate and sulfuric acid is poured
into one beaker, and a solution of potassium iodide is poured into another. A salt bridge is used to
join the beakers. An inactive electrode is suspended into each solution, and the two electrodes are
connected by wires through a voltmeter, detecting the resulting electric current.
a) What is occurring at the anode, and to which species? *I- is being oxidized to I2
b) What is occurring at the cathode, and to which species? *Cr2O72- (or Cr6+) is being reduced to Cr3+
c) What is the oxidizing agent in this reaction? *Cr2O72- (or Cr6+)
d) What is the direction of electron migration? *From anode to cathode, thus from the half-cell holding
the iodide to the half-cell holding the dichromate, via the external wire. The electrodes themselves
are NOT participating in the reaction. They are only helping to transfer the e- from / to the solution
ions.
e) What is the direction of ion migration? *The cations of the salt bridge move towards the half-cell of
the cathode, and the anions of the salt bridge move towards the half-cell of the anode.
f) What are the signs of the electrodes? *The anode is – and the cathode is +
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TRY THIS!!! A classic test question: The following two half-reactions occur in a voltaic cell:
Brown and LeMay p. 868
Zn(s) → Zn2+ (aq) + 2 eClO3-(aq) + 6 H+ (aq) + 6 e- → Cl-(aq) + 3 H2O (ℓ)
(electrode = Zn)
(electrode = Pt)
a) Identify the reaction occurring at the anode: *Zn(s) → Zn2+(aq) + 2 eb) Identify the reaction occurring at the cathode:*ClO3-(aq) + 6 H+(aq) + 6 e- → Cl-(aq) + 3 H2O(ℓ)
c) What happens to the mass of the zinc electrode? increases / rts / decreases
*decreases
d) What happens to the mass of the platinum electrode? increases / rts / decreases *rts …study the
products … a hydrated ion and water is produced … no new metal species is produced.
e) Which electrode is positive? *Pt
f) What is the reducing agent of the reaction? *Zn0
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