Data Table Metal Metal Half reaction that goes Half reaction that goes as at the at the as a reduction (this is (+) Termi Termi terminal)* Voltaic Measured Volt age (V) Cell (metals used) an oxidation (this is for the (-) for the metal at the (+) metal at the (-) terminal)* nal nal # 1 # 2 # 3 Cu-Zn 0.881 0.864 0.850 Cu - Pb 0.486 0.489 0.495 Cu-Ag -0.420 -0.415 -0.407 Cu - Fe 0.544 0.555 0.571
Data Table Metal Metal Half reaction that goes Half reaction that goes as at the at the as a reduction (this is (+) Termi Termi terminal)* Voltaic Measured Volt age (V) Cell (metals used) an oxidation (this is for the (-) for the metal at the (+) metal at the (-) terminal)* nal nal # 1 # 2 # 3 Cu-Zn 0.881 0.864 0.850 Cu - Pb 0.486 0.489 0.495 Cu-Ag -0.420 -0.415 -0.407 Cu - Fe 0.544 0.555 0.571
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![Data Table
Metal Metal Half reaction that goes Half reaction that goes as
at the at the as a reduction (this is
(+)
Termi Termi terminal)*
Voltaic
Measured Volt age (V)
an oxidation (this is for the
for the metal at the (+) metal at the (-) terminal)*
Cell
(metals
used)
(-)
nal
nal
#1
#2
#3
Cu - Zn 0.881 0.864 0.850
Cu – Pb 0.486 0.489 0.495
Cu - Ag -0.420 -0.415 -0.407
Cu – Fe
0.544
0.555 0.571
TO HELP YOU DECIDE THE CHARGE OF EACH TERMINAL OF A HALF CELL.
An oxidation gives up electrons. Electrons are negative. So the half cell
an oxidation is taking place is going to be the negative terminal
negative electrons are coming out of this half cell.
where
because the
A reduction takes electrons. Taking of electrons is essentially making
the half
cell positive because the negative electrons are
half cell that goes as a reduction is going to be the positive
going into the half
cell. So the
electrode.
Results table 1
Metal at the Metal at
(+)
Terminal
Voltaic Cell
Reduction
Adjusted reduction
potential of metal
reacting with Cu
(assumes that Cu
has a reduction
Average
the (-)
Terminal
(metals used)
potential of metal
reacting with Cu
(assumes that Cu
has a reduction
Measured
Voltage from the
Data Table
(V)
voltage of 0.00 V)!
voltage of 0.34 V)2
Cu – Zn
0.865
Cu
Zn
-0.865
-0.525
Cu – Pb
0.49
Cu
Pb
-0.49
-0.150
Cu - Ag
-0.414
Ag
Cu
0.414
0.754
Cu - Fe
0.556
Cu
Fe
-0.556
-0.216](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e392a71-e7bd-4981-938b-8b094393c2fe%2F2549ab16-bc5f-49ac-a0c2-03c77531319c%2F13ko1dw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Data Table
Metal Metal Half reaction that goes Half reaction that goes as
at the at the as a reduction (this is
(+)
Termi Termi terminal)*
Voltaic
Measured Volt age (V)
an oxidation (this is for the
for the metal at the (+) metal at the (-) terminal)*
Cell
(metals
used)
(-)
nal
nal
#1
#2
#3
Cu - Zn 0.881 0.864 0.850
Cu – Pb 0.486 0.489 0.495
Cu - Ag -0.420 -0.415 -0.407
Cu – Fe
0.544
0.555 0.571
TO HELP YOU DECIDE THE CHARGE OF EACH TERMINAL OF A HALF CELL.
An oxidation gives up electrons. Electrons are negative. So the half cell
an oxidation is taking place is going to be the negative terminal
negative electrons are coming out of this half cell.
where
because the
A reduction takes electrons. Taking of electrons is essentially making
the half
cell positive because the negative electrons are
half cell that goes as a reduction is going to be the positive
going into the half
cell. So the
electrode.
Results table 1
Metal at the Metal at
(+)
Terminal
Voltaic Cell
Reduction
Adjusted reduction
potential of metal
reacting with Cu
(assumes that Cu
has a reduction
Average
the (-)
Terminal
(metals used)
potential of metal
reacting with Cu
(assumes that Cu
has a reduction
Measured
Voltage from the
Data Table
(V)
voltage of 0.00 V)!
voltage of 0.34 V)2
Cu – Zn
0.865
Cu
Zn
-0.865
-0.525
Cu – Pb
0.49
Cu
Pb
-0.49
-0.150
Cu - Ag
-0.414
Ag
Cu
0.414
0.754
Cu - Fe
0.556
Cu
Fe
-0.556
-0.216
![Footnote #1: If Cu is the (+) terminal and the other metal (M) is the (-) terminal, the
Cu is going as a reduction and M is going as an oxidation. By definition, the reduction
potential of M is (-). If Cu is (-) and M is (+), then Cu is going as an oxidation and
other metal is going as a reduction. By definition, the reduction potential of M is (+).
Essentially, this means that the sign of each voltage in the first column is to be
reversed when entered in this column.
Footnote #2: In the previous column, Cu was arbitrarily assigned a voltage of 0.00 V.
However, the actual standard reduction potential of Cu is 0.34 V. Thus, to convert the
reduction potential in the previous column to actual reduction potentials, add 0.34 V to
each entry.
Results Table 2
Metal Actual reduction Percent error between your
Reduction potential in
descending order from most
(+) to most (-) reordered
from the last column in
Results Table 1. (Don't
forget to insert Cu at the
+0.34 V position in this
table)
potential of each
measured reduction
metal from the
textbook
voltage (column #1) and
the actual reduction
(Zumdahl, 6th ed, voltage (column #3). (Cu
p 833)*
will, of course, have a 0%
error)**
* Be careful to chose the proper half reactions. These involve the metal and the
ions listed in the materials section on page L–1.
**Students in the past have found that iron gives the largest percent error. It
turns out that the "iron" we are using is really mild steel and is alloyed with other
metals, so a large error is to be expected. We have not found a source of pure iron
at this time.
Is the order of the metals in Results Table 2 the same as that in the table of Standard
Reduction Potentials on p 833 in Zumdahl's text?
Yes
No
(circle one)
If you circled No, check all your calculations. The order should come out the same as
in the standard table.
The formula for percent error:
%
error=yourvalue-book valuex
100% book value
The vertical lines stand for absolute value, meaning that the sign of the number
between the vertical lines is always positive.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e392a71-e7bd-4981-938b-8b094393c2fe%2F2549ab16-bc5f-49ac-a0c2-03c77531319c%2Fo5hnqyd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Footnote #1: If Cu is the (+) terminal and the other metal (M) is the (-) terminal, the
Cu is going as a reduction and M is going as an oxidation. By definition, the reduction
potential of M is (-). If Cu is (-) and M is (+), then Cu is going as an oxidation and
other metal is going as a reduction. By definition, the reduction potential of M is (+).
Essentially, this means that the sign of each voltage in the first column is to be
reversed when entered in this column.
Footnote #2: In the previous column, Cu was arbitrarily assigned a voltage of 0.00 V.
However, the actual standard reduction potential of Cu is 0.34 V. Thus, to convert the
reduction potential in the previous column to actual reduction potentials, add 0.34 V to
each entry.
Results Table 2
Metal Actual reduction Percent error between your
Reduction potential in
descending order from most
(+) to most (-) reordered
from the last column in
Results Table 1. (Don't
forget to insert Cu at the
+0.34 V position in this
table)
potential of each
measured reduction
metal from the
textbook
voltage (column #1) and
the actual reduction
(Zumdahl, 6th ed, voltage (column #3). (Cu
p 833)*
will, of course, have a 0%
error)**
* Be careful to chose the proper half reactions. These involve the metal and the
ions listed in the materials section on page L–1.
**Students in the past have found that iron gives the largest percent error. It
turns out that the "iron" we are using is really mild steel and is alloyed with other
metals, so a large error is to be expected. We have not found a source of pure iron
at this time.
Is the order of the metals in Results Table 2 the same as that in the table of Standard
Reduction Potentials on p 833 in Zumdahl's text?
Yes
No
(circle one)
If you circled No, check all your calculations. The order should come out the same as
in the standard table.
The formula for percent error:
%
error=yourvalue-book valuex
100% book value
The vertical lines stand for absolute value, meaning that the sign of the number
between the vertical lines is always positive.
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