The Table 17 − 1 showing reduction potential values of various elements is given. The reason behind the corrosion of most of the metals and the difficulty in corrosion of noble metals like, gold, platinum, and silver is to be stated. Concept introduction: The metal undergoes oxidation and the oxygen undergoes reduction during corrosion. So, the metal with reduction potential values lower than oxygen easily gets corroded in the air. To determine: The reason of corrosion of most of the metals in air and the reason of difficulty in the corrosion of noble metals like gold, platinum, and silver.
The Table 17 − 1 showing reduction potential values of various elements is given. The reason behind the corrosion of most of the metals and the difficulty in corrosion of noble metals like, gold, platinum, and silver is to be stated. Concept introduction: The metal undergoes oxidation and the oxygen undergoes reduction during corrosion. So, the metal with reduction potential values lower than oxygen easily gets corroded in the air. To determine: The reason of corrosion of most of the metals in air and the reason of difficulty in the corrosion of noble metals like gold, platinum, and silver.
Solution Summary: The author analyzes the relationship between cell potential and Gibbs free energy change by analyzing the Table 17-1 showing reduction potential values of various elements.
The Table
17−1 showing reduction potential values of various elements is given. The reason behind the corrosion of most of the metals and the difficulty in corrosion of noble metals like, gold, platinum, and silver is to be stated.
Concept introduction:
The metal undergoes oxidation and the oxygen undergoes reduction during corrosion. So, the metal with reduction potential values lower than oxygen easily gets corroded in the air.
To determine: The reason of corrosion of most of the metals in air and the reason of difficulty in the corrosion of noble metals like gold, platinum, and silver.
81. a. Propose a mechanism for the following reaction:
OH
CH2=CHCHC=N
b. What is the product of the following reaction?
HO
H₂O
N=CCH2CH2CH
OH
HO
CH3CCH=CH2
H₂O
C=N
82. Unlike a phosphonium ylide that reacts with an aldehyde or a ketone to form an alkene a sulfonium ulia
For each reaction below, decide if the first stable organic product that forms in solution will create a new CC bond, and check
the appropriate box.
Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below.
Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions
- just focus on the first stable product you expect to form in solution.
?
NH2
MgBr
Will the first product that forms in this reaction
create a new CC bond?
○ Yes
○ No
MgBr
?
Will the first product that forms in this reaction
create a new CC bond?
O Yes
O No
Click and drag to start drawing a
structure.
:☐
G
x
c
olo
Ar
HE
Predicting
As the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule
with a new C - C bond as its major product:
H₂N
O
H
1.
?
2. H3O+
If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more
than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for
example to distinguish between major products with different stereochemistry.
0
If the major products of this reaction won't have a new CC bond, just check the box under the drawing area and leave it blank.
فا
Explanation
Check
Click and drag to start drawing a
structure.
Chapter 17 Solutions
Bundle: Chemistry: An Atoms First Approach, Loose-leaf Version, 2nd + OWLv2 with Student Solutions Manual, 4 terms (24 months) Printed Access Card
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