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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Question
![SO 1-
Step 1: Write the "skeleton" half-equation based on the species undergoing oxidation and
reduction.
Oxidation :
Reduction : k
Step 2: Balance each half-equation 'atomically in the order
a) atoms other than H and O
SO
D) O atoms by adding H,0 with the appropriate coefficient; H atoms by adding H* with
the appropriate coefficient.
S + 4 H;0
+ SO + 8H-
2
Step 3:
Examine the charge on both sides of the half-reaction. To the more positive side add
the correct number of electrons to equalize the charge with the other side.
s* • 4 H:0
2 + 2e
+SOP • 8H • 8 e
Step 4: Multiply each entire half-reaction by the lowest factor to equalize the number of
electrons lost and gained.
1 [ S* + 4 H;0
4[ a • 2e-
SO + 8H • 8e ·]
→ 21· ]
Step 5: Obtain the net redox equation by combining the half-equations. Make all possible
cancellations.
4H20
SO + 8H + 8e
4 +
8 e
s* • 4k + 4 H;0
SO + 81.. 8 H-
Step 6: Change from acidic to a basic medium by adding OH to both sides of the net equation;
combining H* and OH to form H;0 and simplify.
8 OH • S • 4 12 + 4 H:0
SO • 8t + 8H 8 OH
8 H;0
8 OH + S*+ 4h + 44.0
The final balanced equation is:
8 OH + S + 4 l-
SO + 81 + 4 H;0
Check:
reactants
products
80
8H
81
charge: -10
80
8 H
81
-10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5233f398-8102-4939-a6a5-fbccd1e64ae9%2F5709c7c0-b9f7-4f9f-a718-129a47cdfc26%2Fz1ce62f_processed.png&w=3840&q=75)
Transcribed Image Text:SO 1-
Step 1: Write the "skeleton" half-equation based on the species undergoing oxidation and
reduction.
Oxidation :
Reduction : k
Step 2: Balance each half-equation 'atomically in the order
a) atoms other than H and O
SO
D) O atoms by adding H,0 with the appropriate coefficient; H atoms by adding H* with
the appropriate coefficient.
S + 4 H;0
+ SO + 8H-
2
Step 3:
Examine the charge on both sides of the half-reaction. To the more positive side add
the correct number of electrons to equalize the charge with the other side.
s* • 4 H:0
2 + 2e
+SOP • 8H • 8 e
Step 4: Multiply each entire half-reaction by the lowest factor to equalize the number of
electrons lost and gained.
1 [ S* + 4 H;0
4[ a • 2e-
SO + 8H • 8e ·]
→ 21· ]
Step 5: Obtain the net redox equation by combining the half-equations. Make all possible
cancellations.
4H20
SO + 8H + 8e
4 +
8 e
s* • 4k + 4 H;0
SO + 81.. 8 H-
Step 6: Change from acidic to a basic medium by adding OH to both sides of the net equation;
combining H* and OH to form H;0 and simplify.
8 OH • S • 4 12 + 4 H:0
SO • 8t + 8H 8 OH
8 H;0
8 OH + S*+ 4h + 44.0
The final balanced equation is:
8 OH + S + 4 l-
SO + 81 + 4 H;0
Check:
reactants
products
80
8H
81
charge: -10
80
8 H
81
-10

Transcribed Image Text:B. Basic Medium
Mno4
So, -
so,-
+
MnO2
Balanced equation:
Reducing Agent:
Oxidizing Agent:
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