4. Using the balanced redox equation from question #3, answer the following: A] A student standardized 36.77 mL of a potassium permanganate solution. They used 0.222 grams of sodium oxalate, Na2C2O4. What is the M (molarity) of the KMnO4 2MnO4 + SC₂0₂² +1GH+ 2Mn²+ +10CO₂ + 8H₂O solution? Molarity= = moler volume(L)
4. Using the balanced redox equation from question #3, answer the following: A] A student standardized 36.77 mL of a potassium permanganate solution. They used 0.222 grams of sodium oxalate, Na2C2O4. What is the M (molarity) of the KMnO4 2MnO4 + SC₂0₂² +1GH+ 2Mn²+ +10CO₂ + 8H₂O solution? Molarity= = moler volume(L)
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter19: Principles Of Chemical Reactivity: Electron Transfer Reactions
Section: Chapter Questions
Problem 114SCQ: (a) Is it easier to reduce water in acid or base? To evaluate this, consider the half-reaction...
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Using the equation provided, please answer question 4
![3. Balance the following redox equation using the half-reaction method:
MnO4¹ + C204² ⇒ Mn²+ + CO2 (acidic medium)
2+
reduction half reachon: MnDy →Mn²+
Oxidation half reaction: C₂04 → CO₂z
->
MnO4 +8H¹+Se- →Mn²+ +41+₂0
2+
1123
C₂04² 200₂ +Se
(MnOy² +8H*Se¯ →Mn²t+41120) x2 (C₂04²-02 CO₂ + 2 €²) x 5
Now combine the two half-reactions
2 MnO4- + SC ₂0₁² - +16H+ → 2 Mn²+ +10 CO₂ + 8H₂0
N
HOND
WORDS
4. Using the balanced redox equation from question #3, answer the following:
A] A student standardized 36.77 mL of a potassium permanganate solution. They used
0.222 grams of sodium oxalate, Na2C2O4. What is the M (molarity) of the KMnO4
solution? Molarity = moles 2 MnO₂ + SC ₂0₁² HIGH+ 2M₁²+10CO₂ + 8H₂O
2
volume(L)
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2827534e-5d68-4ff0-b361-bfdeadbfe949%2F33c506d2-2986-4c6d-ab5c-9ff460dead57%2F9wbzr3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3. Balance the following redox equation using the half-reaction method:
MnO4¹ + C204² ⇒ Mn²+ + CO2 (acidic medium)
2+
reduction half reachon: MnDy →Mn²+
Oxidation half reaction: C₂04 → CO₂z
->
MnO4 +8H¹+Se- →Mn²+ +41+₂0
2+
1123
C₂04² 200₂ +Se
(MnOy² +8H*Se¯ →Mn²t+41120) x2 (C₂04²-02 CO₂ + 2 €²) x 5
Now combine the two half-reactions
2 MnO4- + SC ₂0₁² - +16H+ → 2 Mn²+ +10 CO₂ + 8H₂0
N
HOND
WORDS
4. Using the balanced redox equation from question #3, answer the following:
A] A student standardized 36.77 mL of a potassium permanganate solution. They used
0.222 grams of sodium oxalate, Na2C2O4. What is the M (molarity) of the KMnO4
solution? Molarity = moles 2 MnO₂ + SC ₂0₁² HIGH+ 2M₁²+10CO₂ + 8H₂O
2
volume(L)
-
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