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)

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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)
-
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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