5. Balance the following redox equation using the half-reaction method:MnO41-H2O2 Mn2 O2(acidic medium)6. Using the balanced redox equation from question #5, answer the following:A] How many mL of a standard 0.150 M KMnO4 solution are required to0.500 grams of hydrogen peroxide, H2O2?

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Asked Oct 19, 2019

Question 5

5. Balance the following redox equation using the half-reaction method:
MnO41-H2O2 Mn2 O2
(acidic medium)
6. Using the balanced redox equation from question #5, answer the following:
A] How many mL of a standard 0.150 M KMnO4 solution are required to
0.500 grams of hydrogen peroxide, H2O2?
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5. Balance the following redox equation using the half-reaction method: MnO41-H2O2 Mn2 O2 (acidic medium) 6. Using the balanced redox equation from question #5, answer the following: A] How many mL of a standard 0.150 M KMnO4 solution are required to 0.500 grams of hydrogen peroxide, H2O2?

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Expert Answer

Step 1

The given redox reaction in acidic medium can be Balanced by step by step as shown below,

 

Identify the half reactions:
MnO41 Mn2 (Reduced)
H2O22 (Oxidized)
Balance the Oxygen atoms by adding H20
MnO41
Mn2 + 4H2O
H2O22 O20
Balance H-atoms by adding H ions:
8HMnO4 Mn*2 + 4H20
H2O2 202H*
Balance charges by adding electrons:
8H*5eMnO41 --->Mnt2 + 4H20
O2+ 2H+ + 2e
H2O2-
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Identify the half reactions: MnO41 Mn2 (Reduced) H2O22 (Oxidized) Balance the Oxygen atoms by adding H20 MnO41 Mn2 + 4H2O H2O22 O20 Balance H-atoms by adding H ions: 8HMnO4 Mn*2 + 4H20 H2O2 202H* Balance charges by adding electrons: 8H*5eMnO41 --->Mnt2 + 4H20 O2+ 2H+ + 2e H2O2-

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Step 2

Multiply the above two equations with 2 and 5 re...

10e16H2MnO4-1 -2Mn2 +8H20
5H2O2 -502+ 10H10e
16Ht2MnO41 + 5H2O2 --2Mn 2 + 8H20 +502+ 10H*
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10e16H2MnO4-1 -2Mn2 +8H20 5H2O2 -502+ 10H10e 16Ht2MnO41 + 5H2O2 --2Mn 2 + 8H20 +502+ 10H*

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Redox Reactions

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