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Social Science3. Balance the following redox equation using the half-reaction method:MnO41 C2042 Mn2* +CO2 (acidic medium)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 us0.222 grams of sodium oxalate, Na2C2O4. What is the M (molarity) of the KMsolution?
Question 3
![3. Balance the following redox equation using the half-reaction method:
MnO41 C2042 Mn2* +
CO2 (acidic medium)
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 us
0.222 grams of sodium oxalate, Na2C2O4. What is the M (molarity) of the KM
solution?](https://prod-qna-question-images.s3.amazonaws.com/qna-images/question/8a30a005-d7e5-458b-b4bc-cfc1eedff542/0a027f91-22ea-48a6-8b76-408e0d034b21/7iesw93.jpeg)
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3. Balance the following redox equation using the half-reaction method: MnO41 C2042 Mn2* + CO2 (acidic medium) 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 us 0.222 grams of sodium oxalate, Na2C2O4. What is the M (molarity) of the KM solution?
Expert Answer
Balancing the redox reaction by using the half-reaction method (acidic medium).
In this method, some point is followed during balancing the redox reaction.
1) Identify reduction -half and oxidation half.
2) In both half-reaction balance all the other atom, except oxygen and hydrogen atom.
3) Balance the oxygen atom by adding H2O.
4) Balance the hydrogen atom by adding H+ ion.
5) balance charge only by adding electrons.
Since the given reaction is-
Balancing of reduction half-reaction
1) Since on both sides, Mn is one so it is balanced.
2) on the left-hand side Oxygen atom is 4 and in right- hand side oxygen atom is absent, so add 4 H2O in the right-side. Now to balance hydrogen atom add 8 H+ ions on the left side.
3)now on the left side total charge is +7 whereas on the right side is only+2. So to balance charge add 5 electrons on the left side.
Thus the balanced reduction half-reaction
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Mno+C0M22+ +CO,(Acidic medium oxidation half reaction-: C20C reduction half reaction -: MnO MnO 8Ht + 5 e 4 - Mn2+ + 4 -(1)
Balancing of the oxidation half-reaction-:
1) on the left side 2 carbon atoms are present so to balance the carbon atom on right side multiply CO2 by 2
2) now on both side oxygen atom are...
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c2o22Co2 2 CO, +5 e -(2)
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