Organic Chemistry Chapter 9 Study Guide
Answer # 1:
Oxidizing Agent:
An oxidizing agent is a substance that oxidizes some other substances and gains electrons. Oxidation is basically a gain of oxygen. Oxidizing agents give oxygen to other substances. Oxidation is also defined as loss of hydrogen.
Let consider an example:
CH3CH2OH CH3CHO
Reducing Agent:
A reducing agent reduces other substances and loses electrons. Reduction is basically a loss of oxygen. Reducing agents remove oxygen from other substances. Reduction is also defined as gain of hydrogen.
Let consider an example:
CH3CHO CH3CH2OH
Answer # 2:
Part A)
KMnO4
1 + Mn + (-2)*(4) = 0
+1 + Mn – 8 = 0
Mn = +8 -1
Mn = +7
Hence,
Oxidation number of one Mn atom in KMnO4 = +7
Part B)
…show more content…
Oxidation state of an element is a totally suppositious idea which has got nothing to do with loss or gain of electrons for atom necessity definite quantity of energy to release an electron and they are liberated from the outermost shell simply. It is difficult to eliminate seven electrons. It desires massive energy so it doesn’t occur.
Answer # 3:
In first reaction KMnO4 and Fe2+:
MnO4- + Fe2+ + H+ Mn2+ + Fe3+ + 4H2O
The main thing that I notice is that there is not sufficient Oxygen in the equation; there are 4 on the left and merely 1 on the right. We balance this by putting a 4 next to the H2O on the right to give. Nonetheless, there is now not sufficient H on the left hand side of the equation, so we balance this out by placing an 8 next to the H+ on the left hand side.
MnO4- + Fe2+ + 8H+ Mn2+ + Fe3+ + 4H2O
Similar case will happen, when K2Cr2O7 react with Sn2+, total net ionic equation is:
Cr2O7-+ 3Sn2+ + 14H+ 2Cr3+ + 3Sn4+ +
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