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Identify the oxidizing agent and the reducing agent on the left side of each of the...

Identify the oxidizing agent and the reducing agent on the left side of each of the following reactions. (a) Cr2O72- + 3Sn2+ + 14H+ → 2 Cr3+ + 3 Sn4+ + 7 H2O (b) 4I- + O2 + 4H+ → 2I2 + 2H2O (c)5CH3CHO + 2MnO4- + 6H+ → 5CH3CO2H + 2Mn2+ + 3H2O (d) C8H8 + 2Na →C8H82- + 2Na+
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Answer #1

(a) Cr2O72- + 3Sn2+ + 14H+ → 2 Cr3+ + 3 Sn4+ + 7 H2O

Sn goes from +2 to +4; this is being oxidized, therefore the reducing agent (as well as H+)

Cr goes from +7 to +3; this is being reduced, therefore the oxidizing agent

(b) 4I- + O2 + 4H+ → 2I2 + 2H2O

I goes from -1 --> 0; this is oxidation, therefore the reducing agent (as well as H+)

O2 goes from 0 --> -2; this is reduction, therefore the oxidizing agnet

(c)5CH3CHO + 2MnO4- + 6H+ → 5CH3CO2H + 2Mn2+ + 3H2O

Mn goes from +7 to +2; this is reduction, therefore it is the oxidizing agent

C goes from +4 to +2; therefore, this is oxidation (it also gains Oxygen) therefore it is the reduction agent (as well as H+)

(d) C8H8 + 2Na →C8H82- + 2Na+

Na goes form 0 to +1, this is oxidation, it ist he reducing agent

C goes from +4 to +3 ; this is is reduction, it is th eoxidizing agent

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