Expanation:
Given the followin reactions we have to idenftify the oxidizing and reducing aget.
3 CH₃CHOHCH₃ + Cr₂O₇²⁻ + 8 H⁺ —> 3 (CH₃)₂CO + 2 Cr³⁺ + 7 H₂O
We can split this redox reaction in two half reactions.
3 CH₃CHOHCH₃ ----> 3 (CH₃)₂CO
Cr₂O₇²⁻ ----> 2 Cr³⁺
Let's balance the first half reaction, since it is an acidic meium, we can aDdd H+, molecules of waer and e-.
+ H8 H +⁺ + 8 e⁻
We have 24 atoms of H on the left and and 18 atoms of H of the right side. We have to add 8 H+ on the right side. We have 3 atoms of O on both sides. So we only have to add the electrons to balance the carges. )Also 8 on the right side.
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻----> 2 Cr³⁺ + 7 H₂O
In the second half reaction we have to add 7 molecules of water on the right side to balance the O atoms. We added 14 atoms of H on the right side, so we have to add 14 H+ on the left side. The total charge of the left side is +12 and the total charge of the right ide is is +6. We have to add 6 el- on the left side.
The two balanced half reactions are:
3 CH₃CHOHCH₃ ----> 3 (CH₃)₂CO + 8 H⁺ + 8 e⁻ Oxidation Half-reaction
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻----> 2 Cr³⁺ + 7 H₂O Reduction Half-reaction
If we pay attention to the equations we will see that Cr₂O₇²⁻ is gaining electrons, i is being reduced. So Cr₂O₇²⁻ is our oxidizing agent.
CH₃CHOHCH₃ is losing electrons, it is being oxidized. CH₃CHOHCH₃ is our reducing agent.
Answer:
Cr₂O₇²⁻: oxidizing agent.
CH₃CHOHCH₃: reducing agent.