In the redox reaction 3 NO + Hg → 2 NO2 + HgO, identify the oxidized and reduced elements, the oxidizing agent, and the reducing agent using oxidation numbers.

To analyze the redox reaction 3 NO + Hg → 2 NO2 + HgO, we need to assign oxidation numbers to the elements involved.

Let’s break it down:

  • In nitric oxide (NO), nitrogen (N) has an oxidation number of +2 and oxygen (O) has an oxidation number of -2.
  • In nitric dioxide (NO2), nitrogen (N) has an oxidation number of +4.
  • In mercury(II) oxide (HgO), mercury (Hg) has an oxidation number of +2 and oxygen (O) remains at -2.

Now, let’s summarize the oxidation states before and after the reaction:

  • For NO: N is +2 and O is -2.
  • For NO2: N is +4 (increased from +2 to +4).
  • For Hg: Hg is 0 (elemental state) and becomes +2 in HgO.

From this analysis:

  • The element oxidized is nitrogen (N) because its oxidation number increases from +2 to +4.
  • The element reduced is mercury (Hg) because its oxidation number decreases from 0 to +2.

Now, identify the agents:

  • Since nitrogen is oxidized, it acts as the reducing agent.
  • Since mercury is reduced, it serves as the oxidizing agent.

In conclusion:

  • Oxidized element: Nitrogen (N)
  • Reduced element: Mercury (Hg)
  • Oxidizing agent: Mercury (Hg)
  • Reducing agent: Nitrogen (N)

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