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Use the changes in oxidation numbers to identify which atom is oxidized, reduced, the oxidizing agent, and the reducing agent. 5

A
2H2(g) + O2(g) → 2H2O(l)

B
2KNO3(s) → 2KNO2(s) + O2(g)

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jacob193

Answer:

Reaction A:

  • Hydrogen atoms in H₂ are oxidized.
  • Oxygen atoms in O₂ are reduced.
  • Hydrogen gas H₂ is the reducing agent.
  • Oxygen gas O₂ is the oxidizing agent.

Reaction B:

  • Oxygen atoms in KNO₃ are oxidized.
  • Nitrogen atoms in KNO₃ are reduced.
  • Potassium nitrate (V) KNO₃ is both the oxidizing agent and the reducing agent.

Explanation:

  • When an atom is oxidized, its oxidation number increases.
  • When an atom is reduced, its oxidation number decreases.
  • The oxidizing agent contains atoms that are reduced.
  • The reducing agent contains atoms that are oxidized.

Here are some common rules for assigning oxidation states.

  • Oxidation states on all atoms in a neutral compound shall add up to 0.
  • The average oxidation state on an atom is zero if the compound contains only atoms of that element. (E.g., the oxidation state on O in O₂ is zero.)
  • The oxidation state on oxygen atoms in compounds is typically -2. (Exceptions: oxygen bonded to fluorine, and peroxides.)
  • The oxidation state on group one metals (Li, Na, K) in compounds is typically +1.
  • The oxidation state on group two metals (Mg, Ca, Ba) in compounds is typically +2.
  • The oxidation state on H in compounds is typically +1. (Exceptions: metal hydrides where the oxidation state on H can be -1.)

For this question, only the rule about neutral compounds, oxygen, and group one metals (K in this case) are needed.

Reaction B

Oxidation states in KNO₃:

  • K is a group one metal. The oxidation state on K in the compound KNO₃ shall be +1.
  • The oxidation state on N tend to vary a lot, from -3 all the way to +5. Leave that as [tex]x[/tex] for now.
  • There's no fluorine in KNO₃. The ion NO₃⁻ stands for nitrate. There's no peroxide in that ion. The oxidation state on O in this compound shall be -2.
  • Let the oxidation state on N be [tex]x[/tex]. The oxidation state of all five atoms in the formula KNO₃ shall add up to zero. [tex]1\times (+1) + 1 \times (x) + {\bf 3} \times (-2) = 0\\x = +5[/tex]. As a result, the oxidation state on N in KNO₃ will be +5.

Similarly, for KNO₂:

  • The oxidation state on the group one metal K in KNO₂ will still be +1.
  • Let the oxidation state on N be [tex]y[/tex].
  • There's no peroxide in the nitrite ion, NO₂⁻, either. The oxidation state on O in KNO₂ will still be -2.
  • The oxidation state on all atoms in this formula shall add up to 0. Solve for the oxidation state on N: [tex]1\times (+1) + 1 \times (y) + {\bf 2}\times (-2) = 0\\y = +3[/tex]. The oxidation state on N in KNO₂ will be +3.

Oxygen is the only element in O₂. As a result,

  • The oxidation state on O in O₂ will be 0.

[tex]\rm\stackrel{+1}{K}\stackrel{\bf +5}{N}\stackrel{\bf -2}{O}_3 \to \stackrel{+1}{K}\stackrel{\bf+3}{N}\stackrel{\bf -2}{O}_2 + \stackrel{\bf 0}{O}_2[/tex].

The oxidation state on two oxygen atoms in KNO₃ increases from -2 to 0. These oxygen atoms are oxidized. KNO₃ is also the reducing agent.

The oxidation state on the nitrogen atom in KNO₃ decreases from +5 to +3. That nitrogen atom is reduced. As a result, KNO₃ is also the oxidizing agent.

Reaction A

Apply these steps to reaction A.

H₂:

  • Oxidation state on H: 0.

O₂:

  • Oxidation state on O: 0.

H₂O:

  • Oxidation state on H: +1.
  • Oxidation state on O: -2.
  • Double check: [tex]{\bf 2} \times (+1) + (-2) = 0[/tex].

[tex]\rm \stackrel{}{2}\; \stackrel{\bf 0}{H}_2 + \stackrel{\bf 0}{O}_2\stackrel{}{\to} \stackrel{}{2}\;\stackrel{\bf +1}{H}_2\stackrel{\bf -2}{O}[/tex].

The oxidation state on oxygen atoms decreases from 0 to -2. Those oxygen atoms are reduced. O₂ is thus the oxidizing agent.

The oxidation state on hydrogen atoms increases from 0 to +1. Those hydrogen atoms are oxidized. H₂ is thus the reducing agent.

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