Question

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

Answer 1

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 $x$ 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 $x$. The oxidation state of all five atoms in the formula KNO₃ shall add up to zero. $1\times (+1) + 1 \times (x) + {\bf 3} \times (-2) = 0\\x = +5$. 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 $y$.
• 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: $1\times (+1) + 1 \times (y) + {\bf 2}\times (-2) = 0\\y = +3$. 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.

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

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: ${\bf 2} \times (+1) + (-2) = 0$.

$\rm \stackrel{}{2}\; \stackrel{\bf 0}{H}_2 + \stackrel{\bf 0}{O}_2\stackrel{}{\to} \stackrel{}{2}\;\stackrel{\bf +1}{H}_2\stackrel{\bf -2}{O}$.

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.