Question

question 38 (8 points)
for the unbalanced redox reaction ki(aq) + h₂o(l) + kmno₄(aq) → i₂(s) + mno₂(s) + koh(aq)
which has a balanced coefficient of 3?
which has a balanced coefficient of 2?
which is the spectator ion?
which is the reducing agent?
which is the oxidizing agent?
which has a balanced coefficient of 8?
which has a balanced coefficient of 6?
which has a balanced coefficient of 4?

1. i⁻(aq)
2. h₂o(l)
3. mno₄⁻(aq)
4. i₂(s)
5. oh⁻(aq)
6. k⁺(aq)

Explanation:

To solve this, we first balance the redox reaction. Let's identify oxidation and reduction half - reactions:

Step 1: Identify Oxidation and Reduction

• Oxidation (I⁻ to I₂): The oxidation state of I in KI (I⁻) is - 1, and in I₂ it is 0. So, 2I⁻→I₂ + 2e⁻ (loss of electrons, oxidation).
• Reduction (MnO₄⁻ to MnO₂): The oxidation state of Mn in MnO₄⁻ is + 7, and in MnO₂ it is + 4. So, MnO₄⁻+ 2H₂O+ 3e⁻→MnO₂ + 4OH⁻ (gain of electrons, reduction).

Step 2: Balance Electrons

To balance the electrons, we multiply the oxidation half - reaction by 3 and the reduction half - reaction by 2.

• Oxidation: \(6I^-(aq)

ightarrow3I_2(s)+6e^-\)

• Reduction: \(2MnO_4^-(aq)+4H_2O(l)+6e^-

ightarrow2MnO_2(s)+8OH^-(aq)\)

Step 3: Combine Half - Reactions and Balance the Whole Reaction

Adding the two half - reactions together:
\(6KI(aq)+2KMnO_4(aq)+4H_2O(l)
ightarrow3I_2(s)+2MnO_2(s)+8KOH(aq)\)

Now let's answer each part:

1. Which has a balanced coefficient of 3?

From the balanced reaction, the coefficient of \(I_2(s)\) (option 4) is 3.

2. Which has a balanced coefficient of 2?

From the balanced reaction, the coefficient of \(MnO_4^-(aq)\) (option 3) and \(MnO_2(s)\) is 2. Also, the coefficient of \(KMnO_4\) (which provides \(MnO_4^-\)) is 2.

3. Which is the spectator ion?

In the reaction, \(K^+(aq)\) (option 6) is a spectator ion because it does not participate in the redox process (its oxidation state remains + 1 throughout the reaction).

4. Which is the reducing agent?

The reducing agent is the species that gets oxidized. \(I^-(aq)\) (option 1) is oxidized (loses electrons), so it is the reducing agent.

5. Which is the oxidizing agent?

The oxidizing agent is the species that gets reduced. \(MnO_4^-(aq)\) (option 3) is reduced (gains electrons), so it is the oxidizing agent.

6. Which has a balanced coefficient of 8?

From the balanced reaction, the coefficient of \(OH^-(aq)\) (option 5) is 8.

7. Which has a balanced coefficient of 6?

From the balanced reaction, the coefficient of \(KI\) (which provides \(I^-\)) and the number of electrons transferred (but in terms of species, the coefficient of \(I^-\) from \(KI\) is 6) and the coefficient of \(KOH\) is 8, the coefficient of \(H_2O\) is 4, the coefficient of \(KI\) is 6. The coefficient of \(KI\) (which provides \(I^-\), option 1) is 6.

8. Which has a balanced coefficient of 4?

From the balanced reaction, the coefficient of \(H_2O(l)\) (option 2) is 4.

Answer:

s:

• Coefficient of 3: 4. \(I_2(s)\)
• Coefficient of 2: 3. \(MnO_4^-(aq)\)
• Spectator ion: 6. \(K^+(aq)\)
• Reducing agent: 1. \(I^-(aq)\)
• Oxidizing agent: 3. \(MnO_4^-(aq)\)
• Coefficient of 8: 5. \(OH^-(aq)\)
• Coefficient of 6: 1. \(I^-(aq)\) (from \(KI\))
• Coefficient of 4: 2. \(H_2O(l)\)