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Question
question 37 (5 points)
for the reaction
no(g) + kmno₄(aq) → mno₂(s) + kno₃(aq)
which element is reduced?
- permanganate
which ion is a spectator ion?
- nitrogen
what is the oxidizing agent?
- manganese
what is the reducing agent?
- nitrogen monoxide
- potassium
which element is oxidized?
To solve this, we analyze the oxidation states:
Step 1: Determine Oxidation States
- N in NO: Oxidation state of O is -2. Let N be \( x \): \( x + (-2) = 0 \) → \( x = +2 \).
- Mn in \( \text{KMnO}_4 \): K is +1, O is -2. Let Mn be \( y \): \( +1 + y + 4(-2) = 0 \) → \( y = +7 \).
- Mn in \( \text{MnO}_2 \): O is -2. Let Mn be \( z \): \( z + 2(-2) = 0 \) → \( z = +4 \).
- N in \( \text{KNO}_3 \): K is +1, O is -2. Let N be \( w \): \( +1 + w + 3(-2) = 0 \) → \( w = +5 \).
- K (potassium): In \( \text{KMnO}_4 \) and \( \text{KNO}_3 \), K remains +1 (no change).
1. Which element is reduced?
Reduction = gain of electrons (decrease in oxidation state). Mn goes from \( +7 \) (in \( \text{KMnO}_4 \)) to \( +4 \) (in \( \text{MnO}_2 \)) → manganese (option 3).
2. Which ion is a spectator ion?
Spectator ion: No change in oxidation state. K (potassium) remains +1 in \( \text{KMnO}_4 \) and \( \text{KNO}_3 \) → potassium (option 5).
3. What is the oxidizing agent?
Oxidizing agent: Species that is reduced (causes oxidation of another species). \( \text{KMnO}_4 \) contains \( \text{MnO}_4^- \) (permanganate ion), and Mn is reduced here → permanganate (option 1).
4. What is the reducing agent?
Reducing agent: Species that is oxidized (causes reduction of another species). NO (nitrogen monoxide) has N oxidized from +2 to +5 → nitrogen monoxide (option 4).
5. Which element is oxidized?
Oxidation = loss of electrons (increase in oxidation state). N in NO goes from \( +2 \) to \( +5 \) (in \( \text{KNO}_3 \)) → nitrogen (option 2).
Final Answers:
- Which element is reduced? \(\boldsymbol{3}\) (manganese)
- Which ion is a spectator ion? \(\boldsymbol{5}\) (potassium)
- What is the oxidizing agent? \(\boldsymbol{1}\) (permanganate)
- What is the reducing agent? \(\boldsymbol{4}\) (nitrogen monoxide)
- Which element is oxidized? \(\boldsymbol{2}\) (nitrogen)
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To solve this, we analyze the oxidation states:
Step 1: Determine Oxidation States
- N in NO: Oxidation state of O is -2. Let N be \( x \): \( x + (-2) = 0 \) → \( x = +2 \).
- Mn in \( \text{KMnO}_4 \): K is +1, O is -2. Let Mn be \( y \): \( +1 + y + 4(-2) = 0 \) → \( y = +7 \).
- Mn in \( \text{MnO}_2 \): O is -2. Let Mn be \( z \): \( z + 2(-2) = 0 \) → \( z = +4 \).
- N in \( \text{KNO}_3 \): K is +1, O is -2. Let N be \( w \): \( +1 + w + 3(-2) = 0 \) → \( w = +5 \).
- K (potassium): In \( \text{KMnO}_4 \) and \( \text{KNO}_3 \), K remains +1 (no change).
1. Which element is reduced?
Reduction = gain of electrons (decrease in oxidation state). Mn goes from \( +7 \) (in \( \text{KMnO}_4 \)) to \( +4 \) (in \( \text{MnO}_2 \)) → manganese (option 3).
2. Which ion is a spectator ion?
Spectator ion: No change in oxidation state. K (potassium) remains +1 in \( \text{KMnO}_4 \) and \( \text{KNO}_3 \) → potassium (option 5).
3. What is the oxidizing agent?
Oxidizing agent: Species that is reduced (causes oxidation of another species). \( \text{KMnO}_4 \) contains \( \text{MnO}_4^- \) (permanganate ion), and Mn is reduced here → permanganate (option 1).
4. What is the reducing agent?
Reducing agent: Species that is oxidized (causes reduction of another species). NO (nitrogen monoxide) has N oxidized from +2 to +5 → nitrogen monoxide (option 4).
5. Which element is oxidized?
Oxidation = loss of electrons (increase in oxidation state). N in NO goes from \( +2 \) to \( +5 \) (in \( \text{KNO}_3 \)) → nitrogen (option 2).
Final Answers:
- Which element is reduced? \(\boldsymbol{3}\) (manganese)
- Which ion is a spectator ion? \(\boldsymbol{5}\) (potassium)
- What is the oxidizing agent? \(\boldsymbol{1}\) (permanganate)
- What is the reducing agent? \(\boldsymbol{4}\) (nitrogen monoxide)
- Which element is oxidized? \(\boldsymbol{2}\) (nitrogen)