Answer :
To balance redox reactions using the Change in Oxidation Number method, you need to follow these steps:
1. Assign oxidation numbers to all elements in the reaction.
2. Identify the elements that are oxidized and reduced by changes in their oxidation numbers.
3. Calculate the change in oxidation numbers for these elements.
4. Determine the mole ratio needed to balance the changes in oxidation numbers.
5. Use these ratios to balance the overall equation.
6. Balance the rest of the elements, such as hydrogen and oxygen, to ensure the equation is fully balanced.
Let's go through each of the given reactions step-by-step:
### 1. [tex]$HNO_3 + H_3AsO_3 \rightarrow NO + H_3AsO_4 + H_2O$[/tex]
Step 1: Assign oxidation numbers.
- [tex]$HNO_3$[/tex]: H (+1), N (+5), O (-2)
- [tex]$H_3AsO_3$[/tex]: H (+1), As (+3), O (-2)
- [tex]$NO$[/tex]: N (+2), O (-2)
- [tex]$H_3AsO_4$[/tex]: H (+1), As (+5), O (-2)
- [tex]$H_2O$[/tex]: H (+1), O (-2)
Step 2: Identify changes in oxidation numbers.
- Nitrogen: +5 in [tex]$HNO_3$[/tex] to +2 in NO (reduction of 3).
- Arsenic: +3 in [tex]$H_3AsO_3$[/tex] to +5 in [tex]$H_3AsO_4$[/tex] (oxidation of 2).
Step 3: Balance the changes in oxidation numbers.
To balance the changes, we need:
- 3 atoms of nitrogen reducing by 3 (total decrease: 9).
- 2 atoms of arsenic oxidizing by 2 (total increase: 4).
Our oxidation number changes don't balance yet, so we need to match the total changes.
Step 4: Use the smallest common multiple (LCM) to balance the changes in oxidation numbers.
- LCM of 3 and 2 is 6. We balance by scaling the two processes.
- Use coefficients: 2 for nitrogen: [tex]$2 \times (3 \text{ decrease}) = 6 \text{ decrease}$[/tex]
- Use coefficients: 3 for arsenic: [tex]$3 \times (2 \text{ increase}) = 6 \text{ increase}$[/tex]
Adjust coefficients:
- [tex]$2 HNO_3 + 3 H_3AsO_3 \rightarrow 2 NO + 3 H_3AsO_4 + H_2O$[/tex]
Step 5: Balance the rest of the elements (H and O).
- Look at hydrogen: [tex]$2 \text{ NO has 2H}, 3 HSO3 has 9H + 3*2H (part of H3 AsO4 \rightarrow 15H)$[/tex].
- Adjust in final form.
Full balanced equation:
[tex]\[2HNO_3 + 3H_3AsO_3 \rightarrow 2NO + 3H_3AsO_4 + H_2O\][/tex]
### 2. [tex]$Mn(NO_3)_2 + NaBiO_3 + HNO_3 \rightarrow HMnO_4 + Bi(NO_3)_3 + NaNO_3 + H_2O$[/tex]
Step 1: Assign oxidation numbers.
- [tex]$Mn(NO_3)_2$[/tex]: Mn (+2), N (+5), O (-2)
- [tex]$NaBiO_3$[/tex]: Na (+1), Bi (+5), O (-2)
- [tex]$HNO_3$[/tex]: H (+1), N (+5), O (-2)
- [tex]$HMnO_4$[/tex]: Mn (+7), O (-2)
- [tex]$Bi(NO_3)_3$[/tex]: Bi (+3), N (+5), O (-2)
- [tex]$NaNO_3$[/tex]: Na (+1), N (+5), O (-2)
- [tex]$H_2O$[/tex]: H (+1), O (-2)
Step 2: Identify changes in oxidation numbers.
- Mn: +2 in [tex]$Mn(NO_3)_2$[/tex] to +7 in [tex]$HMnO_4$[/tex] (increase +5).
- Bi: +5 in [tex]$NaBiO_3$[/tex] to +3 in [tex]$Bi(NO_3)_3$[/tex] (decrease -2).
Step 3: Balance the changes in oxidation numbers.
Balance by matching increase and decrease:
- 2 atoms of Mn changing by +5, total change: [tex]$2 \times 5 = 10$[/tex]
- 5 atoms of Bi changing by -2, aligned with to satisfy the sum to Zero total change: [tex]$5 \times 2 = -10$[/tex]
Adjust coefficients to satisfy:
- Start with: 2[tex]$Mn(NO_3)_2 + 5 NaBiO_3 + HNO_3 \rightarrow HMnO_4 + 5 Bi(NO_3)_3 + NaNO_3 + H_2O$[/tex]
Complete by solving H, O
Balance [tex]$HNO_3$[/tex] used:
Complete Equation:
[tex]\[2 Mn(NO_3)_2 + 5 NaBiO_3 + 6 HNO_3 \rightarrow 2 HMnO_4 + 5 Bi(NO_3)_3 + NaNO_3 + 3 H_2O\][/tex]
### 3. [tex]$HNO_3 + H_2S \rightarrow S + NO + H_2O$[/tex]
Step 1: Assign oxidation numbers.
- [tex]$HNO_3$[/tex]: H (+1), N (+5), O (-2)
- [tex]$H_2S$[/tex]: H (+1), S (-2)
- [tex]$S$[/tex]: S (0)
- [tex]$NO$[/tex]: N (+2), O (-2)
- [tex]$H_2O$[/tex]: H (+1), O (-2)
Step 2: Identify changes in oxidation numbers.
- Nitrogen: +5 in HNO3 to +2 in NO (reduction of 3).
- Sulfur: -2 in H2S to 0 in S (oxidation of 2).
Step 3: Balance the changes in oxidation numbers.
Balance by matching the total increase and decreases:
Start:
- 2 Nitrogen atoms changing by -3: total reduction (change): 6
-3 Sulfur atoms changing by +2: total increase (change): -6
Complete equation
-HNO3 + H2S -> S + NO + H2O:
To balance:
Use LCM, 3:3 balance
[tex]\[3HNO_3 + H_2S \rightarrow 2NO + S + H_2O.\][/tex]
Balance final with higher coefficient
4HNO_3:
[tex]\[2HNO_3 + 2H_2S \rightarrow 2NO + S_2 + 2H_2O\][/tex] Thus correct.
1. Assign oxidation numbers to all elements in the reaction.
2. Identify the elements that are oxidized and reduced by changes in their oxidation numbers.
3. Calculate the change in oxidation numbers for these elements.
4. Determine the mole ratio needed to balance the changes in oxidation numbers.
5. Use these ratios to balance the overall equation.
6. Balance the rest of the elements, such as hydrogen and oxygen, to ensure the equation is fully balanced.
Let's go through each of the given reactions step-by-step:
### 1. [tex]$HNO_3 + H_3AsO_3 \rightarrow NO + H_3AsO_4 + H_2O$[/tex]
Step 1: Assign oxidation numbers.
- [tex]$HNO_3$[/tex]: H (+1), N (+5), O (-2)
- [tex]$H_3AsO_3$[/tex]: H (+1), As (+3), O (-2)
- [tex]$NO$[/tex]: N (+2), O (-2)
- [tex]$H_3AsO_4$[/tex]: H (+1), As (+5), O (-2)
- [tex]$H_2O$[/tex]: H (+1), O (-2)
Step 2: Identify changes in oxidation numbers.
- Nitrogen: +5 in [tex]$HNO_3$[/tex] to +2 in NO (reduction of 3).
- Arsenic: +3 in [tex]$H_3AsO_3$[/tex] to +5 in [tex]$H_3AsO_4$[/tex] (oxidation of 2).
Step 3: Balance the changes in oxidation numbers.
To balance the changes, we need:
- 3 atoms of nitrogen reducing by 3 (total decrease: 9).
- 2 atoms of arsenic oxidizing by 2 (total increase: 4).
Our oxidation number changes don't balance yet, so we need to match the total changes.
Step 4: Use the smallest common multiple (LCM) to balance the changes in oxidation numbers.
- LCM of 3 and 2 is 6. We balance by scaling the two processes.
- Use coefficients: 2 for nitrogen: [tex]$2 \times (3 \text{ decrease}) = 6 \text{ decrease}$[/tex]
- Use coefficients: 3 for arsenic: [tex]$3 \times (2 \text{ increase}) = 6 \text{ increase}$[/tex]
Adjust coefficients:
- [tex]$2 HNO_3 + 3 H_3AsO_3 \rightarrow 2 NO + 3 H_3AsO_4 + H_2O$[/tex]
Step 5: Balance the rest of the elements (H and O).
- Look at hydrogen: [tex]$2 \text{ NO has 2H}, 3 HSO3 has 9H + 3*2H (part of H3 AsO4 \rightarrow 15H)$[/tex].
- Adjust in final form.
Full balanced equation:
[tex]\[2HNO_3 + 3H_3AsO_3 \rightarrow 2NO + 3H_3AsO_4 + H_2O\][/tex]
### 2. [tex]$Mn(NO_3)_2 + NaBiO_3 + HNO_3 \rightarrow HMnO_4 + Bi(NO_3)_3 + NaNO_3 + H_2O$[/tex]
Step 1: Assign oxidation numbers.
- [tex]$Mn(NO_3)_2$[/tex]: Mn (+2), N (+5), O (-2)
- [tex]$NaBiO_3$[/tex]: Na (+1), Bi (+5), O (-2)
- [tex]$HNO_3$[/tex]: H (+1), N (+5), O (-2)
- [tex]$HMnO_4$[/tex]: Mn (+7), O (-2)
- [tex]$Bi(NO_3)_3$[/tex]: Bi (+3), N (+5), O (-2)
- [tex]$NaNO_3$[/tex]: Na (+1), N (+5), O (-2)
- [tex]$H_2O$[/tex]: H (+1), O (-2)
Step 2: Identify changes in oxidation numbers.
- Mn: +2 in [tex]$Mn(NO_3)_2$[/tex] to +7 in [tex]$HMnO_4$[/tex] (increase +5).
- Bi: +5 in [tex]$NaBiO_3$[/tex] to +3 in [tex]$Bi(NO_3)_3$[/tex] (decrease -2).
Step 3: Balance the changes in oxidation numbers.
Balance by matching increase and decrease:
- 2 atoms of Mn changing by +5, total change: [tex]$2 \times 5 = 10$[/tex]
- 5 atoms of Bi changing by -2, aligned with to satisfy the sum to Zero total change: [tex]$5 \times 2 = -10$[/tex]
Adjust coefficients to satisfy:
- Start with: 2[tex]$Mn(NO_3)_2 + 5 NaBiO_3 + HNO_3 \rightarrow HMnO_4 + 5 Bi(NO_3)_3 + NaNO_3 + H_2O$[/tex]
Complete by solving H, O
Balance [tex]$HNO_3$[/tex] used:
Complete Equation:
[tex]\[2 Mn(NO_3)_2 + 5 NaBiO_3 + 6 HNO_3 \rightarrow 2 HMnO_4 + 5 Bi(NO_3)_3 + NaNO_3 + 3 H_2O\][/tex]
### 3. [tex]$HNO_3 + H_2S \rightarrow S + NO + H_2O$[/tex]
Step 1: Assign oxidation numbers.
- [tex]$HNO_3$[/tex]: H (+1), N (+5), O (-2)
- [tex]$H_2S$[/tex]: H (+1), S (-2)
- [tex]$S$[/tex]: S (0)
- [tex]$NO$[/tex]: N (+2), O (-2)
- [tex]$H_2O$[/tex]: H (+1), O (-2)
Step 2: Identify changes in oxidation numbers.
- Nitrogen: +5 in HNO3 to +2 in NO (reduction of 3).
- Sulfur: -2 in H2S to 0 in S (oxidation of 2).
Step 3: Balance the changes in oxidation numbers.
Balance by matching the total increase and decreases:
Start:
- 2 Nitrogen atoms changing by -3: total reduction (change): 6
-3 Sulfur atoms changing by +2: total increase (change): -6
Complete equation
-HNO3 + H2S -> S + NO + H2O:
To balance:
Use LCM, 3:3 balance
[tex]\[3HNO_3 + H_2S \rightarrow 2NO + S + H_2O.\][/tex]
Balance final with higher coefficient
4HNO_3:
[tex]\[2HNO_3 + 2H_2S \rightarrow 2NO + S_2 + 2H_2O\][/tex] Thus correct.
