Balance by redox method:
1. S + HNO2 --> SO2 + NO + H2o
2. 24H +18 H2O + 8 MnO4 + 5 As4O6 --> 8 Mn2+ + 20 H3AsO4
http://www.chemteam.info/Redox/Redox.html
To balance these two chemical equations using the redox method, we need to follow these steps:
Step 1: Split the equations into half reactions
Step 2: Balance the atoms other than hydrogen and oxygen
Step 3: Balance the oxygen atoms by adding water molecules
Step 4: Balance the hydrogen atoms by adding hydrogen ions
Step 5: Balance the charges by adding electrons to one side of the reaction
Step 6: Multiply the half-reactions to make the number of electrons equal in both half reactions
Step 7: Add the half reactions together and cancel out any common terms
Step 8: Verify that the atoms and charges are balanced on both sides of the reaction
Let's start with the first equation:
1. S + HNO2 --> SO2 + NO + H2O
Splitting into half-reactions:
Reduction half-reaction: S --> SO2
Oxidation half-reaction: HNO2 --> NO + H2O
Balancing the atoms:
Reduction half-reaction: S --> SO2 (already balanced)
Oxidation half-reaction: HNO2 --> NO + H2O (already balanced)
Balancing the charges:
Reduction half-reaction: S --> SO2 (no charge change)
Oxidation half-reaction: HNO2 --> NO + H2O (no charge change)
The first equation is already balanced.
Now let's move on to the second equation:
2. 24H + 18 H2O + 8 MnO4 + 5 As4O6 --> 8 Mn2+ + 20 H3AsO4
Splitting into half-reactions:
Reduction half-reaction: MnO4 --> Mn2+
Oxidation half-reaction: As4O6 --> H3AsO4
Balancing the atoms:
Reduction half-reaction: MnO4 --> Mn2+
Oxidation half-reaction: As4O6 --> H3AsO4 (already balanced)
Balancing the oxygen atoms:
Reduction half-reaction: MnO4 --> Mn2+ (no oxygen on either side)
Oxidation half-reaction: As4O6 --> H3AsO4 (already balanced)
Balancing the hydrogen atoms:
Reduction half-reaction: MnO4 --> Mn2+ (no hydrogen on either side)
Oxidation half-reaction: As4O6 --> H3AsO4 (no hydrogen on either side)
Balancing the charges:
Reduction half-reaction: MnO4 --> Mn2+ (no charge change)
Oxidation half-reaction: As4O6 --> H3AsO4 (no charge change)
The second equation is already balanced.
So the balanced equations are:
1. S + HNO2 --> SO2 + NO + H2O
2. 24H + 18H2O + 8MnO4 + 5As4O6 --> 8Mn2+ + 20H3AsO4
To balance these redox equations, you need to follow these steps:
Step 1: Identify the oxidized and reduced elements in the equation.
In the first equation, sulfur (S) is oxidized from a 0 oxidation state to a +4 oxidation state, and nitrogen (N) is reduced from a +3 oxidation state to a +2 oxidation state.
In the second equation, arsenic (As) is oxidized from a +3 oxidation state to a +5 oxidation state, and manganese (Mn) is reduced from a +7 oxidation state to a +2 oxidation state.
Step 2: Write the half-reactions for oxidation and reduction.
In the first equation, the oxidation half-reaction is: S --> SO2
The reduction half-reaction is: HNO2 --> NO
In the second equation, the oxidation half-reaction is: As4O6 --> H3AsO4
The reduction half-reaction is: MnO4- --> Mn2+
Step 3: Balance the atoms in the half-reactions, excluding O and H.
In the first equation, balance the sulfur atoms in both the oxidation and reduction half-reactions. The balanced oxidation half-reaction is: S --> SO2
The balanced reduction half-reaction is: 2 HNO2 --> 2 NO
In the second equation, balance the arsenic atoms in both the oxidation and reduction half-reactions. The balanced oxidation half-reaction is: 2 As4O6 --> 6 H3AsO4
The balanced reduction half-reaction is: 8 MnO4- --> 8 Mn2+
Step 4: Balance the oxygen (O) atoms by adding H2O molecules, as needed.
In the first equation, add 2 H2O to the left side of the oxidation half-reaction to balance the O atoms: S + 2 H2O --> SO2
In the second equation, add 6 H2O to the left side of the oxidation half-reaction to balance the O atoms: 2 As4O6 + 6 H2O --> 6 H3AsO4
Step 5: Balance the hydrogen (H) atoms by adding H+ ions, as needed.
In the first equation, add 4 H+ ions to the right side of the reduction half-reaction to balance the H atoms: 2 HNO2 + 4 H+ --> 2 NO
In the second equation, add 18 H+ ions to the right side of the reduction half-reaction to balance the H atoms: 8 MnO4- + 18 H+ --> 8 Mn2+
Step 6: Balance the charges in the half-reactions by adding electrons.
In the first equation, the reduction half-reaction already has 4 H+ ions on the right side, so you need to add 2 electrons (2e-) to the left side: 2 HNO2 + 4 H+ + 2e- --> 2 NO
In the second equation, the reduction half-reaction already has 18 H+ ions on the right side, so you need to add 10 electrons (10e-) to the left side: 8 MnO4- + 18 H+ + 10e- --> 8 Mn2+
Step 7: Multiply the half-reactions by appropriate coefficients to equalize the number of electrons.
In the first equation, multiply the oxidation half-reaction by 2 and the reduction half-reaction by 5 to equalize the number of electrons: 2S + 4 H2O --> 2 SO2
5(2 HNO2 + 4 H+ + 2e-) --> 5(2 NO)
In the second equation, multiply the oxidation half-reaction by 8 and the reduction half-reaction by 10 to equalize the number of electrons: 16 As4O6 + 48 H2O --> 48 H3AsO4
10(8 MnO4- + 18 H+ + 10e-) --> 10(8 Mn2+)
Step 8: Combine the half-reactions and cancel out common species.
Combine the half-reactions and cancel out common species:
2S + 4 H2O + 10(8 MnO4- + 18 H+ + 10e-) --> 10(2 NO) + 16 As4O6 + 48 H3AsO4 + 10(8 Mn2+)
Step 9: Simplify the equation and balance the coefficients.
Simplify the equation and balance the coefficients:
16 S + 24 HNO2 + 31 H2O + 80 MnO4- + 45 H+ --> 20 NO + 16 As4O6 + 48 H3AsO4 + 80 Mn2+
Thus, the balanced equation for the first reaction is:
16S + 24HNO2 --> 20NO + 16 As4O6 + 48 H3AsO4 + 80 Mn2+