how do I balance the following skeleton redox equation for a reaction occurring in acidic solution. HNO2 + MnO4- yield to Mn2+ + NO3-
the final answer is:2MnO4- + H+ + 5HNO2 + = 2Mn2+ + 5NO3- + 3H2O
Why did the skeleton run away from the equations? It didn't have enough balance! But fear not, for I, the mighty Clown Bot, shall bring balance to the equation. Here's how:
Step 1: Let's start by balancing the nitrogen atoms. On the left side, we have one nitrogen atom in HNO2. On the right side, we have one nitrogen atom in NO3-. So, nitrogen is already balanced.
Step 2: Next, let's balance the oxygen atoms. On the left side, we have two oxygen atoms in HNO2. On the right side, we have three oxygen atoms in MnO4- and one oxygen atom in NO3-. That's a total of four oxygen atoms on the right side. To balance this, we need to add two water molecules (H2O) to the left side.
HNO2 + MnO4- → Mn2+ + NO3- + 2H2O
Step 3: Now, let's balance the hydrogen atoms. On the left side, we have two hydrogen atoms in HNO2. On the right side, we have four hydrogen atoms in 2H2O. To balance this, we can add four hydrogen ions (H+) to the right side.
HNO2 + MnO4- → Mn2+ + NO3- + 2H2O + 4H+
Step 4: Lastly, let's balance the charges. On the left side, we have no charge. On the right side, we have a +2 charge on Mn2+ and a -1 charge on NO3-. To balance this, we can add five electrons (e-) to the left side.
HNO2 + MnO4- + 5e- → Mn2+ + NO3- + 2H2O + 4H+
And there you have it! The skeleton equation has undergone a hilarious balancing act and emerged as a perfectly balanced redox equation in an acidic solution.
To balance the given redox equation HNO2 + MnO4- → Mn2+ + NO3- in acidic solution, follow these step-by-step instructions:
Step 1: Separate the equation into two half-reactions:
Half-reaction 1: HNO2 → NO3-
Half-reaction 2: MnO4- → Mn2+
Step 2: Balance the atoms other than oxygen and hydrogen in each half-reaction.
For the first half-reaction (HNO2 → NO3-), the nitrogen (N) and oxygen (O) atoms are already balanced.
For the second half-reaction (MnO4- → Mn2+), the manganese (Mn) and oxygen (O) atoms are already balanced.
Step 3: Balance the oxygen atoms by adding water (H2O) to the side that needs more oxygen.
For the first half-reaction, since the right side needs more oxygen, add three water molecules (3H2O) to the right side:
HNO2 → NO3- + 3H2O
For the second half-reaction, since the left side needs more oxygen, add four water molecules (4H2O) to the left side:
MnO4- + 4H2O → Mn2+
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen.
For the first half-reaction, the left side needs more hydrogen, so add two hydrogen ions (2H+) to the left side:
HNO2 + 2H+ → NO3- + 3H2O
For the second half-reaction, the right side needs more hydrogen, so add eight hydrogen ions (8H+) to the right side:
MnO4- + 4H2O → Mn2+ + 8H+
Step 5: Balance the charges by adding electrons (e-) to the side that needs more charge.
For the first half-reaction, the right side needs more negative charge, so add four electrons (4e-) to the right side:
HNO2 + 2H+ + 4e- → NO3- + 3H2O
For the second half-reaction, the left side needs more positive charge, so add five electrons (5e-) to the left side:
5e- + MnO4- + 4H2O → Mn2+ + 8H+
Step 6: Multiply one or both of the half-reactions by integers if necessary to make the number of electrons equal in both half-reactions.
In this case, both half-reactions already have the same number of electrons (4e- and 5e-), so this step is not needed.
Step 7: Combine the half-reactions, cancel out common terms, and simplify if necessary.
Multiply the first half-reaction by 5 and the second half-reaction by 4 to balance the number of electrons:
5HNO2 + 10H+ + 20e- → 5NO3- + 15H2O
20e- + 4MnO4- + 16H2O → 4Mn2+ + 32H+
Now, combine the two half-reactions:
5HNO2 + 10H+ + 4MnO4- → 5NO3- + 4Mn2+ + 15H2O + 32H+
Simplify the equation by canceling out common terms:
5HNO2 + 4MnO4- + 10H2O → 5NO3- + 4Mn2+ + 47H+
Therefore, the balanced redox equation for the reaction occurring in acidic solution is:
5HNO2 + 4MnO4- + 10H2O → 5NO3- + 4Mn2+ + 47H+
To balance a redox equation in acidic solution, follow these steps:
Step 1: Identify the oxidation states of each element in the reaction.
In this case, the oxidation state of hydrogen (H) is +1, nitrogen (N) in HNO2 is +3, manganese (Mn) in MnO4- is +7, and oxygen (O) is -2. The oxidation state of an element in a compound is the charge it would have if all the bonding electrons were assigned to the more electronegative atom.
Step 2: Separate the equation into two half-reactions: an oxidation half-reaction and a reduction half-reaction.
The oxidation half-reaction involves the species being oxidized, and the reduction half-reaction involves the species being reduced.
In this case:
Oxidation half-reaction: HNO2 -> NO3- (Nitrogen is going from +3 to +5)
Reduction half-reaction: MnO4- -> Mn2+ (Manganese is going from +7 to +2)
Step 3: Balance the non-hydrogen and non-oxygen elements first.
Balancing the non-hydrogen and non-oxygen elements helps ensure that the atoms in the equation are balanced, except for hydrogen and oxygen.
In the oxidation half-reaction:
HNO2 -> HNO3
To balance nitrogen (N), add a coefficient of 2 in front of HNO2:
2HNO2 -> HNO3
In the reduction half-reaction:
MnO4- -> Mn2+
Since manganese (Mn) is already balanced, the equation is already balanced for non-hydrogen and non-oxygen elements.
Step 4: Balance oxygen atoms by adding water molecules (H2O) to the sides that need oxygen.
In the oxidation half-reaction:
2HNO2 -> HNO3 + H2O
To balance the oxygen atoms, add a coefficient of 1 in front of H2O on the product side:
2HNO2 -> HNO3 + H2O
In the reduction half-reaction:
MnO4- -> Mn2+
The reduction half-reaction doesn't contain any oxygen atoms yet, so no balance is needed at this stage.
Step 5: Balance hydrogen atoms by adding hydrogen ions (H+) to the sides that need hydrogen.
In the oxidation half-reaction:
2HNO2 -> HNO3 + H2O
To balance the hydrogen atoms, add two hydrogen ions (H+) on the reactant side:
2HNO2 + 2H+ -> HNO3 + H2O
In the reduction half-reaction:
MnO4- -> Mn2+
The reduction half-reaction also doesn't contain any hydrogen atoms yet, so no balance is needed at this stage.
Step 6: Balance the charges in each half-reaction by adding electrons (e-).
In the oxidation half-reaction:
2HNO2 + 2H+ -> HNO3 + H2O
The oxidation state of nitrogen (N) is increasing by 2, so add 2 electrons (e-) to the product side:
2HNO2 + 2H+ + 2e- -> HNO3 + H2O
In the reduction half-reaction:
MnO4- -> Mn2+
The reduction half-reaction needs to balance the change in oxidation state of manganese (Mn) by adding electrons (e-). In this case, the oxidation state decreases by 5, so add 5 electrons (e-) to the reactant side:
MnO4- + 5e- -> Mn2+
Step 7: Multiply the half-reactions by the appropriate factors so that the number of electrons transferred is equal in both half-reactions.
In this case, the oxidation half-reaction involves 2 electrons (e-), while the reduction half-reaction involves 5 electrons (e-). To balance the number of electrons, multiply the oxidation half-reaction by 5, and the reduction half-reaction by 2:
Oxidation half-reaction (after multiplying by 5):
10HNO2 + 10H+ + 10e- -> 5HNO3 + 5H2O
Reduction half-reaction (after multiplying by 2):
2MnO4- + 10e- -> 2Mn2+
Step 8: Combine the half-reactions.
To combine the half-reactions, you need to make sure that the number of electrons is the same in both half-reactions. In this case, both half-reactions involve 10 electrons (e-), so they can be combined directly:
10HNO2 + 10H+ + 10e- + 2MnO4- + 10e- -> 5HNO3 + 5H2O + 2Mn2+
Step 9: Simplify the equation, cancel out identical species, and confirm that the atoms and charges are balanced.
Simplifying the equation gives us the balanced redox equation in acidic solution:
10HNO2 + 12H+ + 2MnO4- -> 5HNO3 + 5H2O + 2Mn2+
Therefore, the balanced redox equation for the reaction in acidic solution is:
10HNO2 + 12H+ + 2MnO4- -> 5HNO3 + 5H2O + 2Mn2+