Balance the following half-reactions:
(a) (acidic) VO2+ (aq) -------V3+ (aq)
(b) (basic Ni(OH)2 (s)--------Ni2O3 (s)
(c) (acidic) NO3- (aq)-------NO2 (aq)
(d) (basic) Br2 (aq)----------BrO3- (aq)
To balance the half-reactions, you need to follow the steps for balancing redox reactions in acidic or basic solutions.
(a) (acidic) VO2+ (aq) → V3+ (aq)
To balance the atoms:
Step 1: Divide the reaction into two half-reactions.
VO2+ (aq) → V3+
Step 2: Balance oxygen by adding H2O molecules to the side of the equation that needs oxygen.
VO2+ (aq) + H2O → V3+
Step 3: Balance hydrogen by adding H+ ions to the side of the equation that needs hydrogen.
VO2+ (aq) + H2O → V3+ + 2H+
Step 4: Balance charge by adding electrons (e-) to the side of the equation that needs charge balance.
VO2+ (aq) + 2H2O + 2e- → V3+ + 2H+
The balanced half-reaction is:
VO2+ (aq) + 2H2O + 2e- → V3+ + 2H+
(b) (basic) Ni(OH)2 (s) → Ni2O3 (s)
To balance the atoms:
Step 1: Divide the reaction into two half-reactions.
Ni(OH)2 (s) → Ni2O3 (s)
Step 2: Balance oxygen by adding H2O molecules to the side of the equation that needs oxygen.
Ni(OH)2 (s) + H2O → Ni2O3 (s)
Step 3: Balance hydrogen by adding H+ ions to the side of the equation that needs hydrogen.
Ni(OH)2 (s) + 2H2O → Ni2O3 (s)
Step 4: Balance charge by adding electrons (e-) to the side of the equation that needs charge balance.
Ni(OH)2 (s) + 2H2O + 2e- → Ni2O3 (s)
The balanced half-reaction is:
Ni(OH)2 (s) + 2H2O + 2e- → Ni2O3 (s)
(c) (acidic) NO3- (aq) → NO2 (aq)
To balance the atoms:
Step 1: Divide the reaction into two half-reactions.
NO3- (aq) → NO2 (aq)
Step 2: Balance oxygen by adding H2O molecules to the side of the equation that needs oxygen.
NO3- (aq) + H2O → NO2 (aq)
Step 3: Balance hydrogen by adding H+ ions to the side of the equation that needs hydrogen.
NO3- (aq) + H2O → NO2 (aq) + 2H+
Step 4: Balance charge by adding electrons (e-) to the side of the equation that needs charge balance.
NO3- (aq) + H2O + 3e- → NO2 (aq) + 2H+
The balanced half-reaction is:
NO3- (aq) + H2O + 3e- → NO2 (aq) + 2H+
(d) (basic) Br2 (aq) → BrO3- (aq)
To balance the atoms:
Step 1: Divide the reaction into two half-reactions.
Br2 (aq) → BrO3- (aq)
Step 2: Balance oxygen by adding H2O molecules to the side of the equation that needs oxygen.
Br2 (aq) + 6H2O → BrO3- (aq)
Step 3: Balance hydrogen by adding H+ ions to the side of the equation that needs hydrogen.
Br2 (aq) + 6H2O → BrO3- (aq) + 6H+
Step 4: Balance charge by adding electrons (e-) to the side of the equation that needs charge balance.
Br2 (aq) + 6H2O + 12e- → BrO3- (aq) + 6H+
The balanced half-reaction is:
Br2 (aq) + 6H2O + 12e- → BrO3- (aq) + 6H+
To balance the half-reactions, we need to follow a series of steps. Let's go through each of the given examples:
(a) (acidic) VO2+ (aq) -------V3+ (aq)
Step 1: Identify the atoms that are undergoing oxidation and reduction.
In this case, we can see that vanadium is being reduced from +5 to +3.
Step 2: Balance the atoms that are not hydrogen or oxygen.
We start by balancing the vanadium atoms. There is only one vanadium atom in both the reactant and the product side, so it is already balanced.
Step 3: Balance oxygen atoms by adding water molecules (H2O) to the appropriate side.
There are three oxygen atoms on the left side of the reaction (in VO2+) and none on the right side. To balance this, we need to add three water molecules to the right side:
VO2+ (aq) + 3H2O(l) ------- V3+ (aq)
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the appropriate side.
There are already eight hydrogen ions on the right side of the reaction (in 3H2O). So, we add 8 hydrogen ions on the left side:
VO2+ (aq) + 3H2O(l) + 8H+(aq) ------- V3+ (aq)
Step 5: Balance the charges by adding electrons (e-) to the appropriate side.
The charge on the left side is +4 (from VO2+ and H+ ions), while the right side has a charge of +3 (from V3+). To balance the charges, we need to add one electron to the left side:
VO2+ (aq) + 3H2O(l) + 8H+(aq) + e- ------- V3+ (aq)
Now the left side is balanced with respect to atoms and charges.
(b) (basic) Ni(OH)2 (s)--------Ni2O3 (s)
Balancing half-reactions in a basic solution requires an additional step involving hydroxide ions (OH-). We follow a similar stepwise approach as in the acidic solution, but with an extra consideration for the basic conditions.
Step 1: Identify the atoms that are undergoing oxidation and reduction.
In this case, we can see that nickel is being oxidized from +2 to +3.
Step 2: Balance the atoms that are not hydrogen or oxygen.
We start by balancing the nickel atoms. There is one nickel atom on both the reactant and product side, so it is already balanced.
Step 3: Balance oxygen atoms by adding water molecules (H2O) to the appropriate side.
There are three oxygen atoms on the left side of the reaction (in Ni(OH)2) and six oxygen atoms on the right side (in Ni2O3). To balance this, we need to add three water molecules to the left side:
Ni(OH)2(s) + 3H2O(l) ------- Ni2O3(s)
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the appropriate side.
There are already six hydrogen ions on the right side of the reaction (in 3H2O). So, we add six hydroxide ions (OH-) on the left side:
Ni(OH)2(s) + 3H2O(l) + 6OH-(aq) ------- Ni2O3(s)
Step 5: Balance the charges by adding electrons (e-) to the appropriate side.
The charge on the left side is +2 (from Ni(OH)2), while the right side has a charge of +6 (from Ni2O3). To balance the charges, we need to add four electrons to the right side:
Ni(OH)2(s) + 3H2O(l) + 6OH-(aq) ------- Ni2O3(s) + 4e-
Now the reaction is balanced in a basic solution.
(c) (acidic) NO3- (aq)-------NO2 (aq)
Step 1: Identify the atoms that are undergoing oxidation and reduction.
In this case, we can see that nitrogen is being reduced from +5 to +4.
Step 2: Balance the atoms that are not hydrogen or oxygen.
There is one nitrogen atom on both the reactant and product side, so it is already balanced.
Step 3: Balance oxygen atoms by adding water molecules (H2O) to the appropriate side.
There are three oxygen atoms on the left side of the reaction (in NO3-) and two oxygen atoms on the right side (in NO2). To balance this, we need to add a water molecule to the right side:
NO3-(aq) + H2O(l) ------- NO2(aq)
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the appropriate side.
There are already four hydrogen ions on the right side of the reaction (in H2O). So, we add four hydrogen ions on the left side:
NO3-(aq) + H2O(l) + 4H+(aq) ------- NO2(aq)
Step 5: Balance the charges by adding electrons (e-) to the appropriate side.
The charge on the left side is -1 (from NO3- and H+ ions), while the right side has no charge. To balance the charges, we need to add one electron to the right side:
NO3-(aq) + H2O(l) + 4H+(aq) + e- ------- NO2(aq)
Now the reaction is balanced in an acidic solution.
(d) (basic) Br2 (aq)----------BrO3- (aq)
For this example, the given reaction already includes both bromine (Br2) and bromate (BrO3-) ions. So, we don't need to balance any additional atoms or charges. The reaction is already balanced in a basic solution.