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)
(a) (acidic) VO2+ (aq) -------V3+ (aq)
Step 1: Write the equation for the reduction half-reaction:
VO2+ (aq) + 2H+ (aq) + 2e- -------V3+ (aq) + H2O(l)
Step 2: Write the equation for the oxidation half-reaction:
No oxidation half-reaction is given.
Step 3: Determine the number of electrons transferred in the reduction half-reaction (Step 1).
In the reduction half-reaction, 2 electrons are transferred.
Step 4: Multiply the oxidation half-reaction (if given) by the number of electrons in the reduction half-reaction (Step 3) and balance it.
Since no oxidation half-reaction is given, we can skip this step.
Step 5: Ensure mass balance for both substances and charge balance for the overall reaction.
Multiply the reduction half-reaction by 3 in order to balance the charges and ensure mass balance:
3VO2+ (aq) + 6H+ (aq) + 6e- ------- 3V3+ (aq) + 3H2O(l)
Overall balanced equation (acidic conditions):
3VO2+ (aq) + 6H+ (aq) ------- 3V3+ (aq) + 3H2O(l)
(b) (basic) Ni(OH)2 (s)--------Ni2O3 (s)
Step 1: Write the equation for the reduction half-reaction:
No reduction half-reaction is given.
Step 2: Write the equation for the oxidation half-reaction:
Ni(OH)2 (s) ------Ni2O3 (s) + 3H2O(l)
Step 3: Determine the number of electrons transferred in the oxidation half-reaction (Step 2).
In the oxidation half-reaction, 6 electrons are transferred.
Step 4: Multiply the reduction half-reaction (if given) by the number of electrons in the oxidation half-reaction (Step 3) and balance it.
Since no reduction half-reaction is given, we can skip this step.
Step 5: Ensure mass balance for both substances and charge balance for the overall reaction.
Multiply the oxidation half-reaction by 2 in order to balance the charges and ensure mass balance:
2Ni(OH)2 (s) ------2Ni2O3 (s) + 6H2O(l)
Overall balanced equation (basic conditions):
2Ni(OH)2 (s) ------2Ni2O3 (s) + 6H2O(l)
(c) (acidic) NO3- (aq)-------NO2 (aq)
Step 1: Write the equation for the reduction half-reaction:
NO3- (aq) + 2H+ (aq) + 2e- -------NO2 (aq) + H2O(l)
Step 2: Write the equation for the oxidation half-reaction:
No oxidation half-reaction is given.
Step 3: Determine the number of electrons transferred in the reduction half-reaction (Step 1).
In the reduction half-reaction, 2 electrons are transferred.
Step 4: Multiply the oxidation half-reaction (if given) by the number of electrons in the reduction half-reaction (Step 3) and balance it.
Since no oxidation half-reaction is given, we can skip this step.
Step 5: Ensure mass balance for both substances and charge balance for the overall reaction.
Multiply the reduction half-reaction by 2 in order to balance the charges and ensure mass balance:
2NO3- (aq) + 4H+ (aq) + 4e- -------2NO2 (aq) + 2H2O(l)
Overall balanced equation (acidic conditions):
2NO3- (aq) + 4H+ (aq) + 4e- -------2NO2 (aq) + 2H2O(l)
(d) (basic) Br2 (aq)----------BrO3- (aq)
Step 1: Write the equation for the reduction half-reaction:
No reduction half-reaction is given.
Step 2: Write the equation for the oxidation half-reaction:
Br2 (aq) ------BrO3- (aq)
Step 3: Determine the number of electrons transferred in the oxidation half-reaction (Step 2).
In the oxidation half-reaction, 10 electrons are transferred.
Step 4: Multiply the reduction half-reaction (if given) by the number of electrons in the oxidation half-reaction (Step 3) and balance it.
Since no reduction half-reaction is given, we can skip this step.
Step 5: Ensure mass balance for both substances and charge balance for the overall reaction.
Multiply the oxidation half-reaction by 5 in order to balance the charges and ensure mass balance:
5Br2 (aq) ------5BrO3- (aq)
Overall balanced equation (basic conditions):
5Br2 (aq) ------5BrO3- (aq)
To balance these half-reactions, follow these steps:
(a) (acidic) VO2+ (aq) -------V3+ (aq)
1. Start by balancing the elements that are not hydrogen or oxygen.
VO2+ → V3+
There is already one vanadium (V) atom on both sides, so it is balanced.
2. Balance the oxygen atoms by adding water (H2O) molecules to the opposite side of the equation.
VO2+ → V3+ + H2O
There is one oxygen atom on the left and three oxygen atoms on the right. To balance the equation, add two water molecules to the left side.
VO2+ + 2H2O → V3+ + H2O
3. Balance the hydrogen atoms by adding hydrogen ions (H+) to the opposite side of the equation.
VO2+ + 2H2O → V3+ + 4H+
There are four hydrogen ions on the right and none on the left. To balance the equation, add four hydrogen ions to the left side.
VO2+ + 4H2O → V3+ + 4H+
4. Finally, balance the charge by adding electrons (e-) to the side that needs them. In this case, it is the left side.
VO2+ + 4H2O + 3e- → V3+ + 4H+
So, the balanced half-reaction is:
VO2+ + 4H2O + 3e- → V3+ + 4H+
You can follow these same steps to balance the remaining half-reactions: (b), (c), and (d)
You need to learn how to do these. I can help but the most help is by practice. In a, V goes from +4 on the lef to +3 on the right.
Here is a good site on redox including how to balance acidic and basic reactions.
http://www.chemteam.info/Redox/Redox.html