1. Balance the following half reactions in basic solution.
ClO3- (aq) �¨ ClO4- (aq)
Zn (s) �¨ Zn2+ (aq)
BrO3- (aq) �¨ Br- (aq)
IO3- (aq) �¨ I2 (s)
Sn2+ �¨ Sn4+
See your post below.
To balance the half-reactions in basic solution, follow these steps:
1. Write the unbalanced half-reactions for each species involved.
a) ClO3- (aq) -> ClO4- (aq)
b) Zn (s) -> Zn2+ (aq)
c) BrO3- (aq) -> Br- (aq)
d) IO3- (aq) -> I2 (s)
e) Sn2+ -> Sn4+
2. Balance the atoms and charges in each half-reaction.
a) ClO3- (aq) -> ClO4- (aq)
* Balance the oxygens by adding H2O:
ClO3- (aq) -> ClO4- (aq) + H2O
* Balance the charges by adding electrons:
ClO3- (aq) + 6e- -> ClO4- (aq) + H2O
b) Zn (s) -> Zn2+ (aq)
* Balance the charges by adding electrons:
Zn (s) -> Zn2+ (aq) + 2e-
c) BrO3- (aq) -> Br- (aq)
* Balance the oxygens by adding H2O:
BrO3- (aq) + 6H2O -> Br- (aq)
* Balance the charges by adding electrons:
BrO3- (aq) + 6H2O + 6e- -> Br- (aq)
d) IO3- (aq) -> I2 (s)
* Balance the oxygens by adding H2O:
IO3- (aq) -> I2 (s) + 3H2O
* Balance the charges by adding electrons:
IO3- (aq) + 6e- -> I2 (s) + 3H2O
e) Sn2+ -> Sn4+
* Balance the charges by adding electrons:
Sn2+ -> Sn4+ + 2e-
3. Adjust the number of electrons in each half-reaction to equalize the number of electrons transferred.
a) ClO3- (aq) + 6e- -> ClO4- (aq) + H2O
b) Zn (s) -> Zn2+ (aq) + 2e-
c) BrO3- (aq) + 6H2O + 6e- -> Br- (aq)
d) IO3- (aq) + 6e- -> I2 (s) + 3H2O
e) Sn2+ -> Sn4+ + 2e-
4. Multiply each half-reaction by a coefficient to equalize the number of electrons transferred. This step is optional if the number of electrons already matches.
a) ClO3- (aq) + 6e- -> ClO4- (aq) + H2O
Multiply by 2 to balance electrons:
2ClO3- (aq) + 12e- -> 2ClO4- (aq) + 2H2O
b) Zn (s) -> Zn2+ (aq) + 2e-
c) BrO3- (aq) + 6H2O + 6e- -> Br- (aq)
Multiply by 5 to balance electrons:
5BrO3- (aq) + 30H2O + 30e- -> 5Br- (aq)
d) IO3- (aq) + 6e- -> I2 (s) + 3H2O
Multiply by 2 to balance electrons:
2IO3- (aq) + 12e- -> 2I2 (s) + 6H2O
e) Sn2+ -> Sn4+ + 2e-
5. Combine the half-reactions and cancel out the common species.
ClO3- (aq) + 6e- -> ClO4- (aq) + H2O (x2)
Zn (s) -> Zn2+ (aq) + 2e-
BrO3- (aq) + 6H2O + 6e- -> Br- (aq) (x5)
IO3- (aq) + 6e- -> I2 (s) + 3H2O (x2)
Sn2+ -> Sn4+ + 2e-
Multiply each half-reaction to make the number of electrons equal:
2ClO3- (aq) + 12e- -> 2ClO4- (aq) + 2H2O
3Zn (s) -> 3Zn2+ (aq) + 6e-
5BrO3- (aq) + 30H2O + 30e- -> 5Br- (aq)
3IO3- (aq) + 18e- -> 3I2 (s) + 9H2O
10Sn2+ -> 10Sn4+ + 20e-
6. Add the half-reactions together and simplify if necessary.
2ClO3- (aq) + 12e- + 3Zn (s) + 30H2O + 30e- + 3IO3- (aq) + 18e- + 10Sn2+
-> 2ClO4- (aq) + 2H2O + 3Zn2+ (aq) + 6e- + 5Br- (aq) + 3I2 (s) + 9H2O + 10Sn4+ + 20e-
Cancel out the common species and combine:
2ClO3- (aq) + 3Zn (s) + 3IO3- (aq) + 10Sn2+
-> 2ClO4- (aq) + 3Zn2+ (aq) + 5Br- (aq) + 3I2 (s) + 10Sn4+
The balanced overall reaction in basic solution is:
2ClO3- (aq) + 3Zn (s) + 3IO3- (aq) + 10Sn2+ -> 2ClO4- (aq) + 3Zn2+ (aq) + 5Br- (aq) + 3I2 (s) + 10Sn4+
To balance the following half reactions in basic solution, we need to follow a few steps:
Step 1: Write down the unbalanced half reactions.
ClO3- (aq) → ClO4- (aq)
Zn (s) → Zn2+ (aq)
BrO3- (aq) → Br- (aq)
IO3- (aq) → I2 (s)
Sn2+ → Sn4+
Step 2: Balance the atoms other than hydrogen and oxygen.
For ClO3- (aq) → ClO4- (aq), the unbalanced atoms are chlorine.
We can balance the chlorine atoms by adding 4 Cl- on the right side:
ClO3- (aq) + 4 Cl- → ClO4- (aq)
For Zn (s) → Zn2+ (aq), there is only one atom, so it is already balanced.
For BrO3- (aq) → Br- (aq), the unbalanced atoms are bromine.
We can balance the bromine atoms by adding 6 Br- on the left side:
BrO3- (aq) + 6 Br- → Br- (aq)
For IO3- (aq) → I2 (s), the unbalanced atoms are iodine.
We can balance the iodine atoms by adding 6 I- on the left side:
IO3- (aq) + 6 I- → I2 (s)
For Sn2+ → Sn4+, there is only one atom, so it is already balanced.
Step 3: Balance the oxygen atoms using water molecules (H2O).
For ClO3- (aq) + 4 Cl- → ClO4- (aq), there are 6 oxygen atoms on the left side and 8 oxygen atoms on the right side.
We can balance the oxygen atoms by adding 2 H2O on the left side:
ClO3- (aq) + 4 Cl- + 2 H2O → ClO4- (aq)
For BrO3- (aq) + 6 Br- → Br- (aq), there are 9 oxygen atoms on the left side and 1 oxygen atom on the right side.
We can balance the oxygen atoms by adding 4 H2O on the right side:
BrO3- (aq) + 6 Br- → Br- (aq) + 4 H2O
For IO3- (aq) + 6 I- → I2 (s), there are 3 oxygen atoms on the left side and 0 oxygen atoms on the right side.
We can balance the oxygen atoms by adding 3 H2O on the right side:
IO3- (aq) + 6 I- → I2 (s) + 3 H2O
Step 4: Balance the hydrogen atoms using hydrogen ions (H+).
For ClO3- (aq) + 4 Cl- + 2 H2O → ClO4- (aq), there are 4 hydrogen atoms on the left side and 2 hydrogen atoms on the right side.
We can balance the hydrogen atoms by adding 4 H+ on the right side:
ClO3- (aq) + 4 Cl- + 2 H2O → ClO4- (aq) + 4 H+
For BrO3- (aq) + 6 Br- + 4 H2O → Br- (aq) + 4 H2O, there are 8 hydrogen atoms on the left side and 8 hydrogen atoms on the right side, so it is already balanced.
For IO3- (aq) + 6 I- + 3 H2O → I2 (s) + 3 H2O, there are 6 hydrogen atoms on the left side and 6 hydrogen atoms on the right side, so it is already balanced.
Step 5: Balance the charge using electrons.
For ClO3- (aq) + 4 Cl- + 2 H2O → ClO4- (aq) + 4 H+, the total charge on the left side is -3 and the total charge on the right side is -1.
We can balance the charge by adding 2 electrons (e-) on the left side:
ClO3- (aq) + 4 Cl- + 2 H2O + 2 e- → ClO4- (aq) + 4 H+
For BrO3- (aq) + 6 Br- + 4 H2O → Br- (aq) + 4 H2O, there is no charge imbalance, so it is already balanced.
For IO3- (aq) + 6 I- + 3 H2O → I2 (s) + 3 H2O, there is no charge imbalance, so it is already balanced.
Step 6: Combine the half reactions and cancel out common species.
ClO3- (aq) + 4 Cl- + 2 H2O + 2 e- → ClO4- (aq) + 4 H+
BrO3- (aq) + 6 Br- + 4 H2O → Br- (aq) + 4 H2O
IO3- (aq) + 6 I- + 3 H2O → I2 (s) + 3 H2O
Adding these three half reactions together, we get:
ClO3- (aq) + 4 Cl- + 2 H2O + 2 e- + BrO3- (aq) + 6 Br- + 4 H2O + IO3- (aq) + 6 I- + 3 H2O
→ ClO4- (aq) + 4 H+ + Br- (aq) + 4 H2O + I2 (s) + 3 H2O
Cancelling out the common species, we get the final balanced equation:
ClO3- (aq) + 6 e- + 6 H2O + 6 Br- + 6 I-
→ ClO4- (aq) + 4 H+ + 3 H2O + 3 Br- + 3 I2