Balance these reactions using the half-reaction method
Sn2+ + IO3- -> Sn4= + I- (acidic solution)
CrO2 + ClO- -> CrO42- + Cl- (basic solution)
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To balance the first reaction (Sn2+ + IO3- -> Sn4+ + I-), follow the steps below using the half-reaction method in an acidic solution:
Step 1: Identify the oxidation and reduction half-reactions.
In this case, Sn2+ is being oxidized to Sn4+, and IO3- is being reduced to I-.
Step 2: Write the unbalanced half-reactions.
Oxidation half-reaction: Sn2+ -> Sn4+
Reduction half-reaction: IO3- -> I-
Step 3: Balance the atoms other than hydrogen and oxygen in each half-reaction.
Balance the oxidation half-reaction:
2 Sn2+ -> Sn4+
Balance the reduction half-reaction:
6 IO3- -> 5 I-
Step 4: Balance the oxygen atoms by adding water molecules (H2O) to the appropriate side of each equation.
Balance the oxidation half-reaction:
2 Sn2+ -> Sn4+ + 4 H2O
Balance the reduction half-reaction:
6 IO3- + 6 H2O -> 5 I-
Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+) to the appropriate side of each equation.
Balance the oxidation half-reaction:
2 Sn2+ -> Sn4+ + 4 H2O + 8 H+
Balance the reduction half-reaction:
6 IO3- + 6 H2O + 24 H+ -> 5 I- + 30 H+
Step 6: Balance the charges by adding electrons (e-) to the appropriate side of each equation.
Balance the oxidation half-reaction:
2 Sn2+ -> Sn4+ + 4 H2O + 8 H+ + 6e-
Balance the reduction half-reaction:
6 IO3- + 6 H2O + 24 H+ + 6e- -> 5 I- + 30 H+
Step 7: Multiply each half-reaction by a common factor to equalize the number of electrons transferred.
Multiply the oxidation half-reaction by 6:
12 Sn2+ -> 6 Sn4+ + 24 H2O + 48 H+ + 36e-
Multiply the reduction half-reaction by 12:
72 IO3- + 72 H2O + 288 H+ + 72e- -> 60 I- + 360 H+
Step 8: Combine the balanced half-reactions and cancel out any common terms.
The electrons in both reactions will cancel out, so you can add the two half-reactions together to form the balanced overall reaction:
12 Sn2+ + 72 IO3- + 72 H2O + 288 H+ -> 6 Sn4+ + 60 I- + 360 H+ + 24 H2O + 48 H+
Simplifying the reaction gives the final balanced equation:
12 Sn2+ + 72 IO3- + 24 H2O + 240 H+ -> 6 Sn4+ + 60 I- + 312 H2O
To balance these reactions using the half-reaction method, follow these steps:
For Sn2+ + IO3- -> Sn4+ + I- (acidic solution):
Step 1: Split the reaction into two half-reactions, oxidation and reduction.
Oxidation half-reaction: Sn2+ -> Sn4+
Reduction half-reaction: IO3- -> I-
Step 2: Balance the atoms in each half-reaction.
In the oxidation half-reaction, there are 2 Sn atoms on both sides, so it is already balanced.
In the reduction half-reaction, there are 3 O atoms on the left (in IO3-) and only 1 O atom on the right (in I-). To balance the O atoms, add 2 H2O to the right side.
IO3- + 8H+ + 6e- -> I- + 3H2O
Step 3: Balance the charges in each half-reaction.
In the oxidation half-reaction, there is a +2 charge on the left side and a +4 charge on the right side. To balance the charges, add 2 electrons (e-) to the left side.
2Sn2+ -> 2Sn4+ + 4e-
In the reduction half-reaction, there is an 8+ charge on the left side and a 1- charge on the right side. To balance the charges, add 10 electrons (e-) to the left side.
10IO3- + 24H+ + 10e- -> 10I- + 15H2O
Step 4: Multiply the balanced half-reactions by appropriate coefficients to balance the number of electrons.
Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to equalize the number of electrons.
10Sn2+ -> 10Sn4+ + 20e-
20IO3- + 48H+ + 20e- -> 20I- + 30H2O
Step 5: Add the balanced half-reactions.
Combine the balanced half-reactions to get the balanced overall equation.
10Sn2+ + 20IO3- + 48H+ -> 10Sn4+ + 20I- + 30H2O
Finally, check the elements and charges to ensure that both sides of the equation are balanced.
For CrO2 + ClO- -> CrO42- + Cl- (basic solution):
Step 1: Split the reaction into two half-reactions, oxidation and reduction.
Oxidation half-reaction: CrO2 -> CrO42-
Reduction half-reaction: ClO- -> Cl-
Step 2: Balance the atoms in each half-reaction.
In the oxidation half-reaction, there is 1 Cr atom on both sides, so it is already balanced.
In the reduction half-reaction, there is 1 Cl atom on both sides, so it is already balanced.
Step 3: Balance the charges in each half-reaction.
In the oxidation half-reaction, there is a 2- charge on the left side and a 4- charge on the right side. To balance the charges, add 2 hydroxide ions (OH-) to the left side.
CrO2 + 4OH- -> CrO42- + 2H2O
In the reduction half-reaction, there is a 1- charge on the left side and a 1- charge on the right side. The charges are already balanced.
Step 4: Multiply the balanced half-reactions by appropriate coefficients to balance the number of OH- ions.
Multiply the oxidation half-reaction by 4 to equalize the number of OH- ions.
4CrO2 + 16OH- -> 4CrO42- + 8H2O
Step 5: Add the balanced half-reactions.
Combine the balanced half-reactions to get the balanced overall equation.
4CrO2 + 16OH- + ClO- -> 4CrO42- + Cl- + 8H2O
Again, check the elements and charges to ensure that both sides of the equation are balanced.