balance the reaction using the half reaction method.
OsO4 + C2H4 -> Os + CO2
worksheet does not show if it is in a gas and aqueous state.
To balance the given reaction using the Half Reaction Method, follow these steps:
Step 1: Separate the oxidation and reduction half-reactions:
The oxidation half-reaction involves the loss of electrons, while the reduction half-reaction involves the gain of electrons.
Oxidation half-reaction:
OsO4 -> Os
Reduction half-reaction:
C2H4 -> CO2
Step 2: Balance the atoms in each half-reaction except for hydrogen and oxygen:
Start by balancing the least common atoms other than hydrogen and oxygen.
For the oxidation half-reaction:
OsO4 -> 2Os
For the reduction half-reaction:
C2H4 -> 2CO2
Step 3: Balance oxygen atoms by adding water (H2O) molecules:
In the oxidation half-reaction, there are 4 oxygen atoms on the left, so we need to add 4 water molecules (4H2O) on the right side.
OsO4 -> 2Os + 4H2O
In the reduction half-reaction, there are 4 oxygen atoms on the right. To balance them, we need to add 4 water molecules (4H2O) on the left side.
C2H4 + 4H2O -> 2CO2
Step 4: Balance hydrogen atoms by adding hydrogen ions (H+):
In the reduction half-reaction, there are 8 hydrogen atoms on the left side. To balance them, we need to add 8 hydrogen ions (8H+) on the right side.
C2H4 + 4H2O -> 2CO2 + 8H+
Step 5: Balance the charges by adding electrons:
Look at the charges on each side of the half-reactions.
In the oxidation half-reaction, there are no charges, so there is no need to balance them with electrons.
In the reduction half-reaction, there is an excess of positive charge due to the hydrogen ions. To balance the charges, add 8 electrons (8e-) to the left side.
C2H4 + 4H2O + 8e- -> 2CO2 + 8H+
Step 6: Balance the number of electrons in both half-reactions:
Since we have 8 electrons on the left side of the reduction half-reaction, we need 8 electrons on the right side of the oxidation half-reaction as well.
OsO4 + 8e- -> 2Os + 4H2O
Now the number of electrons is balanced in both half-reactions.
Step 7: Multiply the half-reactions so that the number of electrons gained equals the number of electrons lost:
Multiply each half-reaction by a coefficient to ensure that the total number of electrons gained equals the total number of electrons lost. In this case, multiply the oxidation half-reaction by 8 and the reduction half-reaction by 1.
8 OsO4 + 8e- -> 16 Os + 32 H2O
C2H4 + 4H2O + 8e- -> 2CO2 + 8H+
Step 8: Combine the half-reactions, cancel out any species that occur on both sides, and simplify if necessary:
Add the two half-reactions together to obtain the balanced overall reaction:
8 OsO4 + C2H4 + 4H2O -> 16 Os + 2CO2 + 8H+ + 32 H2O
Next, cancel out any species that appear on both sides of the equation, such as water:
8 OsO4 + C2H4 -> 16 Os + 2CO2 + 8H+
Finally, simplify the equation by dividing all coefficients by the greatest common divisor if necessary.
The final balanced equation using the half-reaction method is:
2 OsO4 + C2H4 -> 4 Os + CO2 + 4 H2O