c6+h6+o2=co2+h2o in acidic medium balance by oxidation method
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To balance the given equation, c6+h6+o2=co2+h2o in an acidic medium using the oxidation method, follow these steps:
Step 1: Identify the atoms that are undergoing oxidation and reduction.
In this equation, the carbon atoms are being oxidized from an oxidation state of 0 in C6 to +4 in CO2. The hydrogen atoms are being reduced from an oxidation state of 0 in H6 to +1 in H2O.
Step 2: Write and balance the half-reactions for oxidation and reduction separately.
Oxidation half-reaction: C6 → 6CO2
Reduction half-reaction: H6 → 3H2O
Step 3: Balance atoms other than oxygen and hydrogen.
In the oxidation half-reaction, since there is only carbon, it is already balanced.
In the reduction half-reaction, there are 6 hydrogen atoms on the left side and 6 hydrogen atoms on the right side. Therefore, it is already balanced.
Step 4: Balance oxygen atoms by adding water molecules.
In the oxidation half-reaction, there are 6 oxygen atoms on the right side, so we need 6 water molecules on the left side to balance it.
C6 → 6CO2 + 6H2O
In the reduction half-reaction, there are 6 oxygen atoms on the left side, so we need 3 water molecules on the right side to balance it.
H6 + 3H2O → 3H2O
Step 5: Balance the charges by adding electrons.
In the oxidation half-reaction, each carbon atom loses 8 electrons. As there are 6 carbon atoms, a total of 48 electrons are lost.
C6 → 6CO2 + 6H2O + 48e^-
In the reduction half-reaction, there is no charge on either side. Therefore, no electrons are needed.
H6 + 3H2O → 3H2O
Step 6: Balance the number of electrons transferred in both half-reactions.
To balance the number of electrons, we need to multiply the reduction half-reaction by 48.
48H6 + 144H2O → 144H2O
Now, the equation becomes:
C6 + 6H6 + O2 → 6CO2 + 6H2O + 48e-
48H6 + 144H2O → 144H2O
Step 7: Combine the balanced half-reactions.
To combine the two half-reactions, you need to ensure that the number of electrons lost in the oxidation half-reaction (48e-) is equal to the number of electrons gained in the reduction half-reaction (48e-).
Multiply the oxidation half-reaction by 48:
48C6 + 288H6 + 48O2 → 288CO2 + 288H2O + 48e-
48H6 + 144H2O → 144H2O
Now you can cancel out the electrons:
48C6 + 288H6 + 48O2 → 288CO2 + 288H2O
Finally, simplify the equation:
C6H6 + O2 → CO2 + H2O