Complete and Balance the following equation using the half-reaction method;
(MnO4^-) + (CH3OH) --> (Mn^2+) + (HCO2H)
To balance the given chemical equation using the half-reaction method, we need to follow these steps:
Step 1: Split the equation into two half-reactions:
(MnO4^-) --> (Mn^2+)
(CH3OH) --> (HCO2H)
Step 2: Balance the atoms that aren't oxygen and hydrogen for each half-reaction:
(MnO4^-) --> (Mn^2+)
Here, we only need to balance the manganese (Mn) atoms. Since the Mn atom on the left side is 1, we balance it by putting a coefficient of 1 in front of (Mn^2+).
(CH3OH) --> (HCO2H)
In this half-reaction, we need to balance the carbon (C) and hydrogen (H) atoms. We have 1 carbon (C) atom on both sides, so it is already balanced. However, we have 4 hydrogen (H) atoms on the left side and 2 hydrogen (H) atoms on the right side. To balance the hydrogen (H) atoms, we put a coefficient of 2 in front of (HCO2H).
Step 3: Balance the oxygen (O) atoms by adding water (H2O) molecules to both sides of the half-reactions:
(MnO4^-) --> (Mn^2+)
Here, we have 4 oxygen (O) atoms on the left side and 0 oxygen (O) atoms on the right side. To balance the oxygen (O) atoms, we add 4 water (H2O) molecules to the right side:
(MnO4^-) --> (Mn^2+) + 4(H2O)
(CH3OH) --> (HCO2H)
In this half-reaction, we have 1 oxygen (O) atom on the left side and 2 oxygen (O) atoms on the right side. To balance the oxygen (O) atoms, we add 1 water (H2O) molecule to the left side:
(CH3OH) + (H2O) --> (HCO2H)
Step 4: Balance the hydrogen (H) atoms by adding H ions (H+) to the side with a deficit:
(MnO4^-) + 8(H+) --> (Mn^2+) + 4(H2O)
(CH3OH) + (H2O) --> (HCO2H) + 4(H+)
Step 5: Balance the charges by adding electrons (e-) to the side with an excess charge:
(MnO4^-) + 8(H+) + 5(e-) --> (Mn^2+) + 4(H2O)
(CH3OH) + (H2O) --> (HCO2H) + 4(H+) + 4(e-)
Step 6: Combine the half-reactions by multiplying each half-reaction by appropriate coefficients so that the number of transferred electrons is the same:
Multiply the first half-reaction by 4 and the second half-reaction by 5:
4(MnO4^-) + 32(H+) + 20(e-) --> 4(Mn^2+) + 16(H2O)
5(CH3OH) + 5(H2O) --> 5(HCO2H) + 20(H+) + 20(e-)
Step 7: Combine the half-reactions and cancel out any common terms:
4(MnO4^-) + 32(H+) + 20(CH3OH) + 20(H2O) --> 4(Mn^2+) + 16(H2O) + 5(HCO2H) + 20(H+) + 20(e-)
Simplifying the equation:
4(MnO4^-) + 20(CH3OH) + 12(H+) --> 4(Mn^2+) + 5(HCO2H)
Therefore, the balanced equation using the half-reaction method is:
4(MnO4^-) + 20(CH3OH) + 12(H+) --> 4(Mn^2+) + 5(HCO2H)