Complete and Balance the following equation using the half-reaction method;
(MnO4^-) + (CH3OH) --> (Mn^2+) + (HCO2H)
To balance the given equation using the half-reaction method, we need to divide the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
First, let's write the unbalanced equation:
(MnO4^-) + (CH3OH) --> (Mn^2+) + (HCO2H)
Now let's balance the atoms in each half-reaction separately:
Oxidation Half-Reaction: (MnO4^-) --> (Mn^2+)
- The manganese (Mn) is unchanged on both sides.
- The oxygen (O) atoms are balanced, with 4 on the reactant side and 0 on the product side.
- The charge on the reactant side (4-) needs to be balanced with the charge on the product side (2+).
To balance the charge, we need to add 6 electrons (e^-) to the right side of the equation.
(MnO4^-) --> (Mn^2+) + 6e^-
Reduction Half-Reaction: (CH3OH) --> (HCO2H)
- The carbon (C) atom is balanced, with 1 on each side.
- Hydrogen (H) atom is balanced, with 4 on the reactant side and 4 on the product side.
- Oxygen (O) atoms are balanced, with 1 on the reactant side and 2 on the product side.
- The charge on the reactant side is 0, and the charge on the product side is also 0.
(CH3OH) --> (HCO2H)
Now let's balance the electrons in both half-reactions:
Multiply the oxidation half-reaction by 3 to balance the electrons:
3(MnO4^-) --> 3(Mn^2+) + 18e^-
Now the electrons in both half-reactions are balanced.
To make the electrons in the oxidation half-reaction equal the electrons in the reduction half-reaction, we need to multiply the reduction half-reaction by 3 as well:
3(CH3OH) --> 3(HCO2H)
Now we can rewrite the overall balanced equation:
3(MnO4^-) + 3(CH3OH) --> 3(Mn^2+) + 3(HCO2H)
Therefore, the balanced equation using the half-reaction method is:
3(MnO4^-) + 3(CH3OH) --> 3(Mn^2+) + 3(HCO2H)
To balance the given equation using the half-reaction method, we need to follow these steps:
Step 1: Divide the overall reaction into two half-reactions: oxidation and reduction.
Oxidation half-reaction: (MnO4^-) → (Mn^2+)
Reduction half-reaction: (CH3OH) → (HCO2H)
Step 2: Balance the elements except hydrogen and oxygen in each half-reaction.
For the oxidation half-reaction:
(MnO4^-) → (Mn^2+)
There is only one manganese (Mn) atom on each side, so no need to balance it.
Now let's balance the oxygens:
(MnO4^-) → (Mn^2+) + 4O
Since there are 4 oxygen atoms on the left side, we need to add 4 H2O molecules to balance the oxygens on the right side:
(MnO4^-) + 4H2O → (Mn^2+) + 4O
For the reduction half-reaction:
(CH3OH) → (HCO2H)
There is only one carbon (C) atom on each side, so no need to balance it.
Now let's balance the hydrogens:
(CH3OH) → (HCO2H) + H
Since there are 4 hydrogen atoms on the left side, we need to add 4 H+ ions to balance the hydrogens on the right side:
(CH3OH) + 4H+ → (HCO2H) + H
Step 3: Balance the charge in each half-reaction by adding electrons (e^-).
For the oxidation half-reaction, we have 1 negative charge (from MnO4^-) on the left side and no charge on the right side. Thus, we need to add 5 electrons to the left side of the half-reaction:
(MnO4^-) + 4H2O + 5e^- → (Mn^2+) + 4O
For the reduction half-reaction, we have no charge on the left side and 1 positive charge (from H+) on the right side. Thus, we need to add 1 electron to the left side of the half-reaction:
(CH3OH) + 4H+ + e^- → (HCO2H) + H
Step 4: Multiply each half-reaction by the appropriate factor to equalize the number of electrons.
To equalize the number of electrons (e^-) in both half-reactions, we need to multiply the oxidation half-reaction by 1 and the reduction half-reaction by 5:
5(CH3OH) + 20H+ + 5e^- → 5(HCO2H) + 5H
5(MnO4^-) + 20H2O + 25e^- → 5(Mn^2+) + 20O
Step 5: Add the balanced half-reactions and simplify if necessary.
By adding the two balanced half-reactions together, we get the balanced equation:
5(MnO4^-) + 20H2O + 25(CH3OH) + 100H+ + 25e^- → 5(Mn^2+) + 80O + 5(HCO2H) + 6H
However, if we simplify the equation, we can divide through by 5 to obtain the final balanced equation:
(MnO4^-) + 4H2O + 5(CH3OH) + 20H+ + 5e^- → (Mn^2+) + 16O + (HCO2H) + 6H