Questions: Balance the following redox equation, assuming acidic conditions: HCOOH + MnO4^- -> CO2 + Mn^2+

Transcript text: 32. Balance the following redox equation, assuming acidic conditions (5) \[ \mathrm{HCOOH}+\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{CO}_{2}+\mathrm{Mn}^{2+} \]

Solution

Solution Steps

Step 1: Write the Unbalanced Half-Reactions

First, we write the oxidation and reduction half-reactions separately.

Oxidation half-reaction: \[ \mathrm{HCOOH} \rightarrow \mathrm{CO}_{2} \]

Reduction half-reaction: \[ \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} \]

Step 2: Balance Atoms Other Than Oxygen and Hydrogen

For the oxidation half-reaction, carbon is already balanced: \[ \mathrm{HCOOH} \rightarrow \mathrm{CO}_{2} \]

For the reduction half-reaction, manganese is already balanced: \[ \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} \]

Step 3: Balance Oxygen Atoms

Balance the oxygen atoms by adding \( \mathrm{H_2O} \) molecules.

Oxidation half-reaction: \[ \mathrm{HCOOH} \rightarrow \mathrm{CO}_{2} + \mathrm{H_2O} \]

Reduction half-reaction: \[ \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H_2O} \]

Step 4: Balance Hydrogen Atoms

Balance the hydrogen atoms by adding \( \mathrm{H^+} \) ions.

Oxidation half-reaction: \[ \mathrm{HCOOH} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H^+} + 2\mathrm{e^-} \]

Reduction half-reaction: \[ \mathrm{MnO}_{4}^{-} + 8\mathrm{H^+} + 5\mathrm{e^-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H_2O} \]

Step 5: Balance the Electrons

To balance the electrons, we need to multiply the half-reactions by appropriate factors so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

Oxidation half-reaction (multiplied by 5): \[ 5(\mathrm{HCOOH} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H^+} + 2\mathrm{e^-}) \rightarrow 5\mathrm{HCOOH} \rightarrow 5\mathrm{CO}_{2} + 10\mathrm{H^+} + 10\mathrm{e^-} \]

Reduction half-reaction (multiplied by 2): \[ 2(\mathrm{MnO}_{4}^{-} + 8\mathrm{H^+} + 5\mathrm{e^-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H_2O}) \rightarrow 2\mathrm{MnO}_{4}^{-} + 16\mathrm{H^+} + 10\mathrm{e^-} \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H_2O} \]

Step 6: Combine the Half-Reactions

Combine the balanced half-reactions and cancel out the electrons.

\[ 5\mathrm{HCOOH} + 2\mathrm{MnO}_{4}^{-} + 16\mathrm{H^+} \rightarrow 5\mathrm{CO}_{2} + 2\mathrm{Mn}^{2+} + 8\mathrm{H_2O} + 10\mathrm{H^+} \]

Step 7: Simplify the Equation

Combine and simplify the \( \mathrm{H^+} \) ions on both sides.

\[ 5\mathrm{HCOOH} + 2\mathrm{MnO}_{4}^{-} + 6\mathrm{H^+} \rightarrow 5\mathrm{CO}_{2} + 2\mathrm{Mn}^{2+} + 8\mathrm{H_2O} \]