Complete and balance the following reaction by filling in the missing coefficients. Assume the reaction is occurring in a basic, aqueous solution. CH3CH2OH (aq) + MnO-4 (aq) ⟶CH3COO - (aq) + MnO2 (s) Include all coefficients, even those equal to 1.

4 MnO₄⁻ (aq) + 3 CH₃CH₂OH (aq) ⟶ 4 MnO₂ (s) + 1 OH⁻ (aq) + 3 CH₃COO⁻ (aq) + 4 H₂O (l)

Explanation:

To balance a redox reaction we use the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻ (aq) ⟶ MnO₂ (s)

Oxidation: CH₃CH₂OH (aq) ⟶ CH₃COO⁻ (aq)

Step 2: Balance the mass adding H₂O and OH⁻ where necessary.

2 H₂O (l) + MnO₄⁻ (aq) ⟶ MnO₂ (s) + 4 OH⁻ (aq)

5 OH⁻ (aq) + CH₃CH₂OH (aq) ⟶ CH₃COO⁻ (aq) + 4 H₂O (l)

Step 3: Balance the charge adding eelctrons where necessary.

2 H₂O (l) + MnO₄⁻ (aq) + 3 e⁻ ⟶ MnO₂ (s) + 4 OH⁻ (aq)

5 OH⁻ (aq) + CH₃CH₂OH (aq) ⟶ CH₃COO⁻ (aq) + 4 H₂O (l) + 4 e⁻

Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost are the same.

4 * (2 H₂O (l) + MnO₄⁻ (aq) + 3 e⁻ ⟶ MnO₂ (s) + 4 OH⁻ (aq))

3 * (5 OH⁻ (aq) + CH₃CH₂OH (aq) ⟶ CH₃COO⁻ (aq) + 4 H₂O (l) + 4 e⁻)

Step 5: Add both half-reactions and cancel what is repeated.

8 H₂O (l) + 4 MnO₄⁻ (aq) + 12 e⁻ + 15 OH⁻ (aq) + 3 CH₃CH₂OH (aq) ⟶ 4 MnO₂ (s) + 16 OH⁻ (aq) + 3 CH₃COO⁻ (aq) + 12 H₂O (l) + 12 e⁻

4 MnO₄⁻ (aq) + 3 CH₃CH₂OH (aq) ⟶ 4 MnO₂ (s) + 1 OH⁻ (aq) + 3 CH₃COO⁻ (aq) + 4 H₂O (l)