Answer:
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)
