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)

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