Question 5.6: Balancing the Equation for a Redox Reaction in an Acidic Sol......

Analyze

The reaction occurs in acidic aqueous solution. We can use the method summarized in Table 5.5 to balance it.

Solve

The O.S. of sulfur increases from +4 in SO_{3}^{2−} to +6 in SO_{4}^{2−}. The O.S. of Mn decreases from +7 in MnO_{4}^{−} to +2 in Mn^{2+}. Thus, SO_{3}^{2−} is oxidized and MnO_{4}^{−} is reduced.

Step 1. Write skeleton half-equations based on the species undergoing oxidation and reduction. The half-equations are

SO_{3}^{2−}(aq) → SO_{4}^{2−}(aq)

MnO_{4}^{−}(aq) → Mn^{2+}(aq)

Step 2. Balance each half-equation for numbers of atoms, in this order:

• atoms other than H and O
• O atoms, by adding H_{2}O with the appropriate coefficient
• H atoms, by adding H^{+} with the appropriate coefficient

The other atoms (S and Mn) are already balanced in the half-equations. To balance O atoms, we add one H_{2}O molecule to the left side of the first half-equation and four to the right side of the second.

SO_{3}^{2−} + H_{2}O(l) → SO_{4}^{2−}(aq)

          MnO_{4}^{−}(aq) → Mn^{2+}(aq) + 4 H_{2}O(l)

To balance H atoms, we add two H^{+} ions to the right side of the first half-equation and eight to the left side of the second.

   SO_{3}^{2−} + H_{2}O(l) → SO_{4}^{2−}(aq) + 2  H^{+}(aq)

MnO_{4}^{−}(aq) + 8  H^{+}(aq) → Mn^{2+}(aq) + 4   H_{2}O(l)

Step 3. Balance each half-equation for electric charge. Add the number of electrons necessary to get the same electric charge on both sides of each half-equation. By doing this, you will see that the half-equation in which electrons appear on the right side is the oxidation half-equation. The other half-equation, with electrons on the left side, is the reduction half-equation.

Oxidation:                                           SO_{3}^{2−} + H_{2}O(l) → \underset{(net  charge  on   each  side,  -2)}{SO_{4}^{2−}(aq) + 2  H^{+}(aq) + 2  e^{−}}

Reduction:               MnO_{4}^{−}(aq) + 8  H^{+}(aq) + 5  e^{−} → \underset{(net  charge  on  each  side,  +2)}{Mn^{2+}(aq) + 4   H_{2}O(l)}

Step 4. Obtain the overall redox equation by combining the half-equations. Multiply the oxidation half-equation by 5 and the reduction half-equation by 2. This results in 10 e^{−} on each side of the overall equation. These terms cancel out. Electrons must not appear in the final equation.

Step 5. Simplify. The overall equation should not contain the same species on both sides. Subtract 5 H_{2}O from each side of the equation in step 4. This leaves 3 H_{2}O on the right. Also subtract 10 H^{+} from each side, leaving 6 H^{+} on the left.

5  SO_{3}^{2- } + 2   MnO_{4}^{- }(aq) + 6  H^{+}(aq) → 5  SO_{4}^{2- }(aq) + 2  Mn^{2+}(aq) + 3  H_{2}O(l)

Step 6. Verify. Check the overall equation to ensure that it is balanced both for numbers of atoms and electric charge. For example, show that in the balanced equation from step 5, the net charge on each side of the equation is −6: (5 × −2) + (2 × −1) + (6 × +1) = (5 × −2) + (2 × +2) = −6.

Assess 

The final check completed in step 6 gives us confidence that our result is correct. This is an important step; always take the time to complete it. It is also worth pointing out that, in this example, there was only one atom per formula that was oxidized or reduced. (Refer to the skeleton half equations given in step 1.) Many students have difficulty balancing half-equations in which more than one atom per formula is oxidized or reduced, as is the case when Cr_{2}O_{7}^{2−} is reduced to Cr^{3+}. Had we used Cr_{2}O_{7}^{2−} instead of MnO_{4}^{- } in equation (5.25), the balanced chemical equation for the reaction would have been 3 SO_{3}^{2- } + Cr_{2}O_{7}^{2−} + 8  H^{+} → 3  SO_{4}^{2- } + 2  Cr^{3+} + 4  H_{2}O.