Question 5.7: Balancing the Equation for a Redox Reaction in Basic Solutio......

Analyze

The reaction occurs in basic solution. We can balance it by using the method described in Table 5.6. The half-reactions and the overall reaction are initially treated as though they were occurring in an acidic solution and, finally, the overall equation is adjusted to a basic solution.

Solve

Step 1. Write half-equations for the oxidation and reduction half-reactions, and balance them for Mn, C, and N atoms.

MnO_{4}^{−}(aq) → MnO_{2}(s)

CN^{−}(aq) → OCN^{−}(aq)

Step 2. Balance the half-equations for O and H atoms. Add H_{2}O and/or H^{+} as required. In the MnO_{4}^{−} half-equation, there are four O’s on the left and two on the right. Adding 2 H_{2}O balances the O’s on the right. Since there are now four H’s on the right, it is necessary to add 4 H^{+} on the left side to balance them. In the CN^{−} half-equation, there is one O on the right but none on the left, so H_{2}O must be added to the left side and 2 H^{+} to the right.

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

            CN^{−}(aq) + H_{2}O(l) → OCN^{−}(aq) + 2  H^{+}(aq)

Step 3. Balance the half-equations for electric charge by adding the appropriate numbers of electrons.

Reduction:          MnO_{4}^{−}(aq) + 4  H^{+}(aq) + 3  e^{−} → MnO_{2}(s) + 2  H_{2}O(l)

Oxidation:                               CN^{−}(aq) + H_{2}O(l) → OCN^{−}(aq) + 2  H^{+}(aq) + 2e^{−}

Step 4. Combine the half-equations to obtain an overall redox equation. Multiply the reduction half-equation by two and the oxidation half-equation by three to obtain the common multiple in each half-equation. Make the appropriate cancellations of H^{+} and H_{2}O.

Overall:            \begin{array}{r c} \begin{matrix} 2   MnO_{4}^{−}(aq) & + & 8  H^{+}(aq) & + & \cancel{6  e^{−}} & \longrightarrow & 2  MnO_{2}(s) & + & 4  H_{2}O(l)\\ &3  CN^{−}(aq) & + & 3  H_{2}O(l) & &\longrightarrow & 3  OCN^{−}(aq) &+ & 6   H^{+}(aq) & + & \cancel{6  e^{−}} \\ \hline MnO_{4}^{−}(aq) & + & 3  CN^{−}(aq) & + & 2  H^{+}(aq) & \longrightarrow & 2  MnO_{2}(s) & + & 3  OCN^{−}(aq) & + &  H_{2}O(l)\end{matrix} \end{array}

Step 5. Change from an acidic to a basic medium by adding 2 OH^{−} to both sides of the overall equation; combine 2 H^{+} and 2 OH^{−} to form 2 H_{2}O, and simplify.

2 MnO_{4}^{−}(aq)  +  3  CN^{−}(aq)  +  2  H^{+}(aq)  + 2   OH^{−}(aq) \longrightarrow  2  MnO_{2}(s)  +  3  OCN^{−}(aq)  +  H_{2}O(l) + 2  OH^{−}(aq)

2 MnO_{4}^{−}(aq)  +  3  CN^{−}(aq)  +  2  H_{2}O(l) \longrightarrow 2  MnO_{2}(s)  +  3  OCN^{−}(aq)  +  H_{2}O(l) + 2   OH^{−}(aq)

Subtract one H_{2}O molecule from each side to obtain the overall balanced redox equation for reaction (5.24).

2 MnO_{4}^{−}(aq) + 7  H_{2}O_{2} + 6  H^{+} → 2  Mn^{2+} + 6  O_{2} + 10  H_{2}O        (5.24)

2 MnO_{4}^{−}(aq)  +  3  CN^{−}(aq)  +  H_{2}O(l) \longrightarrow 2  MnO_{2}(s)  +  3  OCN^{−}(aq)  + 2  OH^{−}(aq)

Step 6. Verify. Check the final overall equation to ensure that it is balanced both for number of atoms and for electric charge. For example, show that in the balanced equation from step 5, the net charge on each side of the equation is −5.

Assess

We can use the rules for assigning oxidation states (given in Table 3.2) to deduce that the O.S. of Mn changes from +7 in MnO_{4}^{−} to +4 in MnO_{2}. We conclude that the other substance, CN^{−}, is oxidized. (The rules do not allow us to assign oxidation states to C and N in CN^{−} or NCO^{−}.) Even though we cannot identify the oxidation states of C or N, we could still balance the equation for the reaction. That is one advantage of the methods we presented in Tables 5-5 and 5-6. In Chapter 10, we will discuss another method for assigning oxidation states and learn how to determine the oxidation states of C and N in such species as CN^{−} or NCO^{−}.