Question 21.1: Balancing a Redox Reaction in Basic Solution Problem Permang......

Plan We follow the numbered steps as described in text, and proceed through step 4 as if this reaction occurs in acidic solution. Then, we add the appropriate number of OH^− ions and cancel excess H_2O molecules (see “4 Basic”).

Solution

1. Divide into half-reactions.
   MnO_4^−  ⟶  MnO_2                        C_2O_4^{2−}  ⟶  CO_3^{2−}
2. Balance.
   a. Atoms other than O and H.               a. Atoms other than O and H,
   Not needed.                                            C_2O_4^{2−}  ⟶  2CO_3^{2−}
   b. O atoms with H_2O,                             b. O atoms with H_2O,
   MnO_4^−  ⟶  MnO_2  +  2H_2O                      2H_2O  +  C_2O_4^{2−}  ⟶  2CO_3^{2−}
   c. H atoms with H^+,                                 c. H atoms with H^+,
   4H^+  +  MnO_4^−  ⟶  MnO_2  +  2H_2O         2H_2O  +  C_2O_4^{2−}  ⟶  2 CO_3^{2−}  +  4H^+
   d. Charge with e^{−},                                  d. Charge with e^−,
   3e^−  +  4H^+  +  MnO_4^−  ⟶  MnO_2  +  2H_2O           2H_2O  +  C_2O_4^{2−}  ⟶  2CO_3^{2−}  +  4H^+  +  2e^−
                          \text{[reduction]}                                               \text{[oxidation]}
3. Multiply each half-reaction, if necessary, by some integer to make e^− lost equal e^− gained.
   2(3e^−  +  4H^+  +  MnO_4^−  ⟶  MnO_2  +  2H_2O)              3(2H_2O  +  C_2O_4^{2−}  ⟶   2CO_3^{2−}  +  4H^+  +  2e^−)
   6e^−  +  8H^+  +  2MnO_4^−  ⟶  2 MnO_2  +  4H_2O             6H_2O  +  3C_2O_4^{2−}  ⟶  6CO_3^{2−}  +  12H^+  +  6e^−
4. Add half-reactions, and cancel species appearing on both sides.
   The six e^− cancel, eight H^+ cancel to leave four H^+ on the right, and four H_2O cancel to leave two H_2O on the left:
                \cancel{6e^−}  +  \cancel{8H^+}  +  2MnO_4^−  ⟶  2MnO_2  +  4H_2O
                     2  \cancel{6}H_2O  +  3C_2O_4^{2−}  ⟶  6CO_3^{2−}  +  4  \cancel{12}H^+  +  \cancel{6e^−}
            \overline{2MnO_4^−  +  2H_2O  +  3C_2O_4^{2−}  ⟶  2MnO_2  +  6CO_3^{2−}  +  4H^+}

   4 Basic. Add OH^− to both sides to neutralize the H^+ present, and cancel excess H_2O. Adding four OH^− to both sides forms four H_2O on the right. Two of those cancel the two H_2O on the left and leave two H_2O on the right:
      2MnO_4^−  +  2H_2O  +  3C_2O_4^{2−}  +  4OH^−  ⟶  2MnO_2  +  6CO_3^{2−}  +  [4H^+  +  4OH^−]
      2MnO_4^−  +  \cancel{2H_2O}  +  3C_2O_4^{2−}  +  4OH^−  ⟶  2MnO_2  +  6CO_3^{2−}  +  2  \cancel{4}H_2O
   Including states of matter gives the final balanced equation:
     2MnO_4^−(aq)  +  3C_2O_4^{2−}(aq)  +  4OH^−(aq)  ⟶  2MnO_2(s)  +  6CO_3^{2−}(aq)  +  2H_2O(l)

5. Check that atoms and charges balance.
              (2Mn,  24O,  6C,  4H;  12−)  ⟶  (2Mn,  24O,  6C,  4H;  12−)

Comment As a final step, let’s see how to obtain the balanced molecular equation for this reaction. We note the amount (mol) of each anion in the balanced ionic equation and add the correct amount (mol) of spectator ions (in this case, Na^+, as given in the problem statement) to obtain neutral compounds. The balanced molecular equation is