Question 18.PS.3: Balancing Redox Equations for Reactions in Acidic Solutions ......

10  CO_2(g) + 2  Mn^{2+}(aq) + 14  H_2O(\ell)

Strategy and Explanation   Follow a systematic approach listed in this series of steps to balance the equation for this reaction (and all similar redox reactions).

Step 1:      Recognize whether the reaction is an oxidation-reduction process. If it is, then determine what is reduced and what is oxidized. This is a redox reaction because the oxidation number of Mn changes from +7 in MnO_4^- to +2 in Mn^{2+}, so the Mn in MnO_4^- is reduced. The oxidation number of each C changes from +3 in H_2C_2O_4 to +4 in CO_2, so the C in H_2C_2O_4 is oxidized. The oxidation numbers of H (+1) and O (-2) are unchanged.

Step 2:      Break the overall unbalanced equation into half-reactions.

H_2C_2O_4(aq) → CO_2(g)               (oxidation half-reaction)

MnO_4^-(aq) → Mn^{2+}(aq)                ( reduction half-reaction)

Step 3:      Balance the atoms in each half-reaction. First balance all atoms except for O and H, then balance O by adding H_2O and balance H by adding H^+. (Hydroxide ion, OH^+, cannot be used here because the reaction occurs in an acidic solution and he OH^- concentration is very low.)
Oxalic acid half-reaction: First, balance the carbon atoms in the half-reaction.

H_2C_2O_4(aq) → 2  CO_2(g)

This step balances the O atoms as well (no H_2O needed here), so only H atoms remain to be balanced. Because the product side is deficient by two H, we put 2 H^+ there.

H_2C_2O_4(aq) → 2  CO_2(g) + 2  H^+(aq)                 (oxalic acid half-reaction)

Strictly speaking, we ought to use H_3O^+ instead of H^+, but this would result in adding water molecules to each side of the equation, which is rather cumbersome. It is simpler to add H^+ now and add the water molecules at the end.
Permanganate half-reaction: The Mn atoms are already balanced, but the oxygen atoms are not balanced until H_2O is added. Adding 4  H_2O on the product side takes care of the needed oxygen atoms.

MnO_4^-(aq) → Mn^{2+}(aq) + 4  H_2O(\ell)

Now there are 8 H atoms on the right and none on the left. To balance hydrogen atoms, 8  H^+ are placed on the left side of the half-reaction.

8  H^+(aq) + MnO_4^-(aq) → Mn^{2+}(aq) + 4  H_2O(\ell)               (permanganate half-reaction)

Step 4:      Balance the half-reactions for charge using electrons (e^-). The oxalic acid halfreaction has a net charge of 0 on the left side and 2+ on the right. The reactants have lost two electrons. To show this fact, 2  e^- must appear on the right side.

H_2C_2O_4(aq) → 2  CO_2(g) + 2  H^+(aq) + 2  e^-

This confirms that H_2C_2O_4 is the reducing agent (it loses electrons and is oxidized). The loss of two electrons is also in keeping with the increase in the oxidation number of each of two C atoms by 1, from +3 to +4. The 2  e^- also balance the charge on the product side of the equation.

The MnO_4^- half-reaction has a charge of 7+ on the left and 2+ on the right. Therefore, to achieve a net 2+ charge on each side, 5  e^- must appear on the left. The gain of electrons shows that MnO_4^- is the oxidizing agent; it is reduced.

5  e^- + 8  H^+(aq) + MnO_4^-(aq) → Mn^{2+}(aq) + 4  H_2O(\ell)

Step 5:      Multiply the half-reactions by appropriate factors so that the oxidation halfreaction produces as many electrons as the reduction half-reaction accepts. In this case, one half-reaction involves two electrons, and the other half-reaction involves five electrons. It takes ten electrons to balance each half-reaction. The oxalic acid half-reaction must be multiplied by 5, and the permanganate halfreaction by 2.

5  [H_2C_2O_4(aq) → 2  CO_2(g) + 2  H^+(aq) + 2  e^-]

2 [5  e^- + 8  H^+(aq) + MnO_4^-(aq) → Mn^{2+}(aq) + 4  H_2O(\ell)]

Step 6:      Add the half-reactions to give the net reaction and cancel equal amounts of reactants and products that appear on both sides of the arrow.

5  H_2C_2O_4(aq) → 10  CO_2(g) + 10  H^+(aq) + 10  e^-

10  CO_2(g) + 10  H^+(aq) + 2  Mn^{2+}(aq) + 8  H_2O(\ell)

Since 16  H^+ appear on the left and 10 H^+ appear on the right, 10  H^+ are canceled, leaving 6  H^+ on the left.

10  CO_2(g)  + 2  Mn^{2+}(aq) + 8  H_2O(\ell)

Step 7:      Check the balanced net equation to make sure both atoms and charge are balanced.
Atom balance: Each side of the equation has 2 Mn, 28 O, 10 C, and 16 H atoms.
Charge balance: Each side has a net charge of 4+.
On the left side, (6 × 1+) (2 × 1-) = 4+.
On the right side, 2(2+) = 4+.

Step 8:      Add enough water molecules to both sides of the equation to convert all H^+ to H_3O^+. In this case, six water molecules are needed (6  H_2O + 6  H^+ → H_3O^+). Six water molecules are added to each side of the equation, which increases the total to 14 on the product side.

10  CO_2(g)  + 2  Mn^{2+}(aq) + 14  H_2O(\ell)

Step 9:      Check the final results to make sure both atoms and charges are balanced. The net equation is balanced. The net charges on each side of the reaction are the same, and the numbers of atoms of each kind on each side of the reaction are equal.