Balancing Oxidation-Reduction Reactions A Short Primer

1. BalancingOxidation-ReductionReactionsA Short Primer

2. Oxidation-Reduction Reactions Oxidation-reduction (redox) reactions can be difficult to balance because not only must mass be balanced, so must the electrons passed between species. Presented here is an algorithm for balancing redox reactions quickly using oxidation numbers. The method revolves around balancing electron movement first, then balancing the rest of the equation without changing the species which undergo oxidation or reduction. It is important to apply this algorithm only to redox reactions and only when the equation cannot be balanced by inspection alone. The algorithm is not necessary, nor will it work for non-redox reactions.

3. The Algorithm 1. Write the equation from words (if necessary). Ignore spectator ions (i.e., write the net ionic equation. 2. Assign oxidation numbers to the atoms which undergo a change in oxidation number. Connect identical atoms across the arrow with a line and place on the line the number of electrons gained or lost by the species. 3. If necessary, adjust the stoichiometric ratios of the species. This is not a final balance, just get the ratios correct. An illustration of this “pre-balance” is shown in the example problems. The following steps 4, 5, and 6 must be done in order.

4. The Algorithm 4. Electron Balance: Balance the electrons to their least-common-multiple with multipliers. Multiply the stoichiometric coefficients for the species connected by the lines with the same multipliers. 5. Charge Balance: Balance the ionic charges on the left and right side of the arrow with either H+ or OH- such that the sum of all the ionic charges on the left and right are equal. Unless otherwise specified, usually (but not always), hydrogen ion is used when transition metals are involved in the reaction and hydroxide is used when nonmetals are the only species in the reaction. If there are no ions in the reaction, this step is skipped. 6. Mass Balance: Balance the addition of H+ or OH- with H2O as needed. 7. Check the overall balance. If everything is correct, place the spectator ions back into the equation as necessary.

5. Review of Oxidation Numbers • The oxidation number of an element in its most stable form at room temperature is zero. • The oxidation number of a monatomic ion is its ionic charge. • Except in hydrides, the oxidation number of hydrogen in a compound or polyatomic ion is +1. • Except in peroxides and superoxides, the oxidation number of oxygen in a compound or polyatomic ion is -2.(O22- O = -1; O2- O = -1/2) • The sum of the oxidation numbers in a molecule is zero and is the ionic charge in a polyatomic ion.

6. Example 1 Permanganate ion reacts with iron(II) ion to form manganese(II) ion and iron(III) ion. Write and balance the equation. Step 1. Write the equation from words: MnO4- + Fe2+ Fe3+ + Mn2+ Step 2. No “pre-balance” is necessary, so…. Step 3. Assign oxidation numbers: (+7) (+2) (+3) (+2) MnO4- + Fe2+ Fe3+ + Mn2+

7. (+7) (+2) (+3) (+2) MnO4- + 5 Fe2+ 5 Fe3+ + Mn2+ Example 1 (Step 3 cont’d) And connect with a line identical atoms which undergo an oxidation state change: +5 e- (+7) (+2) (+3) (+2) MnO4- + Fe2+ Fe3+ + Mn2+ -1 e- Step 4. Balance electrons and multiply species coefficients by the same factors: (+5 e-) x 1 (+7) (+2) (+3) (+2) MnO4- + 5 Fe2+ 5 Fe3+ + Mn2+ (-1 e-) x 5

8. Example 1 Step 5. Charge balance with acid (since transition metals are involved) to get equal ionic charges on the left and right: MnO4- + 5 Fe2+ + 8 H+ 5 Fe3+ + Mn2+ Ionic charge: -1 + 5(+2) = +9 (on left) 5(+3) + +2 = +17 (on right) MnO4- + 5 Fe2+ + 8 H+ 5 Fe3+ + Mn2+ MnO4- + 5 Fe2+ + 8 H+ 5 Fe3+ + Mn2+ Step 6. Finally, mass balance with water on the right: MnO4- + 5 Fe2+ + 8 H+ 5 Fe3+ + Mn2+ + 4 H2O

9. Example 1 Step 7. Perhaps the permanganate was prepared from the potassium salt, the iron(II) from the chloride salt, and the reaction was performed in hydrochloric acid solution: KMnO4 + 5 FeCl2 + 8 HCl  5 FeCl3 + MnCl2 + 4 H2O + KCl

10. Example 2 Chromium(III) ion can be converted to dichromate ion by treatment with potassium perchlorate in acid solution. Perchlorate ion is converted to chloride ion. Write and balance the chemical equation. Step 1. Cr3+ + ClO4- Cr2O7-2 + Cl- Step 2. “Pre-balance” the chromium: 2 Cr3+ + ClO4- Cr2O7-2 + Cl- -2(3 e-) = -6 e- Step 3. 2 Cr3+ + ClO4- Cr2O7-2 + Cl- (+3) (+7) (+6)2 (-1) +8 e-

11. -6 e- x 4 = 24 (+3) (+7) (+6)2 (-1) 2 Cr3+ + 3 ClO4- 4Cr2O7-2 + 3 Cl- (+3) (+7) (+6)2 (-1) 8 Cr3+ + 3 ClO4- 4Cr2O7-2 + 3 Cl- (+3) (+7) (+6)2 (-1) 8 Cr3+ + 3 ClO4- 4Cr2O7-2 + 3 Cl- +8 e- x 3 = 24 Example 2 Step 4. Balance electrons: -6 e- x 4 = 24 (+3) (+7) (+6)2 (-1) 8 Cr3+ + 3 ClO4- 4Cr2O7-2 + 3 Cl- +8 e- x 3 = 24 Step 5. Charge balance: Ionic charge: 8 Cr3+ + 3 ClO4- 4 Cr2O7-2 + 3 Cl- Ionic charge: 8(+3) + 3(-1) = +21 (left) 4(-2) + 3(-1) = -11 (right) 8 Cr3+ + 3 ClO4- 4 Cr2O7-2 + 3 Cl- + ?? H+ 8 Cr3+ + 3 ClO4- 4 Cr2O7-2 + 3 Cl- + 32 H+ Step 6. Mass balance: 8 Cr3+ + 3 ClO4- + ?? H2O 4 Cr2O7-2 + 3 Cl- + 32 H+ 8 Cr3+ + 3 ClO4- + 16 H2O 4 Cr2O7-2 + 3 Cl- + 32 H+

12. Example 2 Step 7. Check the balance and replace the spectators. For this example, assume that the anionic spectator is chloride. 8 CrCl3 + 3 KClO4 + 16 H2O  4 K2Cr2O7 + 3 KCl + 32 HCl

13. Problems Solutions are provided on the next page. Resist the temptation to peek at the answers until you have completed balancing the equations. • MnO4- + SO2  Mn2+ + HSO4- • Copper metal reacts in nitric acid to produce copper(II) ion and nitrogen monoxide gas. • C + H2SO4 CO2 + SO2 • Molecular bromine disproportionates into bromate and bromide in basic solution. • Bi(OH)3 + SnO22-  SnO32- + Bi (in basic solution)

14. Solutions to Problems • 2 MnO4- + 5 SO2 + H+ + 2 H2O  2 Mn+ + 5 HSO4- • 3 Cu + 2 HNO3 + 6 H+  3 Cu+2 + 2 NO + 4 H2O • C + 2 H2SO4 CO2 + 2 SO2 + 2 H2O • 6 Br2 + 12 OH- 2 BrO3- + 10 Br- + 6 H2O • 2 Bi(OH)3 + 3 SnO22-  3 SnO32- + 2 Bi + 3 H2O