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Balancing Oxidation-Reduction Reactions A Short Primer.

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Presentation on theme: "Balancing Oxidation-Reduction Reactions A Short Primer."— Presentation transcript:

1 Balancing Oxidation-Reduction Reactions A Short Primer

2 ©D.B.Green, 2000,2002 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 ©D.B.Green, 2000,2002 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 ©D.B.Green, 2000,2002 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 H 2 O as needed. 7.Check the overall balance. If everything is correct, place the spectator ions back into the equation as necessary.

5 ©D.B.Green, 2000,2002 Review of Oxidation Numbers 4 The oxidation number of an element in its most stable form at room temperature is zero. 4 The oxidation number of a monatomic ion is its ionic charge. 4 Except in hydrides, the oxidation number of hydrogen in a compound or polyatomic ion is Except in peroxides and superoxides, the oxidation number of oxygen in a compound or polyatomic ion is -2. (O 2 2- O = -1; O 2 - O = - 1 / 2 ) 4 The sum of the oxidation numbers in a molecule is zero and is the ionic charge in a polyatomic ion.

6 ©D.B.Green, 2000,2002 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: MnO Fe 2+  Fe 3+ + Mn 2+ Step 2. No “pre-balance” is necessary, so…. Step 3. Assign oxidation numbers: MnO Fe 2+  Fe 3+ + Mn 2+ (+7) (+2) (+3) (+2)

7 ©D.B.Green, 2000,2002 Example 1 (Step 3 cont’d) And connect with a line identical atoms which undergo an oxidation state change: (+7) (+2) (+3) (+2) MnO Fe 2+  Fe 3+ + Mn e - -1 e - Step 4. Balance electrons and multiply species coefficients by the same factors: (+5 e - ) x 1 (+7) (+2) (+3) (+2) MnO Fe 2+  5 Fe 3+ + Mn 2+ (-1 e - ) x 5 (+7) (+2) (+3) (+2) MnO Fe 2+  5 Fe 3+ + Mn 2+

8 ©D.B.Green, 2000,2002 Example 1 Step 5. Charge balance with acid (since transition metals are involved) to get equal ionic charges on the left and right: MnO Fe H +  5 Fe 3+ + Mn 2+ Step 6. Finally, mass balance with water on the right: MnO Fe H +  5 Fe 3+ + Mn H2OH2O Ionic charge: (+2) = +9 (on left) 5(+3) + +2 = +17 (on right) MnO Fe H +  5 Fe 3+ + Mn 2+ MnO Fe H +  5 Fe 3+ + Mn 2+

9 ©D.B.Green, 2000,2002 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: KMnO FeCl HCl  5 FeCl 3 + MnCl H 2 O + KCl

10 ©D.B.Green, 2000,2002 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. Cr 3+ + ClO 4 -  Cr 2 O Cl - Step 2. “Pre-balance” the chromium: Step 3. -2(3 e-) = -6 e-e- 2 Cr 3+ + ClO 4 -  Cr 2 O Cl - 2 Cr 3+ + ClO 4 -  Cr 2 O Cl - +8 e - (+3) (+7) (+6)2 (-1)

11 ©D.B.Green, 2000,2002 (+3) (+7) (+6)2 (-1) 2 Cr ClO 4 -  4 Cr 2 O Cl - Example 2 Step 4. Balance electrons: -6 e - x 4 = e - x 3 = 24 Step 5. Charge balance: Ionic charge: 8 Cr ClO 4 -  4 Cr 2 O Cl - 8 Cr ClO 4 -  4 Cr 2 O Cl - + ?? H + Step 6. Mass balance: 8 Cr ClO ?? H 2 O  4 Cr 2 O Cl H+H+ (+3) (+7) (+6)2 (-1) 8 Cr ClO 4 -  4 Cr 2 O Cl - -6 e - x 4 = e - x 3 = 24 (+3) (+7) (+6)2 (-1) 8 Cr ClO 4 -  4 Cr 2 O Cl - (+3) (+7) (+6)2 (-1) 8 Cr ClO 4 -  4 Cr 2 O Cl - Ionic charge: 8(+3) + 3(-1) = +21 (left) 4(-2) + 3(-1) = -11 (right) 8 Cr ClO 4 -  4 Cr 2 O Cl H + 8 Cr ClO H 2 O  4 Cr 2 O Cl H +

12 ©D.B.Green, 2000,2002 Example 2 Step 7. Check the balance and replace the spectators. For this example, assume that the anionic spectator is chloride. 8 CrCl KClO H 2 O  4 K 2 Cr 2 O KCl + 32 HCl

13 ©D.B.Green, 2000,2002 Problems 4 MnO SO 2  Mn 2+ + HSO Copper metal reacts in nitric acid to produce copper(II) ion and nitrogen monoxide gas. 4 C + H 2 SO 4  CO 2 + SO 2 4 Molecular bromine disproportionates into bromate and bromide in basic solution. 4 Bi(OH) 3 + SnO 2 2-  SnO Bi (in basic solution) Solutions are provided on the next page. Resist the temptation to peek at the answers until you have completed balancing the equations.

14 ©D.B.Green, 2000,2002 Solutions to Problems 4 2 MnO SO 2 + H H 2 O  2 Mn HSO Cu + 2 HNO H +  3 Cu NO + 4 H 2 O 4 C + 2 H 2 SO 4  CO SO H 2 O 4 6 Br OH -  2 BrO Br H 2 O 4 2 Bi(OH) SnO 2 2-  3 SnO Bi + 3 H 2 O


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