Balancing Redox Reactions

Transcription

1 Balancing Redox Reactions 1. What is a redox reaction? a. Redox is an abbreviation to say reduction/oxidation is taking place during a chemical reaction. In the old days oxidation was when an element combined with oxygen and reduction was from the Latin word meaning to "to lead back." b. Example 1: The reaction between magnesium oxide and carbon at 2000 o C to form magnesium metal and carbon monoxide is an example of the reduction of magnesium oxide to magnesium metal. c. Example 2: The reaction between magnesium metal and oxygen to form magnesium oxide involves the oxidation of magnesium. d. Today a chemical reaction in which the oxidation numbers of atoms change is a chemist definition of a redox reaction. If an atom oxidation number gets less positive/more negative then the atom has undergone reduction (gained electrons). If an atom oxidation number gets more positive/less negative then the atom has undergone oxidation (lost electrons). Remember the saying, LEO the lion goes GER or OIL RIG e. Practice Problem #1: Determine which element is oxidized and which is reduced when lithium reacts with nitrogen to form lithium nitride in the unbalanced chemical reaction (HINT: Determine the oxidation number of each atom.) Li(s) + N 2 (g) Li 3 N (s) 2. How do I balance a redox reaction? a. A seldom-used method is called the oxidation number technique. This technique work quickly for easy redox reactions, but not at as well for more difficult redox equations. Try the next three practice problems. b. Practice Problem #2: How many electrons moved in the reaction in practice problem #1. In a redox reaction the total number of electrons moved must be equal. Place coefficients to make this be true in the reaction. c. Practice Problem #3: Try balancing this reaction: Al(s) + MnO 2 (s) Al 2 O 3 (s) + Mn(s) d. Practice Problem #4: Try balancing this reaction: Ag + (aq) + Cu (s) Cu 2+ (aq) + Ag (s) e. A method that can be used in all cases is called the half-reaction method. This method can be time consuming, BUT if done correctly ALWAYS gives you a balanced equation! The steps for this method are on the next page for redox reactions in both acidic and basic solutions.

2 f. The steps to this method are listed below, but try and do each step only after you have read each individual step. The ultimate goal is to write a balanced chemical equation for the reaction between oxalic acid and an aqueous solution of potassium dichromate that produces chromium(iii) ions, carbon dioxide and water. For an acidic solution you can stop after STEP #11, for a basic solution you must do step #12 i. STEP #1- Write a skeleton equation first, eliminating spectator ions ii. STEP #2- Determine the oxidation numbers for each atom in the skeleton equation iii. STEP #3- Identify the atoms being reduced and oxidized in the skeleton equation iv. STEP #4- Write a half-reaction for the atoms being oxidized (This means a new equation for just those atoms involved in the oxidation) v. STEP #5- Directly below the oxidation half-reaction write a half-reaction for the atoms being reduced (This means a new equation for just those atoms involved in the reduction) vi. STEP #6- Go back to your oxidation half reaction and balance out all elements, EXCEPT oxygen and hydrogen with coefficients, then to balance the H s and O s do the following two steps A. To the side lacking oxygen(s), add that many waters B. To the side lacking hydrogen(s), add hydrogen ions (H + ) vii. STEP #7- To the oxidation reaction add e - to balance out the charge, since e - are being lost they should be on the product side viii. STEP #8- Repeat step #6 for the reduction reaction ix. STEP #9- To the reduction reaction add e - to balance out the charge, since e - are being gained they should be on the reactant side x. STEP #10- If the number of e - are not equal multiple both equations (this includes all substances in that equation) by a whole number so the number of e - are equal xi. STEP #11- Sum the two half reactions into one final equation A. IF a substance is found on the same side of the arrow for both equations then that total goes in the final equation B. IF a substance is found on both sides of the arrow then the substances cancel each other out and then that total goes in the final equation xii. STEP #12- FOR a basic solution you now must neutralize out the excess H +, with OH - to make that many waters, BUT you must also add that many OH - to the other side of the arrow

3 SOLUTION The steps to this method are listed below, but try and do each step after you have read the step to write a balanced chemical equation for the reaction between oxalic acid and aqueous solution of potassium dichromate that produces chromium(iii) ions, carbon dioxide and water. For an acidic solution you can stop after STEP #11, for a basic solution you must do step #12 STEP #1- Write a skeleton equation first, eliminating spectator ions H 2 C 2 O 4 + Cr 2 O > CO 2 + Cr 3+ + H 2 O Oxidized Reduced (HINT: K+ is a spectator ion) STEP #2- Determine the oxidation numbers for each atom in the skeleton equation STEP #3- Identify the atoms being reduced and oxidized in the skeleton equation STEP #4- Write a half-reaction for the atoms being oxidized (This means a new equation for just those atoms involved in the oxidation) (Cr 2 O H + + 6e - --> 2 Cr H 2 O) 1 (H 2 C 2 O 4 --> 2 CO H + + 2e - ) Cr 2 O H H 2 C 2 O 4 --> 2 Cr H 2 O + 6 CO 2 STEP #5- Directly below the oxidation half-reaction write a half-reaction for the atoms being reduced (This means a new equation for just those atoms involved in the reduction) STEP #6- Go back to your oxidation half reaction and balance out all elements, EXCEPT oxygen and hydrogen with coefficients, then to balance those two atoms do the following two steps A. To the side lacking oxygen(s), add that many waters B. To the side lacking hydrogen(s), add hydrogen ions (H + ) STEP #7- To the oxidation reaction add e - to balance out the charge, since e - are being lost they should be on the product side STEP #8- Repeat step#5 and #6 for the reduction reaction STEP #9- To the reduction reaction add e - to balance out the charge, since e - are being gained they should be on the reactant side STEP #10- If the number of e - are not equal multiple both equations (this includes all substances in that equation) by a whole number so the number of e - are equal STEP #11- Sum the two half reactions into one final equation C. IF a substance is found on the same side of the arrow for both equations then that total goes in the final equation D. IF a substance is found on both sides of the arrow then the substances cancel each other out and then that total goes in the final equation STEP #12- FOR a basic solution you now must neutralize out the excess H +, with OH - to make that many waters, BUT you must also add that many OH - to the other side of the arrow

4 NAME: Practice Sheet Balancing Redox Reactions Complete and balance the following redox reactions (skeleton equations) using the half-equation method: 2-1) SO 3 (aq) + MnO - 4 (aq) ==> Mn SO 4 (aq) (acidic solution) 2) I - (aq) + NO - 2 (aq) ==> I 2 (s) + NO (g) (acidic solution) 3) MnO - 4 (aq) + Cl - (aq) ==> Mn 2+ (aq) + Cl 2 (g) (acidic solution) 4) Br 2 (l) ==> BrO - 3 (aq) + Br - (aq) (basic solution)

5 5) CrO SO 2-3 ===> Cr 3+ + SO 2-4 (acidic solution) 6) H 2 O 2 + NO - 2 ===> H 2 O + NO - 3 (acidic solution) 7) I - (aq) + OCl - (aq) I 2 (s) + Cl - (aq) + H 2 O(l) (basic solution) 8) Cr(OH) 3 (s) + ClO - 3 ( aq) CrO 2-4 (aq) + Cl - (aq) (basic solution)