1 / 13

OXIDATION REDUCTION REACTIONS

OXIDATION REDUCTION REACTIONS. Rules for Assigning Oxidation States

carney
Télécharger la présentation

OXIDATION REDUCTION REACTIONS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. OXIDATION REDUCTION REACTIONS

  2. Rules for Assigning Oxidation States • The oxidation number corresponds to the number of electrons, e-, that an atom loses, gains, or appears to use when joining with other atoms in compounds. When determining the Oxidation State of an atom there are seven guidelines to follow: • The Oxidation State of an individual atom is 0. • The total Oxidation State of all atoms in: a neutral species is 0 and in an ion is equal to the ion charge. • Group 1 metals have an Oxidation State of +1 and group 2 an Oxidation State of +2 • The Oxidation State of fluorine is -1, when in compounds • Hydrogen generally has an Oxidation State of +1 in compounds • Oxygen generally has an Oxidation State of -2 in compounds.(EXCEPTION: the oxygen in peroxide (O22 ) is -1 • In binary metal compounds, group 17 elements have an Oxidation State of -1, group 16 of -2, and group 15 of -3. • (Note: The sum of the oxidation states is equal to zero for neutral compounds and equal to the charge for polyatomic ion species.)

  3. Example 1: Determine the oxidation state of the bold element in each of the following: • 1. Na3PO3 • 2. H2PO4- • Answers to Example 1: • 1. The oxidation numbers of Na and O are +1 and -2. Since sodium phosphite is neutral, the sum of the oxidation numbers must be zero.. Letting x be the oxidation number of phosphorus then, 0= 3(+1) + x + 3(-2). x=oxidation number of P= +3.  • 2. Hydrogen and oxygen have oxidation numbers of +1 and -2. The ion has a charge of -1, so the sum of the oxidation numbers must be -1. Letting y be the oxidation number of phosphorus, -1= y + 2(+1) +4(-2), y= oxidation number of P= +5. • Sample Problems: Determine the oxidation states: • 1. Fe(s) + O2(g) → Fe2O3(g) • 2. Fe2+ • 3. Ag(s) + H2S → Ag2S(g) + H2(g) • Solutions • 1. Fe and O2 are free elements, therefore they have an O.S. of "0" according to Rule #1. The product has a total O.S. equal to "0" and following Rule #6, O3 has an O.S. of -2, which means Fe2 has an O.S. of +2. • 2. The O.S. of Fe corresponds to its charge, therefore the O.S. is +2. • 3. Ag has an O.S. of 0, H2 has an O.S. of +1 according to Rule #5 and S has an O.S. of -2 according to Rule #7.

  4. OXIDATION REDUCTION REACTIONS Redox reactions are comprised of two parts, a reduced half and an oxidized half, that always occur together. The reduced half gains electrons and the oxidation number decreases, while the oxidized half losses electrons and the oxidation number increases. • OXIDATION • Losses electrons • Oxidation number increases. • Electrons are shown on the product side of the half reaction • The species oxidized is called the reducing agent • REDUCTION • Gain of electrons • Oxidation number decreases • Electrons are shown on the reactant side of the half reaction • The species reduced is called the oxidizing agent Simple ways to remember this are the mnemonic devices OIL RIG meaning "oxidation is loss" and "reduction is gain" or LEO says GER meaning "loss of e- = oxidation" and "gain of e- = reduced."

  5. There is no net change in the number of electrons in a redox reaction. Those given off in the oxidation half reaction are taken on by another species in the reduction half reaction. The ion or molecule that accepts electrons is called the oxidizing agent; by accepting electrons it brings about the oxidation of another species. Conversely, the species that donates electrons is called the reducing agent; when reaction occurs it reduces the other species.  Note: the oxidizing and reducing agents can be the same element or compound This will be further discussed under Types of Redox Reactions: Disproportionation).

  6. Example 2: Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the oxidation state of each):  • Zn + 2H+ → Zn2+ + H2 • 2Al + 3Cu2+→2Al3+ +3Cu • CO32- + 2H+→ CO2 + H2O • Answers to Example 2: • Zn is oxidized (Oxidation number: 0 → +2); H+ is reduced (Oxidation number: +1 → 0) • Al is oxidized (Oxidation number: 0 → +3); Cu2+ is reduced (+2 → 0) • This is not a redox type because each element has the same oxidation number in both reactants and products: O= -2, H= +1, C= +4. • (For a more in depth look see oxidation numbers). • Oxidizing and Reducing Agents • An atom is oxidized when it oxidation number increases, the reducing agent, and an atom is reduced when its oxidation number decreases, the oxidizing agent. In other words, what is oxidized is the reducing agent and what is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound).

  7. Example 3. Using the equations from the previous examples determine what is oxidized? • Zn + 2H+ → Zn2+ + H2 • Answer:  • The O.S. of H goes from +1 to 0 and the O.S. of Zn goes from 0 to 2+. Hence, Zn is oxidized and acts as the reducing agent. • Example 4. What is reduced? • Zn + 2H+ → Zn2+ + H2 • Answer:  • The O.S. of H goes from +1 to 0 and the O.S. of Zn goes from 0 to 2+. Hence, H+ ion is reduced and acts as the oxidizing agent.

  8. 2Al + 3Cu2+→2Al3+ +3Cu • Al goes from 0 to 3+ so it is oxidized so Al acts as the reducing agent • Cu goes from 2+ to 0 so it is reduced so Cu2+ acts as the oxidizing agent • CO32- + 2H+→ CO2 + H2O • As stated in a previous slide, C goes from 4+ to 4+ and H+ goes from 1+ to 1+ so this is not a redox reaction b/c there is no gain or loss of electrons.

  9. BALANCING REDOX REACTIONS USING THE HALF REACTION METHOD Half Reactions Before one can balance an overall redox equation, one has to be able to balance two half-equations, one for oxidation (electronloss) and one for reduction (electron gain). Collectively, oxidation and reduction are known as redox, or an electron transfer reaction. After balancing the two half-equations one can determine the total net reaction. Each equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order: 1) Balance the number of atoms of each element.  2) Balance the number of electrons transferred. 3) Balance the total charge on reactants and products (Note: If #1 and #2 are done correctly, #3 will follow. Thus, it serves as a means of checking your work).

  10. Balancing in an Acidic Solution • Balancing in acidic solution is similar to balancing in neutral solutions however, instead of three steps to follow, there are six. These rules are: • Write and balance the half reactions. • Balance oxygen, O, by adding with H2O • Balance hydrogen, H, by adding H+ (acidic) • Balance charge by adding electrons (you should be adding the same number of electrons as H+ ions) • Multiply both half reactions by some integer to cancel out electrons • Add the half reactions together and cancel out what appears on both sides

  11. Example 5:  • Balance the redox reaction in acidic solution: MnO4- + I- --> Mn2+ + I2(s) • Write and balance the half reactions:MnO4- + I- --> Mn2+ + I2(s)O.S:  +7 -2       -1          +2          0        (Mn is reduced and I- is oxidized) • Write and balance the half reactions. • Oxidation Rx:2I-(aq) --> I2(s)Reduction Rx: MnO4- + 5e- --> Mn2+ • Balance oxygen, O, by adding H2OReduction Rx: MnO4- + 5e- --> Mn2+ + 4H2O • Balance hydrogen, H, by adding H+Reduction Rx: MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O • Balance charge by adding electronsOxidation Rx:2I-(aq) --> I2(s) + 2e-Reduction Rx: MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O • Multiply both half reactions by some integer to cancel out electrons(Oxidation Rx:2I-(aq) --> I2(s) + 2e-) * 5(Reduction Rx: MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O) *2Oxidation Rx:10I-(aq) --> 5I2(s) + 10e-Reduction Rx: 2MnO4- + 10e- + 16H+ --> 2Mn2+ + 8H2O • Add the half reactions together and cancel out what appears on both sides:10I-(aq) + 2MnO4-(aq) + 16H+(aq) --> 2Mn2+(aq) + 5I2(s) + 8H2O(l)(Note: Don't forget the states of matter! Generally, anything with a charge is (aq) and H2O is (l))

  12. BALANCING REDOX REACTION THAT TAKE PLACE IN A BASIC SOLUTION FOLLOWS THE SAME STEPS AS IN AN ACIDIC SOLUTION • HOWEVER THERE ARE A FEW MORE EXTRA STEPS • AFTER COMPLETING THE STEPS FOR AN ACID, THE FOLLOWING STEPS ARE ADDED: • ADD OH- TO BOTH SIDES TO BALANCE THE H+ PRESENT • COMBINE THE H+ AND OH- TO FORM WATER • CANCEL THE WATER MOLECULES APPEARING ON BOTH SIDES

  13. Balance the above redox reaction in basic solution: • Balance the reaction in acidic solution10I-(aq) + 2MnO4-(aq) + 16H+(aq) --> 2Mn2+(aq) + 5I2(s) + 8H2O(l) • Add the same amount of OH- ions as H+ ions to both sides of the equation.10I-(aq) + 2MnO4-(aq) + 16H+(aq) + 16OH- --> 2Mn2+(aq) + 5I2(s) + 8H2O(l) + 16OH- • On one side, the OH- and H+ will react to form water (H2O) in a 1:1 ratio.10I-(aq) + 2MnO4-(aq) + 16H2O--> 2Mn2+(aq) + 5I2(s) + 8H2O(l) + 16OH- • Cancel out the water molecules appearing on both sides10I-(aq) + 2MnO4-(aq) + 8H2O(l)--> 2Mn2+(aq) + 5I2(s) + 16OH-(aq)

More Related