chemistry_30_13.2_predicting_redox_reactions.pptx

Section 13.2 p. 568 - 582  Using the Reductions Table Why is this table called a “reduction half reaction” table? – all reactions listed have electrons in them and all reactions show a gain of electrons. What would the table be called if it was written backwards? – a tabl...

Section 13.2 p. 568 - 582  Using the Reductions Table Why is this table called a “reduction half reaction” table? – all reactions listed have electrons in them and all reactions show a gain of electrons. What would the table be called if it was written backwards? – a table showing a loss of electrons would be an oxidation half-reaction table.  Note that as you read the half reactions from left to right, they all gain electrons. That is why the reactions are called reduction half reactions. Within each reduction half reaction, one major species gains electrons. This species is the oxidizing agent. Conversely, any species that loses electrons is the reducing agent.  Given: Pb(s) + 2 Ag+(aq) 2 Ag(s) + Pb2+(aq) The reduction half reaction is: Ag+(aq) + e– Ag(s) (G.E.R.) The oxidation half reaction is: Pb(s) Pb2+(aq) + 2e – (L.E.O.) The species gaining electrons is: Ag+(aq) (oxidizing agent) The species losing electrons is: Pb(s) (reducing agent)  The oxidizing agents are found on the left side and are listed in order from the most reactive or strongest at the upper left-hand corner to the least reactive or weakest near the lower left. Similarly, the most reactive or strongest reducing agents are found near the lower right side of the table and get weaker as they move upward. When a reduction half reaction (showing a gain of electrons) is added to an oxidation half reaction (showing a loss of electrons) we get a REDOX (reduction-oxidation) reaction (net ionic equation).  Determining Whether a Reaction is Spontaneous or Non-Spontaneous A strip of zinc will react when placed in a solution of copper (II) nitrate but a strip of copper will not react in a solution of zinc nitrate. How does this work? Consider the following lab results:  Which species would gain electrons (are OA) in a reduction half reaction? – all the metallic ions Which species would lose electrons (are RA) in an oxidation half reaction? – all the solid metals  Which ion reacts the most? the least? – D3+(aq) (most) - E+(aq) (least) Which solid metal reacts the most? the least? – E(s) (most) - D(s) (least) Rank the OA from the most reactive to the least reactive. – D3+(aq) , A+(aq) , F2+(aq) , B2+(aq) , C3+(aq) , E+  Using reduction half reactions, make a “Reductions Table” similar to the one in your data books. D3+(aq) + 3e– D(s) A+(aq) + e– A(s) F2+(aq) + 2e– F(s) B2+(aq) + 2e– B(s) C3+(aq) + 3e– C(s) E+(aq) + e– E(s)  So, how can you determine if a reaction will take place? In any redox reaction, an OA must react with a RA. (a species from the left side must react with a species from the right side)  If the OA is positioned higher on the table than the RA with which it reacts, the reaction will proceed naturally, therefore, be spontaneous. If the OA is positioned lower on the table than the RA with which it reacts, the reaction will not proceed naturally, therefore, be non‑spontaneous. p. 573 Activity: Practice Questions #11 – 23, p. 573 - 574  Creating Net Ionic (Redox) Equations Using The Reductions Table Example: A strip of tin is placed in a solution of silver nitrate. Creating Net Ionic (Redox) Equations Using The Reductions Table Example: A strip of tin is placed in a solution of silver nitrate. Step 1: Identify all species present. Sn(s) Ag+(aq) NO3–(aq) H2O(l) Step 2: Identify which species are oxidizing agents (OA) and which are reducing agents (RA). The OA are found on the left side of the table and the reducing agents are found on the right side of the table. Sn(s) Ag+(aq) NO3–(aq) H2O(l) RA OA “normal” OA/RA not found on tables  Step 3: Identify the strongest OA and the strongest RA. The top left side of the table contains the strongest OA and the bottom right contains the strongest RA. Sn(s) Ag+(aq) NO3–(aq) H2O(l) SRA SOA normal OA/RA  Step 4: Write down the reduction half reaction (the OA gaining e-). Ag+(aq) + e – Ag(s) Step 5: Write down the oxidation half reaction (the RA losing e-). Sn(s) Sn2+(aq) + 2 e –  Step 6: Add the two half reactions. Ag+(aq) + e – Ag(s) Sn(s) Sn2+(aq) + 2e – Step 7: Balance the electrons. 2 (Ag+(aq) + e– Ag(s) ) Sn(s) Sn2+(aq) + 2 e–  Step 8: Add the two equations to yield the redox reaction. 2 (Ag+(aq) + e– Ag(s) ) Sn(s) Sn2+(aq) + 2 e – 2 Ag+(aq) + Sn(s) 2 Ag(s) + Sn2+(aq)  Writing Net Redox Reactions Assumption: The SOA present will react with the SRA present. 1)A 5¢ piece is left in chlorinated laundry bleach. SOA OA Ni(s) / Cl2(aq) / H2O(l) SRA RA RED: Cl2(aq) + 2 e – 2 Cl –(aq) OX: Ni(s) Ni 2+(aq) + 2 e – REDOX: Cl2(aq) + Ni(s) 2 Cl –(aq) + Ni 2+ (aq) 2) A magnesium ribbon is immersed in an iron (III) chloride solution. SOA OA Mg(s) / Fe 3+(aq) / Cl –(aq) / H2O(l) SRA RA RA RED: 2(Fe 3+(aq) + 1 e - → Fe 2+(aq) ) OX: Mg(s) → Mg2+(aq) + 2 e - REDOX: 2Fe 3+(aq) + Mg(s) → 2 Fe 2+(aq) + Mg2+(aq) 3) Tin (II) bromide solution is poured into acidified sodium dichromate solution. OA OA OA SOA OA Sn2+(aq) /Br –(aq) / H+(aq) / Na+(aq) / Cr2O7 2–(aq) / H2O(l) SRA RA RA RED: Cr2O7 2–(aq) + 14 H +(aq) + 6 e – 2 Cr 3+ (aq) + 7 H2O(l) OX: 3 (Sn 2+(aq) Sn 4+(aq) + 2 e –) REDOX: Cr2O7 2–(aq) + 14 H +(aq) + 3 Sn 2+(aq) 2 Cr 3+(aq) + 7 H2O(l) + 3 Sn 4+(aq) 4) Hydrochloric acid is poured into an aluminum pop can. H+(aq) / Cl – (aq) / Al(s) / H2O(l) RED: OX: REDOX: 5) Acidified potassium permanganate is accidentally spilled onto a "chrome" faucet. OA OA SOA OA H+(aq) / K+(aq) / MnO4–(aq) / Cr(s) / H2O(l) SRA RA RED: 2(MnO4–(aq) + 8 H +(aq) + 5 e –Mn 2+ (aq) + 4 H2O(l)) OX: 5 (Cr(s) Cr 2+(aq) + 2 e –) REDOX: 2 MnO4–(aq) + 16 H + (aq) + 5 Cr(s) 2 Mn 2+ (aq) +8 H2O(l) + 5 Cr 2+(aq) 6) A mixture of sodium hydroxide and sodium sulfite are mixed with a stirrer than churns air into the solution. OA SOA Na +(aq) / OH –(aq) / SO32–(aq) / O2(g) / H2O(l) RA SRA RED: O2(g) + 2 H2O(l) + 4 e – 4 OH –(aq) OX: 2 (SO32–(aq) + 2 OH –(aq) SO42–(aq) + H2O(l) + 2 e –) REDOX: O2(g) + 2 SO32–(aq) 2 SO42–(aq) 7) A steel underground gasoline storage tank is surrounded by moist acidic soil. OA SOA Fe(s) / H2O(l) / H + (aq) / O2(g) SRA RA RED: O2(g) + 4 H + (aq) + 4e – 2 H2O(l) OX: 2 ( Fe(s) Fe 2+ (aq) + 2 e –) REDOX: O2(g) + 4 H + (aq) + 2 Fe(s) 2 H2O(l) + 2 Fe 2+ (aq) 8) Pennies are dropped into concentrated nitric acid. SOA OA Cu(s) / H +(aq) / NO3–(aq) / H2O(l) SRA RA RED: 2NO3–(aq) + 4 H +(aq) + 2e – N2O4(g) + 2H2O(l) OX: Cu(s) Cu 2+(aq) + 2 e – REDOX: 2 NO3–(aq) + 4 H + (aq) + Cu(s) N2O4(g) + 2 H2O(l) + Cu 2+(aq)  Disproportionation Disproportionation is a reaction in which a species is both oxidized and reduced. Example: What happens if two iron(II) ions in a solution collide? SOA OA Fe2+(aq) / H2O(aq) SRA RA Red: Fe2+ (aq) + 2e- → Fe(s) Oxid: 2[Fe2+ (aq) → Fe3+ (aq) + e- ] Redox: 3Fe2+ (aq) → Fe(s) + 2 Fe3+ (aq) Activity: Practice Questions #25 – 30 p. 579 Predicting Redox Reactions by Constructing Half- Reactions Steps to Predict Balanced Redox Equations 1. Use the information provided to start two half- reaction equations. 2. Balance each half-reaction equation (Using the rules from 13.1) 3. Multiply each half-reaction equation by simple whole numbers to balance the electrons lost and gained. 4. Add the two half-reaction equations, cancelling the electrons and anything els that is exactly the same on both sides of the equation. Activity: Practice Questions #31 – 33, p. 581 Section Questions #1 – 18, p. 582