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Oxidation-Reduction (Redox) Reactions Reactions in which electrons are transferred from one reactant to another oxidation – the loss of electrons reduction – the gain of electrons Oxidation of zinc: “LEO the lion says GER” Zn( s ) → Zn 2+ (aq) + 2e − solid zinc has lost 2 electrons to form zinc ions; zinc is oxidized Reduction of copper ions: Cu 2+ (aq) + 2e − → Cu ( s ) copper ions have gained 2 electrons to form solid copper; copper ions are reduced Redox Reactions What happens if solid zinc is placed in a solution containing copper ions? Redox Reactions What happens if solid zinc is placed in a solution containing copper ions? In this case, one species was able to be oxidized while the other was able to be reduced It’s important to have both species present; otherwise where would the electrons come from? Let’s look at the equations again: Zn( s ) → Zn 2+ (aq) + 2e − Cu 2+ (aq) + 2e − → Cu ( s ) oxidizing agent – a species that accepts electrons (Cu2+) reducing agent – a species that donates electrons (Zn) Redox Reactions Oxidation and reduction reactions can be written separately (half-reactions), but they must always occur together The sum of the two half-reactions gives the overall redox equation: Zn( s ) → Zn 2+ (aq) + 2e − Cu 2+ (aq) + 2e − → Cu ( s ) Zn( s ) + Cu 2+ (aq) + 2e − → Zn 2+ (aq) + Cu ( s ) + 2e − Zn( s ) + Cu 2+ (aq) → Zn 2+ (aq ) + Cu ( s ) Redox Reactions What if solid copper was placed in a solution containing zinc ions? Would the reverse reaction occur? No! No reaction would occur at all! Cu ( s ) + Zn 2+ (aq) → no reaction We can use an activity series (see Table 4.5) to predict which redox reactions will occur Any metal in the activity series will only be oxidized by metals appearing below it Zinc can be oxidized by copper, but copper can NOT be oxidized by zinc Redox Reactions Which of the following combinations will result in a redox reaction? Ca ( s ) + Al 3+ (aq) → ? Cr ( s ) + Mn 2+ (aq) → ? Cu ( s ) + Fe 2+ (aq) → ? Ba( s ) + Co 2+ (aq) → ? Pb( s ) + Na + (aq) → ? Balancing Redox Reactions Let’s consider the following reaction: Ba( s ) + Co 2+ (aq) → ? Write each half-reaction and then add them together to generate the net reaction: 2+ Ba( s ) → Ba (aq) + 2e − Co 2+ (aq) + 2e − → Co( s ) Ba( s ) + Co 2+ (aq) + 2e − → Ba 2+ (aq) + Co( s ) + 2e − Ba( s ) + Co 2+ (aq) → Ba 2+ (aq) + Co( s ) Match these terms: oxidation reduction oxidizing agent reducing agent Balancing Redox Reactions Let’s consider the following reaction: Ca ( s ) + Al 3+ (aq) → ? Write each half-reaction and then add them together to generate the net reaction: Ca ( s ) → Ca 2+ (aq) + 2e − Al 3+ (aq ) + 3e − → Al ( s ) Ca ( s ) + Al 3+ (aq) + 3e − → Ca 2+ (aq ) + Al ( s ) + 2e − Electrons must be balanced in redox reactions! Balancing Redox Reactions Let’s consider the following reaction: Ca ( s ) + Al 3+ (aq) → ? 3 2 (Al (Ca(s) → Ca 3+ 2+ (aq) + 2e − (aq) + 3e − → Al ( s ) ) ) Match these terms: oxidation reduction 3Ca ( s ) + 2 Al 3+ (aq) + 6e − → 3Ca 2+ (aq) + 2 Al ( s ) + 6e − 3Ca ( s ) + 2 Al 3+ (aq) → 3Ca 2+ (aq) + 2 Al ( s ) oxidizing agent reducing agent Redox Reactions Sometimes redox reactions involve combination of reactants: Na ( s ) → Na + + e − Cl2 ( g ) + 2e − → 2Cl − Match these terms: Na ( s ) + Cl2 ( g ) + 2e − → Na + + e − + 2Cl − oxidation reduction 2 2 2 Electrons must be balanced in redox reactions! Na ( s ) → Na + + e − Cl2 ( g ) + 2e − → 2Cl − 2 Na ( s ) + Cl2 ( g ) + 2e − → 2 Na + + 2e − + 2Cl − 2 Na ( s ) + Cl2 ( g ) → 2 NaCl ( s ) oxidizing agent reducing agent Oxidation Numbers A complete transfer of electrons never occurs for combination reactions which form molecular compounds H 2 ( g ) + F2 ( g ) → 2 HF ( g ) No ions are shown in this reaction. How can we tell which element is oxidized and which is reduced? Oxidation numbers provide a method of “bookkeeping” by determining the charge an atom would have if electrons were transferred completely The oxidation number of a free element (monatomic or diatomic) is 0 The oxidation number of fluorine is always -1 (except as a free element) The oxidation number of hydrogen is always +1 (when combined with nonmetals) or -1 (when combined with metals) Oxidation Numbers Assign oxidation numbers to all reactants and products: The oxidation number of a free element (monatomic or diatomic) is 0 The oxidation number of fluorine is always -1 (except as a free element) The oxidation number of hydrogen is always +1 (when combined with nonmetals) or -1 (when combined with metals) The oxidation number of a monatomic ion is the charge on the ion H 2 ( g ) + F2 ( g ) → 2 HF ( g ) 0 0 +1 -1 The oxidation number of hydrogen increases from 0 to +1 Æ oxidation The oxidation number of fluorine decreases from 0 to -1 Æ reduction Oxidation Numbers More oxidation number rules: The oxidation number of a Group 1A or 2A metal is +1 or +2 (except as free elements) The oxidation number of oxygen is almost always -2 (exceptions can occur with compounds involving fluorine, hydrogen or Group 1A or 2A metals) NaH SO2 +1 -1 ? -2 ? -4 Use circles to represent the oxidation number of an atom Use squares to represent the total contribution from that type of atom The sum of all oxidation numbers within a compound or ion must be equal to the charge on the compound or ion Oxidation Numbers Assign oxidation numbers for the atoms without rules: SO2 ? -2 To make the overall compound neutral, the oxidation number for sulfur (overall) must be +4 +4 -4 SO2 +4 -2 +4 -4 There is only 1 sulfur atom, so it must be +4 Oxidation Numbers Let’s try another example: CO3−2 Each oxygen is -2 ? -2 There are three oxygen atoms, so the total contribution is -6 ? -6 CO3−2 +4 -2 +4 -6 The ion has an overall charge of -2, so carbon must be +4 (-6 + 4 = -2) Oxidation Numbers Let’s try another example: N 2O5 Each oxygen is -2 ? -2 There are five oxygen atoms, so the total contribution is -10 ? -10 N 2O5 ? -2 The molecule has an overall charge of 0, so nitrogen must contribute a total of +10 +10 -10 N 2O5 +5 -2 +10 -10 There are two nitrogen atoms, so each must contribute +5 Oxidation Numbers Assign the terms oxidation, reduction, oxidizing agent and reducing agent to the following redox reaction: Fe( s ) + PtCl2 (aq) → FeCl2 (aq) + Pt ( s ) 0 +2 -1 +2 -1 0 0 +2 -2 +2 -2 0 The oxidation number of iron increases from 0 to +2 Æ oxidation The oxidation number of platinum decreases from 0 to -2 Æ reduction The oxidation number of chlorine remains constant Æ spectator oxidizing agent – a species that accepts electrons (Pt+2) reducing agent – a species that donates electrons (Fe) Oxidation Numbers Assign the terms oxidation, reduction, oxidizing agent and reducing agent to the following redox reaction: Cr ( s ) + AuCl3 (aq) → CrCl3 (aq) + Au ( s )