Oxidation-reduction (redox) reactions
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1 Oxidation-reduction (redox) reactions Reactions in which there are changes in oxidation state (oxidation number) between reactants and products 2 MnO Br H + 2 Mn Br H 2 O One reactant must be oxidized (lose electrons, the reductant or reducing agent) and another must be reduced (gain electrons, the oxidant or oxidizing agent) In some instances a single reactant will be both oxidized and reduced (disproportionation) in other cases two species containing the same element in different oxidation states will combine to give a single product with an intermediate oxidation state (comproportionation).
2 Oxidation-reduction (redox) reactions, cont d Redox reactions may be separated into oxidation and reduction half-reactions. When half-reaction reagents can be isolated into separate compartments connected by a conduit for ion migration, electron flow through an external circuit can be utilized to perform work (basis for batteries and electrochemical cells). The potential (E o ) of an electrochemical cell is the sum of the potentials of the reduction and oxidation half-reactions The potential for a cell or half-reaction is related to the free energy change for the redox reaction through the relationship ΔG o = -nöe o where n is the number of electrons transferred and ö is the Faraday (9.65 x 10 4 J V -1 mol -1 = 9.65 x 10 4 C mol -1 )
3 An electrochemical cell anode (oxidation) cathode (reduction) Zn (s) = Zn e - E o = -(-0.76) V Cu e - = Cu (s) E o = 0.34 V Zn (s) + Cu 2+ = Zn 2+ + Cu (s) E o (cell) = ( ) =1.10 V anode (oxidation) on left Zn Zn 2+ Cu 2+ Cu cathode (reduction) on right ΔG o = -nöe o
4 Galvanic vs electrolytic cells For a Galvanic Cell, a spontaneous reaction takes place. ΔE > 0 (positive) ΔG < 0 (negative) A positive voltage means a rxn is thermodynamically favored and thus reactants are unstable For the Galvanic cell, the cell performs electrical work on the surroundings (acts as a battery) Zn Zn 2+ Cu 2+ Cu Cu Cu 2+ Zn 2+ Zn For an Electrolytic Cell, a non-spontaneous reaction takes place. A negative voltage means a rxn ΔE < 0 (negative) is not thermodynamically favored and thus reactants are ΔG > 0 (positive) stable For the electrolytic cell, electrical work is performed on the system
5 Conventions relative to electrode potentials All half-reactions are written as reduction reactions Potentials of half-reactions are relative to H + (aq) + e - = ½ H 2(g) with E o = 0.00 V Potentials are recorded for standard conditions: 1.0 M conc. for soluble reactants/products, 100 kpa (1 atm) for gases, pure solids Cell potentials are obtained by adding potentials for halfreactions Sign of potential changes when half-reaction is written as oxidation Changes in half-reaction or cell potentials resulting from nonstandard conditions can be determined using the Nernst equation E = E o - RT ln nf ö [products] [reactants] R = 8.31 V C mol -1 K -1 ö = 9.65 x 10 4 C mol -1
6 Some electrode potentials Most reducing Half-Reaction E (volts) Li + (aq) + e- = Li (s) K + (aq) + e- = K (s) Ca 2+ (aq) + 2e- = Ca (s) Na + (aq) + e- = Na (s) Mg 2+ (aq) + 2e- = Mg (s) Al 3+ (aq) + 3e- = Al (s) H 2 O (liq) + 2e - = H 2(g) + 2 OH - (aq) Zn 2+ (aq) + 2e- = Zn (s) Cr 3+ (aq) + 3e- = Cr (s) Fe 2+ (aq) + 2e- = Fe (s) Cd 2+ (aq) + 2e- = Cd (s) Ni 2+ (aq) + 2e- = Ni (s) H + (aq) + 2e- = H 2(g) 0.00 Most oxidizing Most reducing Half-Reaction E (volts) Sn 4+ (aq) + 2e- = Sn 2+ (aq) 0.15 Cu 2+ (aq) + e- = Cu + (aq) 0.16 Cu 2+ (aq) + 2e- = Cu (s) 0.34 Cu + (aq) + e- = Cu (s) 0.52 I 2(s) + 2e - = 2I - (aq) 0.54 O 2 (g) + 2H + (aq) + 2e - = H 2 O 2 (aq) 0.70 Fe 3+ (aq) + e- = Fe 2+ (aq) 0.77 Ag + (aq) + e- = Ag (s) 0.80 NO - 3 (aq) + 4H + (aq) + 3e- = NO (g) + 2 H 2 O (liq) 0.96 Br 2(liq) + 2e - = 2 Br - (aq) 1.07 O 2(g) + 4 H + (aq) + 4e- = 2 H 2 O (liq) 1.23 Cr 2 O 2-7 (aq) + 14 H + (aq) + 6e- = 2 Cr 3+ (aq) + 7 H 2 O (liq) 1.33 Cl 2(g) + 2e - = 2 Cl - (aq) 1.36 Ce 4+ (aq) + e- = Ce 3+ (aq) 1.44 MnO - 4 (aq) + 8 H + (aq) + 5e- = Mn 2+ (aq) + 4 H 2 O (liq) 1.51 H 2 O 2(aq) + 2 H + (aq) + 2e - = 2 H 2 O (liq) 1.78 Co 3+ (aq) + e- = Co 2+ (aq) 1.82 F 2(g) + 2e - = 2 F - (aq) 2.87 Most oxidizing
7 A is the reductant (reducing agent); gives up electrons, i.e., A = A + + e - B is the oxidant (oxidizing agent); gains electrons, i.e., B + e - = B - A is oxidized by B B is reduced by A Getting the terminology straight A + is the oxidation product B - is the reduction product A + B = A + + B - For this reaction to be favorable as written entries in a table of electrode (half-reaction) potentials would appear as follows: more reducing A + + e - = A B + e - = B - more oxidizing The reduction potential for A + /A will be less positive than that for B/B -
8 Rules for assigning oxidation state The sum of the oxidation states for all atoms in a species must equal the charge on the species. The oxidation state of an atom in an elemental form is zero. The oxidation state of a monatomic ion is equal to its charge. Terminal atoms (or groups of atoms) are assigned an oxidation state consistent with the charge in their monatomic anionic form, i.e., filled valence shell. The most electronegative element is assigned a negative oxidation state in compounds of hydrogen. In compounds such as HO-OH, H 2 N-NH 2, etc. the E-E bond does not contribute to the oxidation state of either atom. However, in compounds such as SSO 3 2- (S 2 O 3 2-, thiosulfate) assignment of a filled valence shell to the terminal sulfur gives it an oxidation state of -2.
9 Balancing oxidation-reduction (redox) reactions HNO 3(aq) + Cu (s) = NO 2(g) + Cu 2+ (aq) Write separate half-reactions for oxidation and reduction Balance each half-reaction for mass utilizing water and protons as necessary, i.e., Get rid of oxygen as water Get rid of hydrogen as H + Get hydrogen from H + Get oxygen from water Balance each half-reaction for charge by adding electrons to the right (oxidation) or left (reduction) as required Multiply half-reactions by the appropriate factors such that the electrons lost in oxidation equal the electrons gained in reduction Add the two half-reactions; if the same reagent appears as both reactant and product, subtract the smaller amount from both sides of the equation
10 Balancing oxidation-reduction (redox) reactions HClO = Cl 2 + ClO 3 - Write separate half-reactions for oxidation and reduction Balance each half-reaction for mass utilizing water and protons as necessary, i.e., Get rid of oxygen as water Get rid of hydrogen as H + Get hydrogen from H + Get oxygen from water Balance each half-reaction for charge by adding electrons to the right (oxidation) or left (reduction) as required Multiply half-reactions by the appropriate factors such that the electrons lost in oxidation equal the electrons gained in reduction Add the two half-reactions; if the same reagent appears as both reactant and product, subtract the smaller amount from both sides of the equation
11 Balancing oxidation-reduction (redox) reactions P 4 = PH 3 + H 2 PO - (basic solution) Write separate half-reactions for oxidation and reduction O Balance each half-reaction for mass utilizing water and protons as necessary, i.e., Get rid of oxygen as water Get rid of hydrogen as H + Get hydrogen from H + Get oxygen from water Balance each half-reaction for charge by adding electrons to the right (oxidation) or left (reduction) as required Multiply half-reactions by the appropriate factors such that the electrons lost in oxidation equal the electrons gained in reduction Add the two half-reactions; if the same reagent appears as both reactant and product, subtract the smaller amount from both sides of the equation If the reaction is conducted in basic medium, add OH - to both sides to eliminate any H + (H + + OH - = H 2 O)
12 Latimer diagrams a way to simplify data Latimer diagrams summarize several electrode potentials (halfreactions) in abbreviated form for a given element Half-reactions are written in sequence, as reductions, with the most highly oxidized species to the left Only the species involving the element are shown; other species (H 2 O, H +, e -, OH - ) must be added using rules for balancing half-reactions When potential of a half-reaction for a given species is more positive than the one to the left the species will disproportionate Separate diagrams are required for acidic and basic conditions
13 Examples of Latimer diagrams Acidic solution Acidic solution Basic solution
14 Examples and interpretation of Latimer diagrams 1.49 Acidic solution MnO H e - = Mn H 2 O Eº = 1.49 Basic solution MnO H 2 O + 3 e - = MnO OH - Eº = 0.59
15 Examples and interpretation of Latimer diagrams Acidic solution HClO e H + = HClO + H 2 O Eº = HClO 2 + H 2 O = ClO H e - Eº = HClO 2 = H + + ClO 3- + HClO Eº = 0.493
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