Announcements. Comprehensive Final Exam: March 24 7:30AM - 9:30 C114 2,9,10,11,13,17,22,29,31,38,40,44,46,50,53,58,62,64,65,70, 72,73,82,85,87

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1 Announcements Exam 3 March 17 Comprehensive Final Exam: March 24 7:30AM - 9:30 C114 Problems Chapter 21: 2,9,10,11,13,17,22,29,31,38,40,44,46,50,53,58,62,64,65,70, 72,73,82,85,87 Up to but not including 21.4 (page 923). Then cover 926 to problem on 927.

2 The Table of Reduction Potentials summarizes and predicts a tremendous amount of redox chemistry. We can use it to predict what will oxidize and be reduced by combining any two reduction reactions. We must learn how to the table to predict redox chemistry and to get the E cell

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4 The standard cell potential, E cell for any half-cell combinations under standard state conditions is obtained by adding the potentials. This can be done in two equivalent ways as shown below. E cell = E red (cathode) - E red (anode) E cell = standard reduction potential of the substance reduced - standard reduction potential of the substance oxidized E cell = E red (cathode) + E oxidation (anode) E cell = standard reduction potential of the substance reduced + standard oxidation potential of the substance oxidized

5 Some simple rules can teach you lots of redox chemistry. 1. The magnitude and sign of a half-reaction are all relative to the the zero point of the hydrogen half-cell. 2. When trying to decide what will be oxidized or what will be reduced remember that the larger the reduction potential is for a half-reaction, the more likely that reaction will occur vs another that has a smaller potential. 3. The species that have the strongest tendency to be reduced (larger E ) the most are the best oxidizing agent. The species that like to be reduced the least are the best reducing agents.

6 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = V Cr 3+ (aq) + 3e - Cr (s) E 0 = V What will be oxidized and what will be reduced? How can we tell?

7 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = V Cd is the stronger oxidizer Cr 3+ (aq) + 3e - Cr (s) E 0 = V Cd will oxidize Cr Anode (oxidation): Cr (s) Cr 3+ (1 M) + 3e - x 2 Cathode (reduction): 2e - + Cd 2+ (1 M) Cd (s) x 3 2Cr (s) + 3Cd 2+ (1 M) 3Cd (s) + 2Cr 3+ (1 M) E cell = E cathode - E anode E 0 = (-0.74) cell E 0 = 0.34 V cell

8 Using the following galvanic cell under standard conditions calculate E cell and show how you got your answer. A. Look at the Reduction Table for Ni +2 and Ag + reduction B. The reaction that will be reduced has the largest postitive reduction potential E. The lower value is oxidized and will have to be reversed. C. Remember that in a galvanic cell the anode is where oxidation occurs, cathode reduction. D. Decide on the shorthand notation for the cell. Ni Ag Ni 2+ Ag + salt bridge or porous partition

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10 The standard reduction potential, Eº shows the tendency to be reduced. Ag + + e! Ag Ni e! Ni Eº(Ag + ) = 0.80 V Eº(Ni 2+ ) = V (not 2 x)! 2Ag + + 2e! 2Ag Eº(Ag + ) = 0.80 V Ni (s)! Ni e Eº(Ni 2+ ) = V Ni + 2Ag +! Ni Ag Eº cell = Eº(Ag + ) Eº(Ni 2+ ) Eº cell = 0.80 V V Eº cell = 1.08 V

11 Voltaic (Galvanic) Cells Cell construction electrical device anode ( ) e e cathode (+) Ni Cl Ni 2+ Ni(NO 3 ) 2 (aq) Ni! Ni e (oxidation) salt bridge KCl in gelatin allows electrolytic conduction without mixing K + Ag + AgNO 3 (aq) Ag + + e! Ag (reduction) Ag Net reaction: Ni + 2Ag +! Ni Ag

12 Example Problem 2: Calculating an unknown E o halfcell from E o cell A voltaic cell under standard conditions houses the reaction between aqueous bromine and zinc metal as described by: Br 2 (aq) + Zn(s) Zn 2+ (aq) + 2Br - (aq) E o cell = 1.83 V Calculate E o Br2 given that E o zinc = V PLAN: E o cell = 1.83V > 0 and therefore the reaction is spontaneous as written. Zinc(s) is being oxidized and is the anode, Br2(aq) is being reduced. E o Br2 can be found using:

13 SOLUTION: E o cell = E o cathode - E o anode = E o Br2 - E o Zn = E o Br2 - (-0.76) using this equation both reduction potentials E from the table (not reversed) therefore E o Br2 = = 1.07 V

14 Writing spontaneous redox reactions and ranking oxidizing and reducing agents by strength PROBLEM: (a) Combine the following three half-reactions into three spontaneous, balanced equations (A, B and C), and calculate E o cell for each. (b) Rank the relative strengths of the oxidizing and reducing agents: (1) NO 3 - (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) E o = 0.96 V (2) N 2 (g) + 5H + (aq) + 4e - N 2 H 5 + (aq) (3) MnO 2 (s) + 4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) PLAN: E o = V E o = 1.23 V Put the equations together in varying combinations so as to produce (+) E o cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E o. Balance the number of electrons gained and lost without changing the E o. In ranking the strengths, compare the combinations in terms of E o cell.

15 SOLUTION: (1) NO - 3 (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) E o = 0.96 V Rev (a) (2) N 2 H + 5 (aq) N 2 (g) + 5H + (aq) + 4e - E o = V E o cell = 1.19 V 4NO 3 - (aq) + 16H + (aq) + 12e - 3N 2 H 5 + (aq) 3N 2 (g) + 15H + (aq) + 12e - 4NO(g) + 8H 2 O(l) E o = 0.96 V E o = V E o cell = 1.19 V 4NO 3 - (aq) + 3N 2 H 5 + (aq) + H + (aq) 4NO(g) + 3N 2 (g) + 8H 2 O(l)

16 SOLUTION: Rev NO(g) + 2H 2 O(l) NO - 3 (aq) + 4H + (aq) + 3e - (MnO 2 (s) + 4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) E o = V E o = 1.23 V E o cell = 0.27 V 2NO(g) + 4H 2 O(l) 2NO 3 - (aq) + 8H + (aq) + 6e - 3MnO 2 (s) + 12H + (aq) + 6e - 3Mn 2+ (aq) + 6H 2 O(l) E o = V E o = 1.23 V E o cell = 0.27 V 2NO(g) + 3MnO 2 (s) + 4H + (aq) 2NO 3 - (aq) + 3Mn 2+ (aq) + 2H 2 O(l)

17 Rev (2) N E o = V 2 H + 5 (aq) N 2 (g) + 5H + (aq) + 4e - (3) MnO E o = 1.23 V 2 (s) + 4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) (2) N 2 H + 5 (aq) N 2 (g) + 5H + (aq) + 4e - E o cell = 1.46 V (3) MnO 2 (s) + 4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) x2 (C) N 2 H 5 + (aq) + 2MnO 2 (s) + 3H + (aq) N 2 (g) + 2Mn 2+ (aq) + 4H 2 O(l)

18 (1) NO 3 - (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) (2) N 2 (g) + 5H + (aq) + 4e - N 2 H 5 + (aq) (3) MnO 2 (s) + 4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) E o = 0.96 V E o = V E o = 1.23 V (b) Ranking oxidizing and reducing agents within each equation: Examine the reduction potentials and remember what that the thing that like to be reduced the most is the best oxidizing agent! (A): oxidizing agents: NO 3 - > N 2 reducing agents: N 2 H 5 + > NO (B): oxidizing agents: MnO 2 > NO 3 - reducing agents: NO > Mn 2+ (C): oxidizing agents: MnO 2 > N 2 reducing agents: N 2 H 5 + > Mn 2+

19 Quiz? Given the two half reactions under standard conditions and the standard reduction potentials (E red) below suppose you make a galvanic cell using the two half reactions: Fe e! Fe Zn e! Zn Eº = 0.04 V Eº = 0.76 V 1. Write the short hand notation for this cell (use KNO3 as a salt bride) 2. Draw a picture showing the anode, cathode, salt bridge and direction of electron flow. 3. Compute the overall E cell

20 The Nernst Equation relates concentration in a galvanic cell to the cell potential. It allows to calculate when conditions are not standard state. E = E cell - RT nf ln Q E = E cell V n ln Q When Q < 1 and thus [reactant] > [product], ln Q < 0, so E cell > E o cell When Q = 1 and thus [reactant] = [product], ln Q = 0, so E cell = E o cell When Q >1 and thus [reactant] < [product], ln Q > 0, so E cell < E o cell

21 Given the following information, compute Ecell of the following reaction under non-standard state conditions? Is the reaction spontaneous? e.g., Ni Ni 2+ (0.05 M) Ag + (0.01 M) Ag

22 Given the following information, compute Ecell of the following reaction under non-standard state conditions? Is the reaction spontaneous? The standard reduction potentials for Ni 2+ and Ag + are -0.25V and 0.80 V respectively. e.g., Ni Ni 2+ (0.05 M) Ag + (0.01 M) Ag Ni (s)! Ni e - 2Ag + + 2e -! 2Ag E = E 0 cell V 2 E = 1.05V V 2 Eº = 0.25 V Eº = 0.80 V Eº = 1.05 V ln [Ni2+ [Ag + ] 2 ln [0.05] [0.01] 2 = 0.97V

23 Using the Nernst equation to calculate E cell PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn 2+ half-cell and an H 2 /H + half-cell under the following conditions: PLAN: SOLUTION: [Zn 2+ ] = M [H + ] = 2.5 M P = 0.30 atm H 2 Calculate E cell at 25 o C. Find E o cell 2H + (aq) + 2e - and Q in order to use the Nernst equation. Determining E o cell : Zn 2+ (aq) + 2e - H 2 (g) Zn(s) E o = 0.00 V E o = V Zn(s) Zn 2+ (aq) + 2e - E o = V E cell = E o V cell - log Q n E cell = (0.0592/2) log (4.8 x 10-4 ) = 0.86 V Q = P x [Zn 2+ ] H 2 [H + ] 2 Q = (0.30)(0.010) (2.5) 2 Q = 4.8 x 10-4