# Red 8 h mno 4 mn 2 4 h 2 o we balance the total

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Red: 8 H + + MnO 4 Mn 2+ + 4 H 2 O We balance the total charge in each half- reaction by adding electrons. In the preceed- ing reduction reaction there is a total charge of +7 on the left and +2 on the right. Five electrons are added to the left: Oxid: 2 Br 1 Br 2 + 2 e Red: 5 e + 8 H + + MnO 4 Mn 2+ + 4 H 2 O The number of electrons gained by Mn must equal the number of electrons lost by Br. We multiply the oxidation reaction by 5 and the reduction reaction by 2 to balance the elec- trons: Oxid: 10 Br 1 5 Br 2 + 10 e Red: 10 e + 16 H + + 2 MnO 4 2 Mn 2+ + 8 H 2 O Adding the half-reactions gives the overall balanced equation 10 Br 1 + 16 H + + 2 MnO 4 5 Br 2 + 2 Mn 2+ + 8 H 2 O 027 10.0 points Oxidation occurs 1. at both anode and cathode. 2. in the electrolyte. 3. at either, depending on whether the cell is electrochemical or electrolytic. 4. at the cathode. 5. at the anode. correct Explanation: 028 10.0 points If 100 mL of 0.040 M NaOH solution is added to 100 mL of solution which is 0.10 M in CH 3 COOH and 0.10 M in NaCH 3 COO, what will the pH of the new solution be? ( K a = 1 . 8 × 10 5 ) 1. 4.89 2. 4.74 3. 5.00 4. 5.11 correct 5. 4.81 Explanation: [CH 3 COOH] = 0.10 M [NaOH] = 0.040 M

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Version – Exam 3 – holcombe – 9 [CH 3 COO ] = 0.10 M K a = 1 . 8 × 10 5 Initial condition (ini): n NaOH = 100 × 0 . 04 = 4 mmol n CH 3 COOH = 100 × 0 . 10 = 10 mmol n Na + = 100 × 0 . 10 = 10 mmol n CH 3 COO - = 100 × 0 . 10 = 10 mmol NaOH +CH 3 COOH Na + +CH 3 COO + H 2 O ini 4 . 0 10 . 0 10 . 0 10 . 0 Δ - 4 . 0 - 4 . 0 4 . 0 4 . 0 fin 0 6 . 0 14 . 0 14 . 0 Na + is a spectator ion. CH 3 COOH / CH 3 COO is a buffer system. pH = p K a + log parenleftBigg bracketleftbig CH 3 COO bracketrightbig [CH 3 COOH] parenrightBigg = - log ( 1 . 8 × 10 5 ) + log parenleftbigg 14 . 0 6 . 0 parenrightbigg = 5 . 1127 029 10.0 points A buffer contains equal concentrations of NH 3 (aq) and NH 4 Cl(aq). What is the pH of the buffer? For NH 3 , K b = 1 . 8 × 10 5 . ) 1. 4.74 2. 7.00 3. 13.00 4. 9.26 correct Explanation: 030 10.0 points If the two half reactions below were used to make an electrolytic cell, what species would be consumed at the anode? Half reaction E Au 3+ ( aq ) + 3 e -→ Au( s ) +1 . 50 I 2 ( s ) + 2 e -→ 2 I ( aq ) +0 . 53 1. I (aq) 2. Au (s) correct 3. I 2 (s) 4. Au 3+ (aq) Explanation: An electrolytic cell must have a negative standard cell potential and therefore the an- odic reaction must consume Au( s ). 031 10.0 points Consider the half-reactions and the bal- anced equation for the cell reaction repre- sented by the skeletal equation Mn(s) + Ti 2+ (aq) Mn 2+ (aq) + Ti(s) . What is the proper cell diagram for this reac- tion? 1. Ti 2+ (aq) | Ti(s) || Mn(s) | Mn 2+ (aq) 2. Mn 2+ (aq) | Mn(s) || Ti(s) | Ti 2+ (aq) 3. Mn(s) | Mn 2+ (aq) || Ti 2+ (aq) | Ti(s) cor- rect 4. Ti(s) | Ti 2+ (aq) || Mn 2+ (aq) | Mn(s) Explanation: The two half-reactions, written as reduc- tions, are Mn 2+ (aq) + 2 e Mn(s) Ti 2+ (aq) + 2 e Ti(s) Equate e : Ti 2+ (aq) + 2 e Ti(s) Mn(s) Mn 2+ (aq) + 2 e Add the balanced half reactions: Mn(s) + Ti 2+ (aq) Mn 2+ (aq) + Ti(s) The cell diagram is Mn(s) | Mn 2+ (aq) || Ti 2+ (aq) | Ti(s) 032 10.0 points Chlorine, bromine, and iodine are good 1. bases.
Version – Exam 3 – holcombe – 10 2. reducing agents. 3. oxidizing agents. correct Explanation: 033 10.0 points Aluminum is more reactive than iron, yet it is used today for a variety of applications in which iron would corrode (cans, rain gutters, etc). The reason for the corrosion durability of aluminum is 1. very unreactive aluminum oxide forms a thin layer on aluminum. correct 2.
• Fall '07
• Holcombe

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