CH19Quiz.pdf

1 Chemistry 120 – Chapter 19 Practice Exams - Dr. Elmo Mawk This practice exam covering Chapter 19 is composed of XX questions, a combination of multiple choice, free response questions. The questions were taken from old exams I gave in CHEM 107 and 120. A formula sheet is provided at the end of thi...

1 Chemistry 120 – Chapter 19 Practice Exams - Dr. Elmo Mawk This practice exam covering Chapter 19 is composed of XX questions, a combination of multiple choice, free response questions. The questions were taken from old exams I gave in CHEM 107 and 120. A formula sheet is provided at the end of this exam. Potentially Useful Information PHYSICAL CONSTANTS NA = 6.02 x 1023 mol-1 c = 3.00 x 108 m/s (in a vacuum) R = 0.08206 L Atm/mol K h = 6.626 x 10-34 Js R = 8.314 J/mol K F = 96,485 C/mol electrons E = 1.60219 x 10-19 C 1 electron = 9.109383 x 10-28 g CONVERSION FACTORS 1 inch = 2.54 cm 1 kg = 2.205 lb 1 gallon = 3.7865 L 760 torr = 1 Atm 1 ft = 12 inches 1 kb = 453.59 g 4 quarts = 1 gallon 1 atm = 101.325 kPa 1 Å = 10-10 m 1 amu = 1.6606 x 10-24 g 1 mL = 1 cm3 1 atm = 1.01325 bar 1 ton = 2000 lb 1 quart = 0.9463 L 1 atm = 14.70 psi 1 torr = 1 mm Hg 1 calorie = 4.184 J o 0 K = -273.15 C 1 J = 1 kg m2/s2 Chapter 19 Practice Exam Questions with Answers, Spring 2024 2 SELECTED STANDARD REDUCTION POTENTIALS (25°C) Acidic Solution Standard Reduction Potential, E° (Volts) + 2+ MnO4¯(aq) + 8H (aq) + 5e¯ → Mn (aq) + 4H2O 1.51 + Ag (aq) + e¯ → Ag(s) 0.7994 Fe3+(aq) + e¯ → Fe2+(aq) 0.771 I2(s) + 2e¯ → 2I¯(aq) 0.535 + Cu (aq) + e ¯ → Cu(s) 0.521 2+ Cu (aq) + 2e¯ → Cu(s) 0.337 2+ + Cu (aq) + e¯ → Cu (aq) 0.153 2H+(aq) + 2e¯ → H2(g) 0.00 2+ Pb (aq) + 2e¯ → Pb(s) − 0.126 2+ Fe (aq) + 2e¯ → Fe(s) − 0.44 2+ Zn (aq) + 2e¯ → Zn(s) − 0.763 2+ Mn (aq) + 2e¯ → Mn(s) − 1.18 Al3+(aq) + 3e¯ → Al(s) − 1.66 2+ Ca (aq) + 2e¯ → Ca(s) − 2.87 + Li (aq) + e¯ → Li(s) − 3.045 Basic Solution MnO4¯(aq) + 2H2O + 3e¯ → MnO2(s) + 4OH¯(aq) 2H2O + 2e¯ → H2(g) + 2OH¯(aq) 0.588 -0.828 Chapter 19 Practice Exam Questions with Answers, Spring 2024 + 2+ 3 19-1) For a voltaic (or galvanic) cell using Ag, Ag (1.0 M) and Zn, Zn (1.0 M) half-cells, which of the following statements is incorrect? A) The zinc electrode is the anode. B) Electrons will flow through the external circuit from the zinc electrode to the silver electrode. C) Reduction occurs at the zinc electrode as the cell operates. D) The mass of the zinc electrode will decrease as the cell operates. E) The concentration of Ag+ will decrease as the cell operates. To answer this question, determine the reduction half-cell, oxidation half-cell, the cell potential, etc. 𝐴𝑔! + 𝑒 " → 𝐴𝑔 & 𝐸#$% = 0.7994 𝑉 𝑍𝑛'! + 2 𝑒 " → 𝑍𝑛 & 𝐸#$% = −0.763 𝑉 & The more positive 𝐸#$% is the reduction half-cell. & The more negative 𝐸#$% is the oxidation half-cell. Flip this half-cell and change the sign on the potential. Multiply half-cells as needed to cancel out the electron transfer, but DO NOT multiply the potentials. Multiplying the half cells is to obtain the overall reaction only. 𝐴𝑔! + 𝑒 " → 𝐴𝑔 & 𝐸#$% = 0.7994 𝑉 𝑍𝑛 → 𝑍𝑛'! + 2 𝑒 " & 𝐸&( = 0.763 𝑉 2 𝐴𝑔! + 𝑍𝑛 → 2 𝐴𝑔 + 𝑍𝑛'! & 𝐸)$** = 1.592 𝑉 A) The zinc electrode is the anode. Oxidation occurs at the anode. Zinc is being oxidized, so the zinc electron is the anode. Correct statement. B) Electrons will flow through the external circuit from the zinc electrode to the silver electrode. The zinc electron is the anode, where oxidation occurs. Oxidation is loss, so the electrons are “lost” into the electrode, flowing toward the silver electrode where the electrons are available for reduction. Correct statement. C) Reduction occurs at the zinc electrode as the cell operates. Oxidation occurs at the anode. Zinc is being oxidized, so the zinc electrode is not being reduced. Incorrect statement! D) The mass of the zinc electrode will decrease as the cell operates. As the cell runs, the zinc electrode is oxidized, releasing zinc cation into solution. The mass of the zinc electrode is being decreasing and the concentration of zinc cation increases. Correct statement. E) The concentration of Ag+ will decrease as the cell operates. The silver electrode is undergoing reduction, plating silver cation out of solution onto the surface of the silver electrode, increasing the mass of the electrode and decreasing the concentration of silver cation in solution. Correct statement. Chapter 19 Practice Exam Questions with Answers, Spring 2024 4 Answer C & Another way to determine 𝐸)$** is to subtract the reduction potential of the anode from the reduction & & & potential of the cathode, 𝐸)$** = 𝐸#$% − 𝐸#$% = 0.7994 𝑉 − −0.763 𝑉 = 1.592 𝑉 )+,-&%$ +.&%$ Chapter 19 Practice Exam Questions with Answers, Spring 2024 5 19-2) In the standard notation for a voltaic cell, the double vertical line "||" represents: A) a phase boundary B) gas electrode C) a wire (metal) connection D) a salt bridge E) a standard hydrogen electrode || represents the salt bridge. | represents the interface between the electrode and electrolyte. Answer D 19-3) Which of the following statements about electrodes is false? A) The cathode is the electrode at which reduction occurs. B) The anode can be either positive or negative. C) Electrodes are surfaces on which oxidation or reduction half-reactions occur. D) Electrodes that do not react are called inert electrodes. E) Electrons are gained by some species at the anode. A) The cathode is the electrode at which reduction occurs. This is a true statement. Reduction always occurs at the cathode. Oxidation always occurs at the anode. (Red Cat, An Ox) B) The anode can be either positive or negative. This is a true statement. The type of cell, electrolytic or galvanic, determines the sign on the electrode. The same is true for the cathode. C) Electrodes are surfaces on which oxidation or reduction half-reactions occur. This is a true statement. Electrons are gained at the surface of the electrode or lost at the surface of the electrode. D) Electrodes that do not react are called inert electrodes. This is a true statement. Active electrodes change mass while the cell operates. Inert electrodes do not change mass while the cell operates. E) Electrons are gained by some species at the anode. This is a false statement. Oxidation occurs at the anode and oxidation is the loss of electrons (OIL RIG.) Answer E! Chapter 19 Practice Exam Questions with Answers, Spring 2024 6 19-4) How long would a constant current of 4.5 amperes be required to flow in order to plate out 15 g of chromium from a chromium(III) sulfate solution? A) 268 hr B) 309 hr C) 5.15 hr D) 23.2 hr E) 1.72 hr Faraday’s Law states that the moles electrons passed through a cell equals the current multiplied by the time in seconds divided by Faraday’s Constant, 96485 C/mol e. Once the moles electrons that flowed through the cell is calculated, stoichiometry is used to determine mass/moles material plated/removed from solution. Faraday’s Law can also be used in the opposite direction, use the stoichiometry to determine moles electrons required to remove/plate a given mass from solution, and for a fixed current, determine the time required. 𝐶𝑟 /! + 3 𝑒 " → 𝐶𝑟 15 𝑔 𝐶𝑟 × 1 𝑚𝑜𝑙 𝐶𝑟 3 𝑚𝑜𝑙 𝑒 " × = 0.865 𝑚𝑜𝑙 𝑒 " 52.00 𝑔 𝐶𝑟 1 𝑚𝑜𝑙 𝐶𝑟 𝐶 " 𝐴×𝑡 𝑛 × 𝐹 (0.865 𝑚𝑜𝑙 𝑒 ) B96485 𝑚𝑜𝑙 𝑒 " C 3600 𝑠 𝑛= 𝑜𝑟 𝑡 = = = 18,547 𝑠 × = 5.15 𝐻𝑟 𝐹 𝐴 4.5 𝐴 1 ℎ𝑟 Answer C 19-5) Which of the following statements about the salt bridge in a voltaic cell is false? A) It allows electrical contact between the two solutions in the half-cells. B) The ions from the salt bridge participate in the half-reactions. C) It prevents mixing of the electrode solutions. D) It maintains the electrical neutrality in each half-cell. E) Ions flow into and out of the salt bridge. Properties of a Salt Bridge - complete the circuit, allowing the reaction to occur - prevent the contents of the half cells from mixing - maintain charge balance by allowing cations and anions in the salt bridge to flow into half cells - ions from the salt bridge do not participate in the redox reactions. Answer B Chapter 19 Practice Exam Questions with Answers, Spring 2024 7 19-6) Which of the following is the strongest oxidizing agent in 1 M solutions? A) MnO4– (in aqueous acidic solution) B) Ag+(aq) C) Cu2+(aq) D) Ca2+(aq) E) Li+(aq) & Oxidizing agents are the reduced species in a redox reaction. The species with the largest positive 𝐸#$% is the strongest oxidizing agent. Using a standard reduction potential table, the reduction potentials for each answer choice are 𝑀𝑛𝑂0" & 𝐸#$% = 1.51 𝑉 𝐴𝑔! & 𝐸#$% = 0.7994 𝑉 𝐶𝑢'! & 𝐸#$% = 0.337 𝑉 𝐶𝑎'! & 𝐸#$% = −2.87 𝑉 𝐿𝑖 ! & 𝐸#$% = −3.045 𝑉 𝑀𝑛𝑂0" has the most positive reduction potential and is the strongest oxidizing agent of the 5 answer choices. Answer A 19-7) In an electrolytic cell the electrode at which the electrons enter the solution is called the ______ ; the chemical change that occurs at this electrode is called _______. A) anode, oxidation B) anode, reduction C) cathode, oxidation D) cathode, reduction E) cannot tell unless we know the species being oxidized and reduced. In electrolytic cells, the electrons are on the surface of the cathode, facilitating reduction. Reduction always occurs at the cathode. Answer D Chapter 19 Practice Exam Questions with Answers, Spring 2024 8 19-8) Given the standard electrode potentials below, calculate Kc at 25°C for the following reaction. (F = 96,500 J/V mol e– and R = 8.314 J/mol K) 2 𝐹𝑒 /! (𝑎𝑞) + 2 𝐼 " (𝑎𝑞) → 2 𝐹𝑒 '! (𝑎𝑞) + 𝐼' (𝑠) 𝐸& 𝐹𝑒 /! (𝑎𝑞) + 𝑒 " → 𝐹𝑒 '! (𝑎𝑞) +0.771 V 𝐼' (𝑠) + 2 𝑒 " → 2 𝐼 " (𝑎𝑞) A) 1.6 x 1012 +0.535 V B) 1.1 x 10–8 C) 1.0 x 10–4 𝐹𝑒 /! (𝑎𝑞) + 𝑒 " → 𝐹𝑒 '! (𝑎𝑞) + 0.771 𝑉 2 𝐼 " (𝑎𝑞) → 𝐼' (𝑠) + 2 𝑒 " − 0.535 𝑉 2 𝐹𝑒 /! (𝑎𝑞) + 2 𝐼 " (𝑎𝑞) → 2 𝐹𝑒 '! (𝑎𝑞) + 𝐼' (𝑠) D) 9.6 x 107 0.236 𝑉 To calculate the Kc value for the reaction, the equation 𝐸 = 𝐸 & − 1.134'. potential decreases to 0 V, the cell is at equilibrium, and Q becomes K. 0 𝑉 = 0.236 𝑉 − −0.236 𝑉 = − E) 9.7 x 103 0.0592 log (𝐾) 2 0.0592 log (𝐾) 2 𝐿𝑜𝑔(𝐾) = 7.97 𝐾 = 105.45 = 9.6 × 105 Answer D Chapter 19 Practice Exam Questions with Answers, Spring 2024 log (𝑄) must be used. When the cell 9 19-9) A voltaic cell is constructed by immersing a strip of copper metal in 1.0 M CuSO4 solution and a strip of aluminum in 0.50 M Al2(SO4)3 solution. A wire and a salt bridge complete the circuit. The aluminum strip loses mass, and the concentration of aluminum ions in the solution increases. The copper electrode gains mass, and the concentration of copper ions decreases. Which of the following are applicable to the copper electrode? I. The anode II. The cathode III. The positive electrode IV. The electrode at which electrons are produced V. The negative electrode VI. The electrode at which electrons are used up A) I, III, and V B) I, IV, and V C) II, IV, and V D) II, III, and VI E) None of the first four responses contains all the correct choices and no others. The only way the aluminum electrode can lose mass is if the aluminum metal is oxidized to aluminum cation. 𝐴𝑙 → 𝐴𝑙 /! + 3 𝑒 " The copper electrode gains mass by reducing copper(II) to copper metal. 𝐶𝑢'! + 2 𝑒 " → 𝐶𝑢 As the mass of the Al electrode decreases, the concentration of Al3+ will increase. As the mass of the Cu electrode increases, the concentration of Cu2+ will decrease. The Al electrode is the anode. Oxidation always occurs at the anode. The Cu electrode is the cathode. Reduction always occurs at the cathode. The Al electrode is the negative electrode. Electrons flow away from the negative electrode. The Cu electrode is the positive electrode. Electrons flow toward the positive electrode. True statements are II. The cathode III. The positive electrode VI. The electrode at which electrons are used up Answer D Chapter 19 Practice Exam Questions with Answers, Spring 2024 19-10) Calculate the Pb2+ concentration in the following cell if Ecell = 0.95 V. 10 Mn(s)|Mn2+(1.0 M)||Pb2+(x M)|Pb Oxidation occurs at the anode and the anode is written first in cell notation. Mn is the anode. Reduction occurs at the cathode and the cathode is written second in cell notation. Pb is the cathode. (I used the fact that anode and cathode, from right to left, is in alphabetical order.) & To determine the lead concentration, the 𝐸)$** , is needed, as well as the Nernst equation, 𝐸 = 𝐸 & − 1.134'. log (𝑄). Looking up the reduction potentials from a table of standard reduction potentials, the reduction potentials are 𝑀𝑛'! (𝑎𝑞) + 2 𝑒 " → 𝑀𝑛(𝑠) & 𝐸#$% = −1.18 𝑉 𝑃𝑏 '! (𝑎𝑞) + 2 𝑒 " → 𝑃𝑏(𝑠) & 𝐸#$% = −0.126 𝑉 & & & The cell potential can be calculated with 𝐸)$** = 𝐸#$% − 𝐸#$% or by flipping the reduction half)+,-&%$ +.&%$ reaction for the anode, and adding the reduction half-reaction and the oxidation half-reaction. & 𝐸)$** = −0.126 𝑉 − −1.18 𝑉 = +1.054 𝑉 or 𝑃𝑏 '! (𝑎𝑞) + 2 𝑒 " → 𝑃𝑏(𝑠) & 𝐸#$% = −0.126 𝑉 𝑀𝑛(𝑠) → 𝑀𝑛'! (𝑎𝑞) + 2 𝑒 " & 𝐸&( = +1.18 𝑉 𝑃𝑏 '! (𝑎𝑞) + 𝑀𝑛(𝑠) → 𝑃𝑏(𝑠) + 𝑀𝑛'! (𝑎𝑞) & 𝐸)$** = +1.054 𝑉 The number of moles electrons transferred equals 2 moles. The reaction quotient is 𝑄 = 67.!" 8 [:; !" ] &. 𝐸)$** = 0.95 𝑉 and 𝐸)$** = 1.054 𝑉. [Mn2+]= 1.0 M 0.0592 log (𝑄) 𝑛 0.0592 1.0 𝑀 0.95 𝑉 = 1.054 𝑉 − log W Y 2 𝑚𝑜𝑙 𝑒 " 𝑥 1.0 𝑀 −0.104 = −0.0296 log W Y 𝑥 1.0 𝑀 3.51 = log W Y 𝑥 𝐸 = 𝐸& − =.1 7 10/.3= = 10>?@A ( C 1 3265.9 = 𝑥 𝑥 = 3.06 × 10"0 𝑜𝑟 [𝑃𝑏 '! ] = 3.06 × 10"0 𝑀 Chapter 19 Practice Exam Questions with Answers, Spring 2024 11 19-11) Calculate the cell potential of the following voltaic cell at 25°C. 𝑀𝑔|𝑀𝑔'! (1.0 × 10"D 𝑀) ∥ 𝐴𝑔! (1.0 × 10"' 𝑀)|𝐴𝑔 A) +3.17 V B) +3.29 V C) +3.21 V D) +3.11 V E) +1.63 V To calculate the cell potential at nonstandard conditions, the standard cell potential must be calculated, and the number of electrons transferred determined from the balanced redox reaction. The cell diagram shows Mg is oxidized, and Ag is reduced. The standard reduction potentials for Mg cation, and Ag cation are looked up from a table of standard reduction potentials. 𝑀𝑔'! (𝑎𝑞) + 2𝑒 " → 𝑀𝑔(𝑠) & 𝐸#$% = −2.372 𝑉 𝐴𝑔! + 𝑒 " → 𝐴𝑔(𝑠) & 𝐸#$% = 0.7796 𝑉 The magnesium half-cell is an oxidation (anode listed first). The reduction half reaction is written as an oxidation. 𝑀𝑔(𝑠) → 𝑀𝑔'! (𝑎𝑞) + 2𝑒 " & 𝐸&( = +2.372 𝑉 𝐴𝑔! (𝑎𝑞) + 𝑒 " → 𝐴𝑔(𝑠) & 𝐸#$% = 0.7796 𝑉 𝑀𝑔(𝑠)+2 𝐴𝑔! (𝑎𝑞) → 𝑀𝑔'! (𝑎𝑞) + 2 𝐴𝑔(𝑠) & 𝐸)$** = 3.152 𝑉 The number of electrons transferred is 2. The reaction quotient is 𝑄 = & 𝐸 = 𝐸)$** − 0.0592 𝑙𝑜𝑔(𝑄) 𝑛 [𝑀𝑔'! ] 0.0592 𝐸 = 3.152 − 𝑙𝑜𝑔 ^ _ [𝐴𝑔! ]' 2 𝐸 = 3.152 − 0.0592 1.0 × 10"D 𝑙𝑜𝑔 ^ _ (1.0 × 10"' )' 2 𝐸 = 3.152 − 0.0592 𝑙𝑜𝑔(0.01) = 3.152 − −0.0592 = 3.211 𝑉 2 Answer C Chapter 19 Practice Exam Questions with Answers, Spring 2024 67E!" 8 [FE" ]!. 12 19-12) Which transformation could take place at the cathode of an electrochemical cell? A) 𝑀𝑛𝑂' → 𝑀𝑛𝑂0" B) 𝐵𝑟' → 𝐵𝑟𝑂/" C) 𝑁𝑂 → 𝐻𝑁𝑂' D) 𝐻𝑆𝑂0" → 𝐻' 𝑆𝑂/ E) 𝑀𝑛'! → 𝑀𝑛𝑂0" Reduction occurs at the cathode. (red cat, an ox) Assign oxidation numbers to each species and determine which oxidation number decreased. A) 𝑀𝑛𝑂' → 𝑀𝑛𝑂0" 𝑀𝑛 = +4, 𝑂 = −2; 𝑀𝑛 = +7, 𝑂 = −2, 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑎𝑛 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 B) 𝐵𝑟' → 𝐵𝑟𝑂/" 𝐵𝑟 = 0; 𝐵𝑟 = +5, 𝑂 = −2, 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑎𝑛 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 C) 𝑁𝑂 → 𝐻𝑁𝑂' 𝑁 = +2, 𝑂 = −2; 𝐻 = +1, 𝑁 = +3, 𝑂 = −2, 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑎𝑛 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 D) 𝐻𝑆𝑂0" → 𝐻' 𝑆𝑂/ 𝐻 = +1, 𝑆 = +6, 𝑂 = −2; 𝐻 = +1, 𝑆 = +5, 𝑂 = −2, 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑎 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛. E) 𝑀𝑛'! → 𝑀𝑛𝑂0" 𝑀𝑛 = +2; 𝑀𝑛 = +7, 𝑂 = −2, 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑎𝑛 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 Answer D Chapter 19 Practice Exam Questions with Answers, Spring 2024 13 19-13) How many grams of Ca metal are produced by the electrolysis of molten CaBr2 using a current of 30.0 amp for 10.0 hours __________? A) 22.4 B) 448 C) 0.0622 D) 224 E) 112 In molten CaBr2, Ca2+ and Br- are flowing past each other, allowing current to flow between the electrode placed in the molten salt. When current passes through the cell, copper is reduced the cathode and the bromide is oxidized at the anode. The copper is the species of interest. 𝐶𝑎'! (𝑎𝑞) + 2 𝑒 " → 𝐶𝑎(𝑠) For every mole of calcium cation reduced, 2 moles of electrons must be supplied, 1 Ca:2 e-. The relationship between moles electrons/copper/etc. oxidized or reduced is Faradays law. Faradays law says that the moles electrons that passes through the cell is proportional to the product of current, A, and time, s. 𝑛 ∝ 𝐴𝑡 To remove the proportionality, the constant F, Faraday’s constant is introduced. 𝑛= 𝐴𝑡 𝐹 Faraday’s constant, F, has a value of 96485 C/mol e-. A is the amperage, C/s, and t is in seconds. First calculate the moles electrons. 𝐴𝑡 (30.0 𝐶/𝑠)(10.0 ℎ𝑜𝑢𝑟𝑠 × 𝑛= = 96485 𝐶 𝐹 𝑚𝑜𝑙 𝑒 " 3600 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 1 ℎ𝑜𝑢𝑟 = 11.19 𝑚𝑜𝑙𝑒𝑠 𝑒 " Now use the stoichiometry between moles electrons and moles copper, and ultimately grams copper. 11.19 𝑚𝑜𝑙𝑒𝑠 𝑒 " × 1 𝑚𝑜𝑙𝑒 𝑐𝑎𝑙𝑐𝑖𝑢𝑚 40.08 𝑔 𝑐𝑎𝑙𝑐𝑖𝑢𝑚 × = 224 𝑔𝑟𝑎𝑚𝑠 𝑐𝑎𝑙𝑐𝑖𝑢𝑚 2 𝑚𝑜𝑙𝑒𝑠 𝑒 " 1 𝑚𝑜𝑙𝑒 𝑐𝑎𝑙𝑐𝑖𝑢𝑚 Answer D Chapter 19 Practice Exam Questions with Answers, Spring 2024 14 19-14) Consider an electrochemical cell based on the reaction: 2 𝐻! (𝑎𝑞) + 𝑆𝑛(𝑠) → 𝑆𝑛'! (𝑎𝑞) + 𝐻' (𝑔) Which of the following actions would not change the measured cell potential? A) lowering the pH in the cathode compartment B) addition of more tin metal to the anode compartment C) increasing the tin (II) ion concentration in the anode compartment D) increasing the pressure of hydrogen gas in the cathode compartment E) Any of the above will change the measured cell potential. The electrochemical cell is the SHE coupled with a Sn/Sn2+ half-cell. The Sn is oxidized at the anode, and the hydronium is reduced at the cathode. 𝑆𝑛(𝑠) → 𝑆𝑛'! (𝑎𝑞) + 2 𝑒 " 2 𝐻! (𝑎𝑞) + 2 𝑒 " → 𝐻' (𝑔) The Nernst equation for each half reaction is 𝐸 = 𝐸& − 𝑃G 0.0592 log W !! ' Y [𝐻 ] 𝑛 We can see that changing the partial pressure hydrogen or the hydronium concentration will change the cell potential. (When a cell is allowed to “run” the potential will slowly decrease as the concentrations and partial pressures change.) 0.0592 𝐸 = 𝐸& − log([𝑆𝑛'! ]) 𝑛 If the Sn2+ concentration changes, the cell potential will change. Now to the answer choices. A) lowering the pH in the cathode compartment The SHE is the cathode. The SHE requires H+ to function. Changing the pH will change the cell potential. B) addition of more tin metal to the anode compartment The Sn/Sn2+ half-reaction occurs at the anode. Adding more tin metal to the half-cell does not impact the cell potential because metal is a solid, which does not change in concentration. C) increasing the tin (II) ion concentration in the anode compartment Increasing the tin(II) concentration will change the cell potential. D) increasing the pressure of hydrogen gas in the cathode compartment Increasing the hydrogen gas partial pressure will change the cell potential. F) Any of the above will change the measured cell potential Any means anyone, not all. Answer B Chapter 19 Practice Exam Questions with Answers, Spring 2024 15 19-15) The standard cell potential (E°) of a voltaic cell constructed using the cell reaction below is 0.76 V: 𝑍𝑛(𝑠) + 2 𝐻! (𝑎𝑞) → 𝑍𝑛'! (𝑎𝑞) + 𝐻' (𝑔) With PH2 = 1.0 atm and [Zn2+] = 1.0 M, the cell potential is 0.66 V. The concentration of H+ in the cathode compartment is __________ M. A) 2.0 x 10-2 B) 4.2 x 10-4 C) 1.4 x 10-1 D) 4.9 x 101 E) 1.0 x 10-12 This is a Nernst equation problem where all concentrations, but one is given, and you are required to find that concentration. The calculation is not possible unless the cell potential is given; and can look up and calculate the cell potential under standard conditions. 2 𝐻! (𝑎𝑞) + 2 𝑒 " → 𝐻' (𝑔), 𝐸#$% = 0.00 𝑉 𝑍𝑛'! (𝑎𝑞) + 2 𝑒 " → 𝑍𝑛(𝑠), 𝐸#$% = −0.763 𝑉 The SHE remains the reduction because the SHE has the more positive reduction potential. Zinc is oxidized. & & The cell potential under standard conditions, if not given, is 𝐸 = 𝐸)+,-&%$ − 𝐸+.&%$ = 0 𝑉 − −0.763 𝑉 = 0.763 𝑉. Next, plug the measured potential, the standard reduction potential, moles electrons, and concentrations into the Nernst equation. & 𝐸)$** = 𝐸)$** − [𝑍𝑛'! ]𝑃G! 0.0592 log ^ _ [𝐻! ]' 𝑛 0.66 𝑉 = 0.76 𝑉 − [1.0 𝑀]1 𝑎𝑡𝑚 0.0592 log ^ _ [𝐻! ]' 2 1 −0.1 = −0.0296 log W ! ' Y [𝐻 ] 3.378378 = log W 1 Y [𝐻! ]' = >?@I " ! J [G ] 10/./5H/5H = 10 2389.89 = [𝐻 ! ] ' = 1 [𝐻! ]' 1 = 0.000418 2389.89 h[𝐻! ]' = √0.0000418 [𝐻! ] = 0. 020 𝑀 Answer A Chapter 19 Practice Exam Questions with Answers, Spring 2024 16 19-16) Balance the following half-reaction occurring in acidic solution. 𝑁𝑂/" (𝑎𝑞) → 𝑁𝑂(𝑔) A) 𝑁𝑂/" (𝑎𝑞) + 4 𝐻! (𝑎𝑞) + 3 𝑒 " → 𝑁𝑂(𝑔) + 2 𝐻' 𝑂(𝑙) B) 𝑁𝑂/" (𝑎𝑞) + 2 𝐻' 𝑂(𝑙) + 3 𝑒 " → 𝑁𝑂(𝑔) + 4 𝐻! (𝑎𝑞) C) 𝑁𝑂/" (𝑎𝑞) + 4 𝐻! (𝑎𝑞) → 𝑁𝑂(𝑔) + 2 𝐻' 𝑂(𝑙) + 3 𝑒 " D) 𝑁𝑂/" (𝑎𝑞) + 3 𝑒 " → 𝑁𝑂(𝑔) + 4 𝐻! (𝑎𝑞) + 2 𝐻' 𝑂(𝑙) E) 𝑁𝑂/" (𝑎𝑞) + 4 𝐻! (𝑎𝑞) → 𝑁𝑂(𝑔) + 2 𝐻' 𝑂(𝑙) When balancing redox reactions in acid, all hydrogens are balanced with H+ and all oxygens are balanced with H2O. To being with, the oxidation numbers must be balanced using electrons. Each oxygen has an oxidation number of -2. For nitrate the total oxidation contributed by O is -6. The overall charge is -1, so the nitrogen has an oxidation number of +5. (+5+-6=-1). For Nitrogen monoxide, the oxygen has a -2, the overall charge is 0, so the oxidation on nitrogen is +2. To go from +5 on nitrogen in the reactant, to +2 on nitrogen in the product, a total of 3 electrons must be added to the reactants. The nitrate is being reduced. 𝑁𝑂/" (𝑎𝑞) + 3 𝑒 " → 𝑁𝑂(𝑔) The reaction as written does not have any hydrogens. It does have oxygens, so they must be balanced first. The nitrate has 3 oxygens, and the nitrogen monoxide only 1. To bring the oxygens in balance, add 2 waters to the products. 𝑁𝑂/" (𝑎𝑞) + 3 𝑒 " → 𝑁𝑂(𝑔) + 2 𝐻' 𝑂(𝑙) The balancing of the oxygen adds 4 hydrogen atoms to the products, which must be balanced by adding 4 hydronium to the reactant side. 𝑁𝑂/" (𝑎𝑞) + 4 𝐻! (𝑎𝑔) + 3 𝑒 " → 𝑁𝑂(𝑔) + 2 𝐻' 𝑂(𝑙) To verify a balanced equation, the type and number of each type of element must be the same on both sides of the reaction, AND the total charge for the reactants must equal the total charge for the products. For the reactants, there is 1 N, 3 O, and 4 H. For the products, there is 1 N, 3 O, and 4 H. For the reactants the total charge is -1++4+-3=0. For the products, the total charge is 0+0=0. The reaction is balanced. The correct answer is A. Chapter 19 Practice Exam Questions with Answers, Spring 2024 17 19-17) Balance the following half-reaction occurring in basic solution. 𝑀𝑛𝑂' (𝑠) → 𝑀𝑛(𝑂𝐻)' (𝑠) A) B) C) D) E) 𝑀𝑛𝑂' (𝑠) + 2 𝐻' 𝑂(𝑙) + 2 𝑒 " → 𝑀𝑛(𝑂𝐻)' (𝑠) + 2 𝑂𝐻" (𝑎𝑞) 𝑀𝑛𝑂' (𝑠) + 2 𝐻' 𝑂(𝑙) + 4 𝑒 " → 𝑀𝑛(𝑂𝐻)' (𝑠) + 2 𝑂𝐻" (𝑎𝑞) 𝑀𝑛𝑂' (𝑠) + 𝐻''! (𝑎𝑞) + 2 𝑒 " → 𝑀𝑛(𝑂𝐻)' (𝑠) 𝑀𝑛𝑂' (𝑠) + 𝐻' (𝑔) → 𝑀𝑛(𝑂𝐻)' (𝑠) + 2 𝑒 " 𝑀𝑛𝑂' (𝑠) + 2 𝐻' 𝑂(𝑙) → 𝑀𝑛(𝑂𝐻)' (𝑠) + 2 𝑂𝐻" (𝑎𝑞) Question 18 describes how a redox reaction must be balanced in terms of changes in oxidation number. Unlike question 18, this question balances in base. Base is tricker to balance than acid. There are two sources of oxygen, water and hydroxide, and two sources of hydrogen, water and hydroxide. To balance O, for each oxygen needed, add 2 hydroxides for each oxygen needed, and to balanced out the extra oxygen and hydrogen, add a water to the opposite side. 2 𝑂𝐻" 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 1 𝐻' 𝑂 For each hydrogen needed, add 1 water for each hydrogen needed and a hydroxide to the other side of the reaction. 𝐻' 𝑂 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑂𝐻" Oxygen has an oxidation number of -2 and hydrogen has a +1 oxidation. For manganese(VI) oxide, the total oxidation from oxygen is -4. The oxidation on Mn is +4. Checking work, -4++4=0. For the manganese(II) hydroxide, hydroxide has a -1 charge, there are 2 hydroxides, so the charge on Mn is 2+. Remember oxidation equals charge for monoatomic cations. To change oxidation from +4 to +2, 2 electrons must be gained by the manganese(VI) oxide. 𝑀𝑛𝑂' (𝑠) + 2 𝑒 " → 𝑀𝑛(𝑂𝐻)' (𝑠) To balance the oxygens, there are 2 oxygens in the reactant, and 2 oxygens and 2 hydrogens on the product. The reactant and product each have a single manganese. The hydrogens are not balanced. To the product side of the reaction, add 2 waters, one for each hydrogen needed. 𝑀𝑛𝑂' (𝑠) + 2 𝐻' 𝑂(𝑙) + 2 𝑒 " → 𝑀𝑛(𝑂𝐻)' (𝑠) To balance the 2 additional oxygens and 2 additional hydrogens, add 2 hydroxides to the opposite side of the reaction, one hydroxide for each extra hydrogen and oxygen. 𝑀𝑛𝑂' (𝑠) + 2 𝐻' 𝑂(𝑙) + 2 𝑒 " → 𝑀𝑛(𝑂𝐻)' (𝑠) + 2 𝑂𝐻" (𝑎𝑞) Now to verify mass balance. There is 1 Mn, 4 O, and 4 H in the reactants. There is 1 Mn, 4 O and 4 H in the products. Verify charge balance: -2 charge in the reactants, and -2 charge in the products, balanced! Answer A! Chapter 19 Practice Exam Questions with Answers, Spring 2024 '! 19-18) For the redox reaction 2 Fe /! + Cl' → 2 Fe 18 " + 2 Cl which of the following are the correct half- reactions? I Cl' + 2 e" → 2 Cl" II Cl → Cl" + e" III Cl' → 2 Cl" + 2 e" IV Fe'! → Fe/! + e" V Fe'! + e" → Fe/! A) I and IV B) I and V C) II and IV D) II and V E) III and IV Assign oxidation numbers to each element in the reactants and products. Elements have 0 as their oxidation state. (Not in compounds) Monoatomic ions, the oxidation state is the charge O is always -2 in compounds, unless a peroxide, -1, or a super oxide, -1/2. H is always +1 in compounds unless bonded to a metal, -1. F is always -1 in compounds. Cl, Br, I is -1 in compounds, unless bonded to a F or O. The sum of the oxidation numbers must equal the overall charge. Pair species that change oxidation number. Determine if an oxidation or a reduction is occurring and add the appropriate number of electrons to account for the electron transfer. 2 𝐹𝑒 '! + 𝐶𝑙' → 2 𝐹𝑒 /! + 2 𝐶𝑙 " Except for chlorine, all species are monatomic ions, so the charge equals the oxidation number. Chlorine is an element in its free standard state with an oxidation number of 0. Iron goes from +2 to +3 and chlorine goes from 0 to -1. 𝐹𝑒 '! → 𝐹𝑒 /! 𝐶𝑙' → 2 𝐶𝑙 " The oxidation number on iron increased, requiring an electron to be released (oxidation). The oxidation number on chlorine decreases, requiring two electrons to be gained. (For Cl2, each chlorine atom is gaining an electron, or 2 total electrons for Cl2.) 𝐹𝑒 '! → 𝐹𝑒 /! + 𝑒 " 𝐶𝑙' + 2 𝑒 " → 2 𝐶𝑙 " I and IV are the correct half reactions. Answer A Chapter 19 Practice Exam Questions with Answers, Spring 2024 19 19-19) Given the following reaction in acidic media: Fe2+ + Cr2O72– → Fe3+ + Cr3+ answer the following question: The coefficient for water in the balanced reaction is: A) 1 B) 3 C) 5 D) 7 E) none of these To balance this reaction, separate the given reaction into half reactions and assign oxidation numbers. 𝐹𝑒 '! → 𝐹𝑒 /! Both species are monoatomic cations. The oxidation number is the charge. Fe2+ must lose 1 electron to give Fe3+. This is oxidation. 𝐹𝑒 '! → 𝐹𝑒 /! + 1 𝑒 " The second half reaction must be a reduction. Oxidation cannot occur without a reduction. Assign oxidation numbers to determine how many electrons are gained. 𝐶𝑟' 𝑂5'" → 𝐶𝑟 /! The oxidation number on Cr in dichromate is (-2*7=-14+2x=-2, x=6) +6 per chromium. The oxidation state for chromium(III) is +3. On a per chromium basis, each chromium gains 3 electrons, or a total of 6 electrons considering dichromate has 2 Cr’s. Also balance the chromium. 𝐶𝑟' 𝑂5'" + 6𝑒 " → 2 𝐶𝑟 /! Now balance the oxygen deficiency in the products. The rule for acidic media is add a water for each oxygen needed. Balance the additional hydrogens with H+ on the opposite side. There are 7 oxygens in dichromate. Add 7 H2O to the products. The addition of the 7 H2O now balances the oxygens but added 14 hydrogens. 𝐶𝑟' 𝑂5'" + 6𝑒 " → 2 𝐶𝑟 /! + 7 𝐻' 𝑂 To balance the hydrogens, add 14 H+ to the reactant side of the reaction. 14 𝐻! + 𝐶𝑟' 𝑂5'" + 6𝑒 " → 2 𝐶𝑟 /! + 7 𝐻' 𝑂 Verify balance by adding up the charges. For the reactants there is +14+-2+-6=+6. For the products, there is 6+ only. The reaction is balanced in terms of charge as well as by mass. Multiply each half reaction by the appropriate number so that the number of electrons lost equals the number of electrons again. The oxidation half reaction releases 1 electron. The reduction half reaction gains 6 electrons. Multiply the oxidation half reaction by 6 and add to the reduction half reaction. 6 𝐹𝑒 '! → 6 𝐹𝑒 /! + 6 𝑒 " Chapter 19 Practice Exam Questions with Answers, Spring 2024 ! 14 𝐻 + 𝐶𝑟' 𝑂5'" " + 6𝑒 → 2 𝐶𝑟 /! 20 + 7 𝐻' 𝑂 14 𝐻! + 𝐶𝑟' 𝑂5'" + 6𝑒 " + 6 𝐹𝑒 '! → 2 𝐶𝑟 /! + 7 𝐻' 𝑂 + 6 𝐹𝑒 /! + 6 𝑒 " Cancel out common species, in this case only the electrons, to get the overall balanced equation. 14 𝐻! + 𝐶𝑟' 𝑂5'" + 6 𝐹𝑒 '! → 2 𝐶𝑟 /! + 6 𝐹𝑒 /! + 7 𝐻' 𝑂 Verify balance by adding up charges. For the reactants, +14+-2+12=+24. For the products, +6+18=+24. The reaction is balanced. The coefficient on water is 7, answer D! Chapter 19 Practice Exam Questions with Answers, Spring 2024 21 19-20) The following reaction occurs in basic solution: F 2 + H2 O → O 2 + F – When the equation is balanced, the sum of the coefficients is: A) 10 B) 11 C) 12 D) 13 E) None of these First separate the reaction into half reactions and then assign oxidation numbers. 𝐹' → 𝐹 " 𝐻' 𝑂 → 𝑂' The oxidation number for fluorine is 0, and -1 for fluoride, a gain of 1 electron per F. Add 2 e- to the reactants, and balance fluorine. 𝐹' + 2 𝑒 " → 2 𝐹 " The oxidation numbers for water are +1 for each H, and -2 for each oxygen. For molecular oxygen in the products, the oxidation number is 0. Oxygen loses 2 electrons per oxygen, or a total of 4 electrons when balancing oxygen. 2 𝐻' 𝑂 → 𝑂' + 4 𝑒 " After balancing the oxygen to consider the total number of electrons that must be transferred to give molecular oxygen, the hydrogens must be balanced. For basic solution, for each hydrogen needed add a water, and a hydroxide to the opposite side. Water supplies 2 hydrogens and an O. The extra hydrogen and oxygen are then balanced with a hydroxide. 4 𝑂𝐻" + 2 𝐻' 𝑂 → 𝑂' + 4 𝑒 " + 4 𝐻' 𝑂 There are now 8 hydrogens and 6 oxygens in the reactants, and 8 hydrogens and 6 oxygens in the products. The sum of the charges is -4 in the reactants and -4 in the products. Balanced! Before adding the half reactions together, the reduction half reaction, 𝐹' + 2 𝑒 " → 2 𝐹 " , must be multiplied by 2 so the electrons gained equals the electrons lost in the oxidation half reaction, 4 𝑂𝐻" + 2 𝐻' 𝑂 → 𝑂' + 4 𝑒 " + 4 𝐻' 𝑂. 2 𝐹' + 4 𝑒 " → 4 𝐹 " 4 𝑂𝐻" + 2 𝐻' 𝑂 → 𝑂' + 4 𝑒 " + 4 𝐻' 𝑂 4 𝑂𝐻" + 2 𝐹' → 𝑂' + 4 𝐹 " + 2 𝐻' 𝑂 The sum of the coefficients is 4+2+1+4+2=13 Answer D! Chapter 19 Practice Exam Questions with Answers, Spring 2024 22 19-21) How many electrons are transferred in the following reaction when it is balanced in acidic solution? SO32–(aq) + MnO4–(aq) → SO42–(aq) + Mn2+(aq) A) 6 B) 2 C) 10 D) 5 E) 3 Break the reactants and products into half reactions. 𝑆𝑂/'" (𝑎𝑞) → 𝑆𝑂0'" (𝑎𝑞) 𝑀𝑛𝑂0'" (𝑎𝑞) → 𝑀𝑛'! (𝑎𝑞) Calculate the oxidation numbers and whether the difference in oxidation number is an oxidation or reduction. The sulfur in sulfite is +4. The sulfur in sulfate is +6. Sulfite is being oxidized to sulfate, losing 2 electrons. The manganese in permanganate is +7. The oxidation on manganese(II) is +2. Permanganate is being reduced to manganese(II) by gaining 5 electrons. 𝑆𝑂/'" (𝑎𝑞) → 𝑆𝑂0'" (𝑎𝑞) + 2𝑒 " 𝑀𝑛𝑂0" (𝑎𝑞) + 5𝑒 " → 𝑀𝑛'! (𝑎𝑞) Since the question is asking for the number of electrons being transferred, the oxygens and hydrogens do not need balancing. Sulfite loses 2 electrons, permanganate gains 5 electrons. Multiply sulfite half reaction by 5 and permanganate half reaction by 2 to get 10 electrons transferred. Answer C! As practice, lets completely balance the reaction. 𝑆𝑂/'" (𝑎𝑞) → 𝑆𝑂0'" (𝑎𝑞) + 2𝑒 " Sulfate has 1 more O than sulfite, and the reaction is in acid, add a water to the reactant side of the reaction, and 2 H+ to the product side to balance the two H in water. 𝐻' 𝑂(𝑙) + 𝑆𝑂/'" (𝑎𝑞) → 𝑆𝑂0'" (𝑎𝑞) + 2𝑒 " + 2 𝐻! (𝑎𝑞) Check the charge balance, 2- for reactants, and 2- for products. 𝑀𝑛𝑂0" (𝑎𝑞) + 5𝑒 " → 𝑀𝑛'! (𝑎𝑞) Permanganate has 4 O’s, and Manganese(II) has none. Add 4 waters to the products. To balance the added 10 hydrogens, add 10 H+ to the reactants. 8 𝐻! (𝑎𝑞) + 𝑀𝑛𝑂0" (𝑎𝑞) + 5𝑒 " → 𝑀𝑛'! (𝑎𝑞) + 4 𝐻' 𝑂(𝑙) Check the charge balance. 2+ in the reactants, and 2+ in the products. Multiply the sulfite half reaction by 5 and the permanganate half reaction by 2 and add. Chapter 19 Practice Exam Questions with Answers, Spring 2024 5 𝐻' 𝑂(𝑙) + 5 𝑆𝑂/'" (𝑎𝑞) →5 𝑆𝑂0'" (𝑎𝑞) " ! + 10 𝑒 + 10 𝐻 (𝑎𝑞) 16 𝐻! (𝑎𝑞) + 2 𝑀𝑛𝑂0" (𝑎𝑞) + 10 𝑒 " → 2 𝑀𝑛'! (𝑎𝑞) + 8 𝐻' 𝑂(𝑙) 5 𝐻' 𝑂(𝑙) + 5 𝑆𝑂/'" (𝑎𝑞) + 16 𝐻! (𝑎𝑞) + 2 𝑀𝑛𝑂0" (𝑎𝑞) + 10 𝑒 " → 2 𝑀𝑛'! (𝑎𝑞) + 8 𝐻' 𝑂(𝑙) + 5 𝑆𝑂0'" (𝑎𝑞) + 10 𝑒 " + 10 𝐻! (𝑎𝑞) 6 𝐻! (𝑎𝑞) + 2 𝑀𝑛𝑂0" (𝑎𝑞) + 5 𝑆𝑂/'" (𝑎𝑞) → 2 𝑀𝑛'! (𝑎𝑞) + 5 𝑆𝑂0'" (𝑎𝑞) + 3 𝐻' 𝑂(𝑙) Check charge balance: +6+-2+-10=-6 for reactants, and +4+-10=-6 for the products. Chapter 19 Practice Exam Questions with Answers, Spring 2024 23 24 19-22) __________ is the oxidizing agent in the reaction below. Cr2O72- + 6S2O32- + 14H+ → 2Cr3+ + 3S4O62- + 7H2O A) Cr2O72- B) S2O32- C) H+ D) Cr3+ E) S4O62- The reducing agent is the oxidized species, and the oxidizing agent is the reduced species. The oxidizing and reducing agents are always reactants. Oxidation cannot occur without a reduction and vice versa. The reactant being reduced gains the electrons from the oxidizing species, and the reactant being oxidized loses the electrons gained by the reducing species. For the reaction between dichromate and thiosulfate, whether dichromate or thiosulfate is gaining or losing electrons must be determined before assigning as oxidizing or reducing agents. The oxidation number on each Cr atom in dichromate is +6. The total oxidation provided by the 7 O atoms is -14. The oxidation number on each chromium must add to -14 to give the overall charge if -1. +12/2=+6 on each chromium. The oxidation number for chromium(III) in the products is +3, oxidation equals charge for monoatomic ions. The chromium half reaction is 𝐶𝑟' 𝑂5'" + 6 𝑒 " → 2 𝐶𝑟 /! which is a reduction. This means dichromate is the oxidizing agent. By the process of elimination, thiosulfate is oxidized and the reduction half reaction. Answer A! For completeness’ sake, lets balance the reaction. 𝐶𝑟' 𝑂5'" + 6 𝑒 " → 2 𝐶𝑟 /! To balance the oxygens, add 7 waters to the products, and to balance the additional hydrogens, add 14 hydroniums to the reactants. 14 𝐻! + 𝐶𝑟' 𝑂5'" + 6 𝑒 " → 2 𝐶𝑟 /! + 7 𝐻' 𝑂(𝑙) 2 𝑆' 𝑂/'" (𝑎𝑞) → 𝑆0 𝑂D'" The oxidation on the sulfur in thiosulfate is +2 and the oxidation on the sulfur in S4O62- is 0 on two sulfurs and +5 on two sulfurs. (You will not have a question where an element has mixed oxidation. I had to google it.) The total oxidation on sulfur in thiosulfate is +4. To balance the S and O in the half reaction, two thiosulfates are required, which gives a total oxidation of +8. The total oxidation on S in S4O62- is +10. This means thiosulfate loses a total of 2 electrons. 2 𝑆' 𝑂/'" (𝑎𝑞) → 𝑆0 𝑂D'" + 2 𝑒 " To balance the reaction, the thiosulfate half reaction must be multiplied by 3, so 6 electrons are being released for the 6 electrons gained by dichromate. 14 𝐻! + 𝐶𝑟' 𝑂5'" + 6 𝑒 " → 2 𝐶𝑟 /! + 7 𝐻' 𝑂(𝑙) Chapter 19 Practice Exam Questions with Answers, Spring 2024 6 𝑆' 𝑂/'" (𝑎𝑞) →3 𝑆0 𝑂D'" +6𝑒 " 25 Add the half reactions together and cancel out common species, ie electrons, water, hydronium, hydroxide, etc. 14 𝐻! + 𝐶𝑟' 𝑂5'" + 6 𝑒 " + 6 𝑆' 𝑂/'" (𝑎𝑞) → 2 𝐶𝑟 /! + 3 𝑆0 𝑂D'" + 6 𝑒 " + 7 𝐻' 𝑂(𝑙) Cancel out the electrons gives the balanced equation. 14 𝐻! + 𝐶𝑟' 𝑂5'" + 6 𝑆' 𝑂/'" (𝑎𝑞) → 2 𝐶𝑟 /! + 3 𝑆0 𝑂D'" + 7 𝐻' 𝑂(𝑙) To check the balancing, add up the charges on both sides of the reaction. Reactants 14*1+2-+6*2- = 0 Products 2*3+3*2-=0 Chapter 19 Practice Exam Questions with Answers, Spring 2024 26 19-23) What is the oxidation number on potassium in KMnO4? A) 0 B) +1 C) +2 D) -1 E) +3 Assign oxidation numbers to each element in the reactants and products. Elements have 0 as their oxidation state. (Not in compounds) Monoatomic ions, the oxidation state is the charge O is always -2 in compounds, unless a peroxide, -1, or a super oxide, -1/2. H is always +1 in compounds unless bonded to a metal, -1. F is always -1 in compounds. Cl, Br, and I are -1 in compounds, unless bonded to a F or O. The sum of the oxidation numbers must equal the overall charge. KMnO4 is an ionic compound. To properly assign oxidation numbers to ionic compounds, separate into the cation and anion. Do Not break the ionic compound into K, Mn and O. K+ is a monoatomic cation, its charge is its oxidation number. MnO4- is a polyatomic ion and the oxidation on each element sums to give the overall charge. The question asks for the oxidation on K, so we can stop here. Answer B! However, if the question asked for Mn or O, we could continue to assign oxidation numbers. There is no oxidation number rule for Mn. The rules to use are the oxidation rule for oxygen, O is 2-, and the sum of the oxidation numbers equals the overall charge rule If Mn is x, O is -2 and the overall charge is -1. 𝑥 + 4 × −2 = −1 𝑥 + −8 = −1 𝑥 = +7 Chapter 19 Practice Exam Questions with Answers, Spring 2024 3+ 27 2+ 19-24) The reduction potentials for Au and Ni are as follows: Au/! + 3 e" → Au, Ni'! + 2 e" → Ni, ? EKLM = +1.50 V ? EKLM = −0.227 V Calculate DGo (at 25 oC) for the reaction: 2 Au/! + 3 Ni → Au + 3 Ni'! A) 1.00 x 103 kJ B) -7.37 x 103 kJ C) 7.37 x 102 kJ D) -1.67 x 102 kJ E). -1.00 x 103 kJ The relationship between DG and Eo is ∆𝐺 & = −𝑛𝐹𝐸 &. To solve the problem the standard cell potential must be calculated, and the number of electrons transferred. The chemical equation shows Au3+ being reduced, and Ni is oxidized. The gold half reaction occurs at the cathode, and the nickel half reaction occurs at the anode. The cell potential is 𝐸)$** = 𝐸)+,-&%$ − 𝐸+.&%$. 𝐸)$** = 1.50 𝑉 − −0.227 𝑉 = 1.73 𝑉 Au3+ gains 3 electrons, Ni2+ gains 2 electrons. The total number of electrons transferred is 6. 96485 𝐹 ∆𝐺 & = −(6 𝑚𝑜𝑙𝑒𝑠 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠) W Y (1.73 𝑉) = −1001514.3 𝑜𝑟 − 1.00 × 10/ 𝑘𝐽 𝑚𝑜𝑙𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 Answer E 19-25) For a certain reaction, ΔH° = –77.8 kJ and ΔS° = –228 J/K. If n = 2, calculate E° for the reaction at 25°C. A) 0.0511 V B) 0.755 V C) 0.433 V D) 0.102 V E) 0.374 V This question is like the previous question, but instead of calculating DG, E is calculated instead. To calculate E, the DGo is calculated using the DHo and DSo provided. ∆𝐺 & = ∆𝐻& − 𝑇∆𝑆 = ^(−77.8 𝑘𝐽) W 1000 𝐽 𝐽 Y_ − ^(298.15 𝐾) W−228 Y_ = −9821.8 𝐽 1 𝑘𝐽 𝐾 Now calculate the standard cell potential using the formula ∆𝐺 & = −𝑛𝐹𝐸 &. The number of electrons transferred must be known and is provided as part of the question. 𝐸& = −∆𝐺 & −(−9821.8 𝐽) = = 0.0510 𝑉 96485 𝐶 𝑛𝐹 (2 𝑚𝑜𝑙𝑒𝑠 𝑒 " ) B 𝑚𝑜𝑙 𝑒 " C Answer A Chapter 19 Practice Exam Questions with Answers, Spring 2024 2+ 28 2+ 19-26) The reduction potentials for Ni and Sn are ? Ni'! (aq) + 2 e" → Ni(s), EKLM = −0.232 V ? Sn'! (aq) + 2 e" → Sn(s), EKLM = −0.140 V Calculate the equilibrium constant at 25 °C for the reaction: Sn'! + Ni ⇌ Sn + Ni'! A) 3.6 x 1012 B) 35 C) 5.9 D) 8.3 x 10-4 E) 1.2 x 103 The relationship between Eo and K is derived from DGo = -RTlnK and DGo=-nFEo. 𝑅𝑇𝑙𝑛𝐾 = 𝑛𝐹𝐸 & 𝑛𝐹𝐸 & 𝑙𝑛𝐾 = 𝑅𝑇 Before calculating K, Eo must be calculated for the cell and the number of moles electrons transferred. Sn2+ is reduced (cathode) and the Ni is oxidized (anode), Eocell = Ecathode – Eanode= -0.140 V - -0.232 V = 0.092 V. The n of electrons transferred is 2 moles. 96485 𝐶 " 𝑛𝐹𝐸 & (2 𝑚𝑜𝑙 𝑒 ) B 𝑚𝑜𝑙 𝑒 " C (0.092 𝑉) 𝑙𝑛𝐾 = = = 7.166 𝐽 𝑅𝑇 (298 𝐾) B8.314 𝑚𝑜𝑙 𝐾 C 𝑒 *.N = 𝑒 5.=DD 𝐾 = 1.2 × 10/ Answer E! Chapter 19 Practice Exam Questions with Answers, Spring 2024