U6 Hw1

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lowe (mcl2357) – U6 HW1 – stokes – (CeHS-S13) This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points These two reactions are combined in a redox reaction (I) (II) Cu → Cu2+ Cr2 O2− → Cr3+ 7 1. one proton to the right 2. one electron to the right 3. one proton to the left 4. one electron to the left 5. two electrons to the left 6. two protons to the right 7. two protons to the left 8. two electrons to the right correct Explanation: The balanced half reaction is 2 I− → I2 + 2 e−

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Choose the correct statement below. Recall that the oxidation state of oxygen is usually −2. 1. The Cr is reduced from +6 to +3; the Cu is oxidized from 0 to +2. correct 2. The Cr is oxidized from −6 to +3; the Cu is reduced from 0 to −2. 3. The Cr is oxidized from −6 to −3; the Cu is reduced from 0 to −2. 4. The Cr is reduced from +12 to +3; the Cu is oxidized from 0 to +2. 5. The Cr is reduced from +7 to +3; the Cu is oxidized from 0 to +2. Explanation: The net charge in a molecule is equal to the sum of the oxidation numbers of the atoms within that molecule. 2− The oxidation number of Cr in Cr2 O7 is +6 since each O is −2 and the net charge on the molecule is −2. The oxidation number of Cr in Cr3+ is equal to the charge on the ion, 3+. Since the oxidation number is getting smaller from +6 to +3, this is a reduction. The oxidation number of Cu(s) is zero. Since the oxidation number is increasing from 0 to +2, Cu is being oxidized. 002 10.0 points To correctly balance the half reaction 2 I → I2 , −

003 10.0 points To correctly balance the half reaction Pu4+ → Pu3+ , what would you need to add? 1. one electron to the right 2. one proton to the left 3. one proton to the right 4. four protons to the left and three protons to the right 5. one electron to the left correct 6. three protons to the left and four protons to the right 7. four electrons to the left and three electrons to the right 8. three electrons to the left and four electrons to the right Explanation:

what would you need to add?

lowe (mcl2357) – U6 HW1 – stokes – (CeHS-S13) The balanced half reaction is Pu4+ + e− → Pu3+ , 004 10.0 points Given the unbalanced equation Ca(s) + Ag+ (aq) → Ca2+ (aq) + Ag(s) , for a reaction, what is being reduced? 1. calcium 2. silver correct 3. This is not an oxidation-reduction reaction. Explanation: 005 10.0 points What is the oxidation number of selenium, Se, in SeO2− ? 4 1. +8 2. +2 3. +6 correct 4. +4 Explanation: Let the oxidation number for Se = x x + 4(−2) = −2 x = +6 006 10.0 points What is the oxidation number of the Fe in FeO? 1. +3 Mn2 O7 : 2. +2 correct 3. +1 4. −3 MnO2 :

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Explanation: The sum of the numbers must equal the total charge on the molecule or in the case of neutral species, zero. Start with the oxygen: Fex + O2− = 0 x−2 =0 x=2 So the oxidation state of the iron is +2. 007 10.0 points Manganese (Mn) exhibits a larger number of oxidation numbers in its compounds than any other Period 4 transition metal. The most important oxidation numbers of manganese are illustrated by the oxides MnO, MnO2 , and Mn2 O7 . What oxidation numbers of manganese are illustrated here? 1. +2, +4, and +1 2. +2 and +7 3. +2; +4, and +7 correct 4. −2, −4, and −7 5. +2, +3, and +9 Explanation: To calculate oxidation numbers, remember that the sum of the numbers must equal the total charge on the molecule or in the case of neutral species, zero. Start with the oxygen: MnO : Mnx + O2− = 0 x−2=0 x=2 Mnx + 2(O2− ) = 0 x + 2(−2) = 0 x−4=0 x=4 2(Mnx ) + 7(O2− ) = 0 2x + 7(−2) = 0 2x − 14 = 0 2x = 14 x=7

lowe (mcl2357) – U6 HW1 – stokes – (CeHS-S13) 008 10.0 points The oxidation numbers of nitrogen in NH3 , NO− and NO are 3 1. −3, +5, and +2, respectively. correct 2. +3, +6, and +2, respectively. 1. −2, −1, and −6, respectively. 3. −3, +5, and +1, respectively. 2. −3, −2, and −8, respectively. 4. −3, +6, and +2, respectively. Explanation:...
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