Unformatted text preview: ion state, so: 1(x)+ 1(‐2)=0 x‐2=0 x=+2 *Therefore (D) is the lowest oxidation state. 6. Na2Cr2O7: Sodium has an oxidation number of +1 because it is an alkali metal. Oxygen is not in peroxide form, so its oxidation state is ‐2. For chromium then: 2(+1)+2(x)+7(‐2)=0 2+2x‐14=0 2x=12 x=+6 7. Aluminum is in group 3A, which is always +3 in oxidation state. Magnesium is in group 2A, which is always +2 in oxidation state. Manganese is a transition metal that has many transition states. Therefore, (C) is the correct answer. Mercury is also a transition metal, but contains only 3 oxidation states: Hg(0), Hg2+(+2) and Hg22+(+1). 8. Cu2+Cu *Copper goes from a +2 oxidation state to one of zero. It therefore gains electrons and is reduced (answer choice (C) is correct). 9. To require an oxidizing agent, something must be oxidized. Look for the answer choice in which an element is oxidized: (A) Sulfur goes from a +2 to a +2.5 oxidation state. Since the oxidation number increases, sulfur is oxidized. Thus (A) is correct. (B) Zinc goes from a +2 to a 0 oxidation state and is therefore reduced. (C) Chlorine goes from a +1 to a ‐1 oxidation state and is therefore reduced. (D) Sulfur goes from a +6 to a +6 oxidation state (no change). 10. ZnZn2+: Metallic zinc is therefore oxidized (loses electrons), and is thus the reducing agent (A). 11. Remember, the oxidizing agent is reduced: (A) PbPbSO4: Lead goes from 0+2 (Pb is oxidized). (B) PbSO4 is on the products side and is neither an oxidizing nor a reducing agent. Dr. Fus Electrochemistry Problem Set Solutions Chemistry 123 (C) PbO2PbSO4: Lead goes from +4+2 (lead in PbO2 is reduced, PbO2 is an oxidizing agent) (D) Neither sulfur, oxygen, nor hydrogen changes oxidation state in this reaction. 12. (D)‐this is the only answer that explains how a single compound can gain or lose electrons depending on the conditions. 13. The cell potential, E°, will be positive if the reaction occurs spontaneously. Potentials are written for reductions, and the more positive the value, the more that atom/ion wants to be reduced, and therefore oxidize other species. The more negative the E°, the more that atom/ion wants to be oxidized, and therefore reduce other species. According to the given values of E°, Sn4+ and Br2 are both reduced favo...
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- Spring '06
- Electrochemistry, Oxidation state, Dr. Fus, Electrochemistry Problem Set Solutions, mol e‐