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III. Solve all the following problems.1. Which of the following reactions are spontaneous at standard state? (Consult table ofelectrode potentials.) A) I2 + 2 Fe2+ 2 I + 2 Fe3+ B) 8 H+ + 2 NO3 + 3 Cu(s) 2 NO(g) + 3 Cu2++ 4 H2O C) 4 MnO4 + 12 H+ 4 Mn2+ + 5 O2 + 6 H2O D) Cu(s) + Cl2(g) Cu2+(aq) + 2 Cl(aq)2. A standard electrode X when coupled with a standard H2 electrode to make a cell gives a cellvoltage of 0.34. What is the potential of the standard electrode X?3. How much time is required to plate a 0.01 mm thick layer of Cr on an object whose area is200 cm2 if the current used is 1.00 A and the density of Cr is 7.1 g/cm3?4. The same quantity of electricity that caused 10.0 g of Ag to plate out from a solution ofAgNO3 is passed through a solution of CuSO4. How much Cu will plate out?5. A current of 0.01 A is passed through a solution of iridium bromide. The only reaction at thecathode is the deposition of Ir. After 3.00 hr, 0.072 g of Ir has been deposited. What is thecharge on the iridium ion?6. A. Calculate the potential of the cell where the reaction: 2 HClO2 HClO + ClO3 + H+ occursif: [HClO2] = 0.01 M; [HClO] = 0.10 M; [ClO3] = 0.10 M and pH = 5.0 B. Taking all species to be1 M except HClO2, calculate the concentration of HClO2 at which the reaction would no longerbe spontaneous.7. Show that the oxidation of Cl to Cl2(g) by Cr2O72 in acidic solution will NOT occur withreactants and products in their standard states. If, however, this method is to be used for theproduction of Cl2 in the laboratory, how would you go about it?8. Design a cell that will enable you to measure the Ksp of PbCl2.9. A compound contains 21.35 % Cr, 28.70 % N, 6.209 % H and 43.68 % Cl by mass. (a) Itdoes not react with HCl. (b) On reaction with AgNO3, it gives 2 moles of AgCl per mole of thecompound. (c) It has an electrical conductivity corresponding to 3 moles of ions per mol ofcompound. A) Give the inferences of (a), (b) and (c). B) Write the formula and give the name ofthe compound.10. Explain why [Fe(CN)6]3 is poisonous while [Fe(CN)6]4 is not.11. Describe how the diamagnetic behavior of [Co(en)3]3+ is explained by: A) Valence BondTheory B) Crystal–Field Theory12. A certain coordination compound has the simplest formula PtN2H6Cl2. It has a molar massof about 600 g and contains both a complex cation and a complex anion. What is its structure?13. When Cr2+ is linked with a strong–field ligand to produce an octahedral complex, the ionhas two unpaired electrons. When it is linked with a weak–field ligand, the ion has four unpairedelectrons. In contrast, the complex resulting from Cr3+ always has three unpaired electronsregardless of the nature of the ligand. Explain.14. Draw a structure to represent the complex ion, [Co(en)3]3+.15. When 6.0 M NaOH solution is added to a solution containing equimolar concentrations ofCd(NO3)2 and Zn(NO3)2, the precipitate that first forms contains both Zn and Cd. However,upon the addition of an additional base with stirring, part of the precipitate dissolves. Theundissolved precipitate contains Cd but no Zn. Describe what happened in terms of net ionicequations.

Question

III. Solve all the following problems.1. Which of the following reactions are spontaneous at standard state? (Consult table ofelectrode potentials.) A) I2 + 2 Fe2+ 2 I + 2 Fe3+ B) 8 H+ + 2 NO3 + 3 Cu(s) 2 NO(g) + 3 Cu2++ 4 H2O C) 4 MnO4 + 12 H+ 4 Mn2+ + 5 O2 + 6 H2O D) Cu(s) + Cl2(g) Cu2+(aq) + 2 Cl(aq)2. A standard electrode X when coupled with a standard H2 electrode to make a cell gives a cellvoltage of 0.34. What is the potential of the standard electrode X?3. How much time is required to plate a 0.01 mm thick layer of Cr on an object whose area is200 cm2 if the current used is 1.00 A and the density of Cr is 7.1 g/cm3?4. The same quantity of electricity that caused 10.0 g of Ag to plate out from a solution ofAgNO3 is passed through a solution of CuSO4. How much Cu will plate out?5. A current of 0.01 A is passed through a solution of iridium bromide. The only reaction at thecathode is the deposition of Ir. After 3.00 hr, 0.072 g of Ir has been deposited. What is thecharge on the iridium ion?6. A. Calculate the potential of the cell where the reaction: 2 HClO2 HClO + ClO3 + H+ occursif: [HClO2] = 0.01 M; [HClO] = 0.10 M; [ClO3] = 0.10 M and pH = 5.0 B. Taking all species to be1 M except HClO2, calculate the concentration of HClO2 at which the reaction would no longerbe spontaneous.7. Show that the oxidation of Cl to Cl2(g) by Cr2O72 in acidic solution will NOT occur withreactants and products in their standard states. If, however, this method is to be used for theproduction of Cl2 in the laboratory, how would you go about it?8. Design a cell that will enable you to measure the Ksp of PbCl2.9. A compound contains 21.35 % Cr, 28.70 % N, 6.209 % H and 43.68 % Cl by mass. (a) Itdoes not react with HCl. (b) On reaction with AgNO3, it gives 2 moles of AgCl per mole of thecompound. (c) It has an electrical conductivity corresponding to 3 moles of ions per mol ofcompound. A) Give the inferences of (a), (b) and (c). B) Write the formula and give the name ofthe compound.10. Explain why [Fe(CN)6]3 is poisonous while [Fe(CN)6]4 is not.11. Describe how the diamagnetic behavior of [Co(en)3]3+ is explained by: A) Valence BondTheory B) Crystal–Field Theory12. A certain coordination compound has the simplest formula PtN2H6Cl2. It has a molar massof about 600 g and contains both a complex cation and a complex anion. What is its structure?13. When Cr2+ is linked with a strong–field ligand to produce an octahedral complex, the ionhas two unpaired electrons. When it is linked with a weak–field ligand, the ion has four unpairedelectrons. In contrast, the complex resulting from Cr3+ always has three unpaired electronsregardless of the nature of the ligand. Explain.14. Draw a structure to represent the complex ion, [Co(en)3]3+.15. When 6.0 M NaOH solution is added to a solution containing equimolar concentrations ofCd(NO3)2 and Zn(NO3)2, the precipitate that first forms contains both Zn and Cd. However,upon the addition of an additional base with stirring, part of the precipitate dissolves. Theundissolved precipitate contains Cd but no Zn. Describe what happened in terms of net ionicequations.

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Solution

  1. To determine which of the given reactions are spontaneous at standard state, we need to consult the table of electrode potentials. By comparing the standard reduction potentials of the species involved in each reaction, we can determine if the reaction will occur spontaneously or not.

  2. The potential of the standard electrode X can be calculated by subtracting the potential of the standard H2 electrode from the cell voltage of 0.34 V.

  3. To calculate the time required to plate a 0.01 mm thick layer of Cr on an object with an area of 200 cm2, we need to use Faraday's law of electrolysis. By knowing the current (1.00 A), the density of Cr (7.1 g/cm3), and the thickness of the layer, we can calculate the time required.

  4. To determine how much Cu will plate out when the same quantity of electricity that caused 10.0 g of Ag to plate out is passed through a solution of CuSO4, we need to use Faraday's law of electrolysis. By knowing the charge of Ag and the ratio of the charges of Ag and Cu, we can calculate the mass of Cu plated out.

  5. The charge on the iridium ion can be calculated by using Faraday's law of electrolysis. By knowing the current, the time, and the mass of Ir deposited, we can calculate the charge.

  6. A. To calculate the potential of the cell where the given reaction occurs, we need to use the Nernst equation. By plugging in the concentrations of the species involved and the pH value, we can calculate the cell potential. B. To determine the concentration of HClO2 at which the reaction would no longer be spontaneous, we need to calculate the cell potential using the Nernst equation and find the concentration at which the potential becomes zero.

  7. To show that the oxidation of Cl to Cl2(g) by Cr2O72- in acidic solution will not occur with reactants and products in their standard states, we need to compare the standard reduction potentials of the species involved. If the reaction is not spontaneous at standard state, we need to find an alternative method for the production of Cl2 in the laboratory.

  8. To design a cell that will enable us to measure the Ksp of PbCl2, we need to construct a cell with a Pb electrode and a reference electrode. By measuring the cell potential and using the Nernst equation, we can calculate the Ksp value.

  9. A. By analyzing the composition and the reactions of the compound, we can infer that it does not react with HCl, it forms 2 moles of AgCl per mole of the compound when reacted with AgNO3, and it has an electrical conductivity corresponding to 3 moles of ions per mole of the compound. B. Based on the composition and the inferences, we can write the formula and give the name of the compound.

  10. The toxicity of [Fe(CN)6]3- compared to [Fe(CN)6]4- can be explained by the ability of [Fe(CN)6]3- to release CN- ions, which can bind to cytochrome oxidase and inhibit cellular respiration.

  11. A. The diamagnetic behavior of [Co(en)3]3+ can be explained by Valence Bond Theory, which states that the complex has a low-spin configuration with all the electrons paired. B. Crystal-Field Theory can also explain the diamagnetic behavior by considering the splitting of d orbitals in the presence of ligands.

  12. The structure of the coordination compound PtN2H6Cl2 can be determined by analyzing its formula and molar mass.

  13. The change in the number of unpaired electrons in Cr2+ and Cr3+ complexes when linked with different ligands can be explained by the Crystal-Field Theory. The strong-field ligands cause a greater splitting of the d orbitals, resulting in fewer unpaired electrons, while the weak-field ligands cause less splitting and more unpaired electrons.

  14. The structure of the complex ion [Co(en)3]3+ can be represented by drawing a central cobalt ion surrounded by three ethylenediamine ligands.

  15. The addition of 6.0 M NaOH solution to a solution containing equimolar concentrations of Cd(NO3)2 and Zn(NO3)2 results in the formation of a precipitate containing both Zn and Cd. However, upon the addition of an additional base with stirring, part of the precipitate dissolves, leaving behind a precipitate containing only Cd. This can be explained by net ionic equations involving the formation and dissolution of the precipitates.

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