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Daily Quiz

Daily Quiz. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells:. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells:. 1. Cr in a solution of Cr 3+ ions; Cu in a solution of Cu 2+ ions a. 0.5600 volts

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Daily Quiz

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  1. Daily Quiz

  2. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells:

  3. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 1. Cr in a solution of Cr3+ ions; Cu in a solution of Cu2+ ions a. 0.5600 volts b. 0.7489 volts c. 0.8970 volts d. 1.0859 volts

  4. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 2. zinc in a solution of Zn2+ ions; platinum in a solution of Pt2+ ions a. 0.4182 volts b. 0.7618 volts c. 1.1800 volts d. 1.9418 volts

  5. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 3. HgCl2 and Hg2Cl2; lead in a solution of Pb2+ ions a. 0.031 volts b. 0.398 volts c. 1.000 volts d. 1.046 volts e. 1.977 volts

  6. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 4. tin in a solution of Sn2+ ions; iodine in a solution of I- ions a. 0.3845 volts b. 0.6730 volts c. 0.6865 volts d. 1.0050 volts e. 1.2935 volts .

  7. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 1. Cr in a solution of Cr3+ ions; Cu in a solution of Cu2+ ions a. 0.5600 volts b. 0.7489 volts c. 0.8970 volts d. 1.0859 volts

  8. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 1. Cr in a solution of Cr3+ ions; Cu in a solution of Cu2+ ions Cr3+ + 3e- → Cr -0.913 Cu2+ + 2e- → Cu 0.153 -(-0.913) 1.066 V

  9. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 1. Cr in a solution of Cr3+ ions; Cu in a solution of Cu2+ ions Cr3+ + 3e- → Cr -0.744 Cu2+ + 2e- → Cu 0.3419 -(-0.913) 1.066 V

  10. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 1. Cr in a solution of Cr3+ ions; Cu in a solution of Cu2+ ions Cr3+ + 3e- → Cr -0.744 Cu2+ + 2e- → Cu 0.3419 -(-0.744) 1.066 V

  11. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 1. Cr in a solution of Cr3+ ions; Cu in a solution of Cu2+ ions Cr3+ + 3e- → Cr -0.744 Cu2+ + 2e- → Cu 0.3419 -(-0.744) 1.0859 V

  12. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 1. Cr in a solution of Cr3+ ions; Cu in a solution of Cu2+ ions a. 0.5600 volts b. 0.7489 volts c. 0.8970 volts d. 1.0859 volts

  13. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 2. zinc in a solution of Zn2+ ions; platinum in a solution of Pt2+ ions a. 0.4182 volts b. 0.7618 volts c. 1.1800 volts d. 1.9418 volts

  14. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 2. zinc in a solution of Zn2+ ions; platinum in a solution of Pt2+ ions Zn2+ + 2e- → Zn -0.7618 Pt2+ + 2e- → Pt 1.18 -(-0.7618) 1.94 V

  15. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 2. zinc in a solution of Zn2+ ions; platinum in a solution of Pt2+ ions Zn2+ + 2e- → Zn -0.7618 Pt2+ + 2e- → Pt 1.18 -(-0.7618) 1.94 V

  16. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 2. zinc in a solution of Zn2+ ions; platinum in a solution of Pt2+ ions Zn2+ + 2e- → Zn -0.7618 Pt2+ + 2e- → Pt 1.18 -(-0.7618) 1.94 V

  17. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 2. zinc in a solution of Zn2+ ions; platinum in a solution of Pt2+ ions Zn2+ + 2e- → Zn -0.7618 Pt2+ + 2e- → Pt 1.18 -(-0.7618) 1.94 V

  18. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 2. zinc in a solution of Zn2+ ions; platinum in a solution of Pt2+ ions a. 0.4182 volts b. 0.7618 volts c. 1.1800 volts d. 1.9418 volts

  19. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 3. HgCl2 and Hg2Cl2; lead in a solution of Pb2+ ions a. 0.031 volts b. 0.398 volts c. 1.000 volts d. 1.046 volts e. 1.977 volts

  20. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 3. HgCl2 and Hg2Cl2; lead in a solution of Pb2+ ions 2Hg2+ + 2e- → 2Hg22+ 0.920 Pb2+ + 2e- → Pb -(-0.1262) 1.046 V

  21. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 3. HgCl2 and Hg2Cl2; lead in a solution of Pb2+ ions 2Hg2+ + 2e- → 2Hg22+ 0.920 Pb2+ + 2e- → Pb -(-0.1262) 1.046 V

  22. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 3. HgCl2 and Hg2Cl2; lead in a solution of Pb2+ ions 2Hg2+ + 2e- → 2Hg22+ 0.920 Pb2+ + 2e- → Pb -(-0.1262) 1.046 V

  23. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 3. HgCl2 and Hg2Cl2; lead in a solution of Pb2+ ions 2Hg2+ + 2e- → 2Hg22+ 0.920 Pb2+ + 2e- → Pb -(-0.1262) 1.046 V

  24. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 3. HgCl2 and Hg2Cl2; lead in a solution of Pb2+ ions a. 0.031 volts b. 0.398 volts c. 1.000 volts d. 1.046 volts e. 1.977 volts

  25. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 4. tin in a solution of Sn2+ ions; iodine in a solution of I- ions a. 0.3845 volts b. 0.6730 volts c. 0.6865 volts d. 1.0050 volts e. 1.2935 volts

  26. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 4. tin in a solution of Sn2+ ions; iodine in a solution of I- ions Sn2+ + 2e- → Sn-0.1375 I2+ 2e- → 2I- 0.5355 -(-0.1375) 0.6730 V

  27. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 4. tin in a solution of Sn2+ ions; iodine in a solution of I- ions Sn2+ + 2e- → Sn-0.1375 I2+ 2e- → 2I- 0.5355 -(-0.1375) 0.6730 V

  28. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 4. tin in a solution of Sn2+ ions; iodine in a solution of I- ions Sn2+ + 2e- → Sn-0.1375 I2+ 2e- → 2I- 0.5355 -(-0.1375) 0.6730 V

  29. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 4. tin in a solution of Sn2+ ions; iodine in a solution of I- ions Sn2+ + 2e- → Sn-0.1375 I2+ 2e- → 2I- 0.5355 -(-0.1375) 0.6730 V

  30. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 4. tin in a solution of Sn2+ ions; iodine in a solution of I- ions a. 0.3845 volts b. 0.6730 volts c. 0.6865 volts d. 1.0050 volts e. 1.2935 volts .

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