1. Daily Quiz

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

3. 1.Cr in a solution of Cr 3+ ions; Cu in a solution of Cu 2+ 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 Zn 2+ ions; platinum in a solution of Pt 2+ 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. HgCl 2 and Hg 2 Cl 2 ; lead in a solution of Pb 2+ 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 Sn 2+ 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 Cr 3+ ions; Cu in a solution of Cu 2+ 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 Cr 3+ ions; Cu in a solution of Cu 2+ ions Cr 3+ + 3e - → Cr-0.913 Cu 2+ + 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 Cr 3+ ions; Cu in a solution of Cu 2+ ions Cr 3+ + 3e - → Cr-0.744 Cu 2+ + 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 Cr 3+ ions; Cu in a solution of Cu 2+ ions Cr 3+ + 3e - → Cr-0.744 Cu 2+ + 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 Cr 3+ ions; Cu in a solution of Cu 2+ ions Cr 3+ + 3e - → Cr-0.744 Cu 2+ + 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 Cr 3+ ions; Cu in a solution of Cu 2+ 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 Zn 2+ ions; platinum in a solution of Pt 2+ 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 Zn 2+ ions; platinum in a solution of Pt 2+ ions Zn 2+ + 2e - → Zn-0.7618 Pt 2+ + 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 Zn 2+ ions; platinum in a solution of Pt 2+ ions Zn 2+ + 2e - → Zn-0.7618 Pt 2+ + 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 Zn 2+ ions; platinum in a solution of Pt 2+ ions Zn 2+ + 2e - → Zn-0.7618 Pt 2+ + 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 Zn 2+ ions; platinum in a solution of Pt 2+ ions Zn 2+ + 2e - → Zn-0.7618 Pt 2+ + 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 Zn 2+ ions; platinum in a solution of Pt 2+ 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. HgCl 2 and Hg 2 Cl 2 ; lead in a solution of Pb 2+ 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. HgCl 2 and Hg 2 Cl 2 ; lead in a solution of Pb 2+ ions 2Hg 2+ + 2e - → 2Hg 2 2+ 0.920 Pb 2+ + 2e - → Pb -(-0.1262) 1.046 V

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

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

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

24. Calculate the cell potential of voltaic cells that contain the following pairs of half-cells: 3. HgCl 2 and Hg 2 Cl 2 ; lead in a solution of Pb 2+ 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 Sn 2+ 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 Sn 2+ ions; iodine in a solution of I - ions Sn 2+ + 2e - → Sn-0.1375 I 2 + 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 Sn 2+ ions; iodine in a solution of I - ions Sn 2+ + 2e - → Sn-0.1375 I 2 + 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 Sn 2+ ions; iodine in a solution of I - ions Sn 2+ + 2e - → Sn-0.1375 I 2 + 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 Sn 2+ ions; iodine in a solution of I - ions Sn 2+ + 2e - → Sn-0.1375 I 2 + 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 Sn 2+ 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.