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 Cr3+ ions; Cu in a solution of Cu2+ ionsa. 0.5600 voltsb. 0.7489 voltsc. 0.8970 voltsd. 1.0859 volts

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 voltsb. 0.7618 voltsc. 1.1800 voltsd. 1.9418 volts

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+ ionsa. 0.031 voltsb. 0.398 voltsc. 1.000 voltsd. 1.046 voltse. 1.977 volts

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- ionsa. 0.3845 voltsb. 0.6730 voltsc. 0.6865 voltsd. 1.0050 voltse. 1.2935 volts .

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+ ionsa. 0.5600 voltsb. 0.7489 voltsc. 0.8970 voltsd. 1.0859 volts

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 -0.913Cu2+ + 2e- → Cu 0.153 -(-0.913)

1.066 V

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 -0.744Cu2+ + 2e- → Cu 0.3419 -(-0.913)

1.066 V

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 -0.744Cu2+ + 2e- → Cu 0.3419 -(-0.744)

1.066 V

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 -0.744Cu2+ + 2e- → Cu 0.3419 -(-0.744)

1.0859 V

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+ ionsa. 0.5600 voltsb. 0.7489 voltsc. 0.8970 voltsd. 1.0859 volts

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 voltsb. 0.7618 voltsc. 1.1800 voltsd. 1.9418 volts

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+ ionsZn2+ + 2e- → Zn -0.7618Pt2+ + 2e- → Pt 1.18

-(-0.7618) 1.94 V

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+ ionsZn2+ + 2e- → Zn -0.7618Pt2+ + 2e- → Pt 1.18

-(-0.7618) 1.94 V

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+ ionsZn2+ + 2e- → Zn -0.7618Pt2+ + 2e- → Pt 1.18

-(-0.7618) 1.94 V

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+ ionsZn2+ + 2e- → Zn -0.7618Pt2+ + 2e- → Pt 1.18

-(-0.7618) 1.94 V

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 voltsb. 0.7618 voltsc. 1.1800 voltsd. 1.9418 volts

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+ ionsa. 0.031 voltsb. 0.398 voltsc. 1.000 voltsd. 1.046 voltse. 1.977 volts

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

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

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

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

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+ ionsa. 0.031 voltsb. 0.398 voltsc. 1.000 voltsd. 1.046 voltse. 1.977 volts

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- ionsa. 0.3845 voltsb. 0.6730 voltsc. 0.6865 voltsd. 1.0050 voltse. 1.2935 volts

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

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

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

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

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- ionsa. 0.3845 voltsb. 0.6730 voltsc. 0.6865 voltsd. 1.0050 voltse. 1.2935 volts .