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Transcript 
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;
2+
Cu in a solution of Cu ions
a.
b.
c.
d.
0.5600 volts
0.7489 volts
0.8970 volts
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;
2+
platinum in a solution of Pt ions
a.
b.
c.
d.
0.4182 volts
0.7618 volts
1.1800 volts
1.9418 volts
Calculate the cell potential of voltaic cells that
contain the following pairs of half-cells:
3. HgCl2 and Hg2Cl2;
2+
lead in a solution of Pb ions
a.
b.
c.
d.
e.
0.031 volts
0.398 volts
1.000 volts
1.046 volts
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 ions
a.
b.
c.
d.
e.
0.3845 volts
0.6730 volts
0.6865 volts
1.0050 volts
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;
2+
Cu in a solution of Cu ions
a.
b.
c.
d.
0.5600 volts
0.7489 volts
0.8970 volts
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;
2+
Cu in a solution of Cu ions
Cr3+ + 3e- → Cr -0.913
2+
Cu + 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;
2+
Cu in a solution of Cu ions
Cr3+ + 3e- → Cr -0.744
2+
Cu + 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;
2+
Cu in a solution of Cu ions
Cr3+ + 3e- → Cr -0.744
2+
Cu + 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;
2+
Cu in a solution of Cu ions
Cr3+ + 3e- → Cr -0.744
2+
Cu + 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;
2+
Cu in a solution of Cu ions
a.
b.
c.
d.
0.5600 volts
0.7489 volts
0.8970 volts
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;
2+
platinum in a solution of Pt ions
a.
b.
c.
d.
0.4182 volts
0.7618 volts
1.1800 volts
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;
2+
platinum in a solution of Pt ions
Zn2+ + 2e- → Zn -0.7618
2+
Pt + 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;
2+
platinum in a solution of Pt ions
Zn2+ + 2e- → Zn -0.7618
2+
Pt + 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;
2+
platinum in a solution of Pt ions
Zn2+ + 2e- → Zn -0.7618
2+
Pt + 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;
2+
platinum in a solution of Pt ions
Zn2+ + 2e- → Zn -0.7618
2+
Pt + 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;
2+
platinum in a solution of Pt ions
a.
b.
c.
d.
0.4182 volts
0.7618 volts
1.1800 volts
1.9418 volts
Calculate the cell potential of voltaic cells that
contain the following pairs of half-cells:
3. HgCl2 and Hg2Cl2;
2+
lead in a solution of Pb ions
a.
b.
c.
d.
e.
0.031 volts
0.398 volts
1.000 volts
1.046 volts
1.977 volts
Calculate the cell potential of voltaic cells that
contain the following pairs of half-cells:
3. HgCl2 and Hg2Cl2;
2+
lead in a solution of Pb 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;
2+
lead in a solution of Pb 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;
2+
lead in a solution of Pb 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;
2+
lead in a solution of Pb 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;
2+
lead in a solution of Pb ions
a.
b.
c.
d.
e.
0.031 volts
0.398 volts
1.000 volts
1.046 volts
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 ions
a.
b.
c.
d.
e.
0.3845 volts
0.6730 volts
0.6865 volts
1.0050 volts
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
I2 + 2e- → 2I-
-0.1375
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
I2 + 2e- → 2I-
-0.1375
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
I2 + 2e- → 2I-
-0.1375
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
I2 + 2e- → 2I-
-0.1375
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
a.
b.
c.
d.
e.
0.3845 volts
0.6730 volts
0.6865 volts
1.0050 volts
1.2935 volts
.
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