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Study Quiz 1
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1
What do galvanic cells do?
A)They ensure reactants in redox reactions come into contact with each other.
B)They ensure reactants in redox reactions flow through an internal circuit.
C)They convert electrical energy to chemical energy.
D)They convert chemical energy to electrical energy.
2
Which statement is true about galvanic cells?
A)Oxidation occurs at the cathode, which is positively charged.
B)Reduction occurs at the anode, which is negatively charged.
C)Oxidation occurs at the anode, which is negatively charged.
D)Reduction occurs at the anode, which is positively charged.
3
What is the reduction half-reaction for the cell described by the following shorthand representation?

Cd(s)|Cd2+(aq)||H+(aq)|H2(g) |Pt(s)?

A)2H+(aq) + 2e-à H2(g)
B)Cd(s)à Cd2+(aq) + 2e-

C)H+(aq) + e-à H2(g)

D)
2H+(aq)à H2(g)
4
What is the shorthand representation of a cell that uses an inert electrode for the following reaction?

Ni(s) + 2Fe3+à Ni2+ + 2Fe2+

A)Ni(s)|Ni2+(aq)|| Fe3+(aq)| Fe(s)

B)Ni(s)|Ni2+(aq)|| Fe2+(aq)| Fe(s)

C)Ni(s)|Ni2+(aq)|| Fe3+(aq), Fe2+(aq)| Pt(s)

D)Ni(s)|| Fe3+(aq), Fe2+(aq)| Pt(s)
5
If a battery contains eight 1.5-V dry cells, what is the voltage of the battery?
A)12 V
B)6 V
C)8 V
D)24 V
6
How do you calculate the standard cell potential of a cell?
A)Subtract the standard reduction potential of the cathode from that of the anode.
B)Subtract the standard reduction potential of the anode from that of the cathode.
C)Subtract the standard oxidation potential from the overall standard reduction potential.
D)Subtract the standard reduction potential from the overall standard oxidation potential.
7
Which considerations are important when calculating a standard cell potential?
A)the balanced chemical equation
B)the amounts of reactants
C)the standard reduction potentials of the reactants
D)the balanced chemical equation and the standard reduction potentials of the reactants
8
Given the following standard cell potentials, what is the standard cell potential for the reaction below?

Al3+(aq) + 3e-à Al(s)  E = -1.662 V

Pb2+(aq) + 2e-à Pb(s)  E = -0.126 V

2Al(s) + 3Pb2+(aq)à 3Pb(s) + 2Al3+(aq)

A)0.209 V
B)1.536 V
C)1.788 V
D)-1.788





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