4 . D
Remember that in opposition to free energy, the more positive the E is for a reaction, the more
favorable it is. Since the reduction of Cl 2 has a more positive E than Cr3+, it will be more easily
reduced, meaning that it is a stronger oxidizing agent.5 . C
The two processes are reverses of each other, and thus their potentials are the negative of
each other.
6 . C
The half-reaction with the greater reduction potential will proceed forward as written, while
the half-reaction with the smaller reduction potential will proceed in the opposite direction
(i.e., as oxidation). F 2 has a greater tendency to be reduced than Ca2+, because it has the
greater reduction potential. Therefore, Equation 1 will proceed as written, while Equation 2
will proceed in the opposite direction. As a result, F 2 is reduced to 2 F− and Ca (s) is oxidized to
Ca2+.
7 . A
From the values of the reduction potentials, it is evident that Equation 1 will be the reduction
half-reaction, since it has the larger reduction potential, and Equation 2 will be reversed for
the oxidation half-reaction. The EMF of a reaction is determined by adding the standard
potentials of the reduced and oxidized species. The standard reduction potential of Co3+ is
+1.82 V. The standard oxidation potential of Na (s) is +2.71 V, which is equivalent, but opposite
in sign, to the standard reduction potential of Na+ (aq).
EMF = 1.82 V + 2.71 V = 4.53 V
8 . D
Use the relationship ∆G° = −nFE°, which can be rearranged to E° = −∆G°/(nF). ∆G° = −553.91 kJ,
n = 2 , and F = 96,487 C/mol e−. The next step is to convert the value of ∆G° from kJ to J. When
this is done, ∆G° is equal to −553,910 J. Now, the final step is to substitute the values into the
formula, to get the final answer of E° = 2.87 V.9 . D
First, determine the EMF (E°) by adding the potentials of the reduced and oxidized species.