Cliffs AP Chemistry, 3rd Edition

8. Given that

Zn2+(aq) + 2e–→Zn(s) E°red= −0.76 V

Cr3+(aq) + 3e–→Cr(s) E°red= −0.74 V

calculate the equilibrium constant K for the following balanced reaction:

3 Zn(s) + 2 Cr3+(aq) →3 Zn2+(aq) + 2 Cr(s)

A. K= e–0.02

B. K= e0.02

C. K= e4.7

D. K= e8.0

E. cannot be determined from the information provided

Answer: C

Step 1:Determine the oxidation and reduction half-reactions and E°cell.

+

+

+

:.+

:.

.

saq

aq s

saq aqs

ox Zn Zn e E V

red Cr e Cr E V

Zn Cr Zn Cr E V

2

3

232

3076

2074

3002

ox

red

cell

2

3

32

"

"

"

+

+

++

=+

=-

=

• 
  


• 
  

%

%

%

^ _

_ ^

^ __^

h i

i h

h iih

9

9

C

C

Step 2:Use the equation.

..

lnK nE. V.

0 0257 0 0257

== =: %cell^6002 : 47

K= e4.7

9. An electric current is applied to an aqueous solution of FeCl 2 and ZnCl 2. Which of the
   following reactions occurs at the cathode?

A. Fe2+(aq) + 2 e–(aq) →Fe(s) E°red= −0.44 V

B. Fe(s) →Fe2+(aq) +2 e– E°ox= 0.44 V

C. Zn2+(aq) + 2 e–(aq) →Zn(s) E°red= −0.76 V

D. Zn(s) →Zn2+(aq) + 2 e– E°ox= 0.76 V

E. 2 H 2 O(l) →O 2 (g) + 4 H+(aq) + 4 e– E°ox= −1.23 V

Answer: A

Reduction occurs at the cathode. You can eliminate choices (B), (D), and (E) because these re-
actions are oxidations. E°redfor Fe2+(aq) is −0.44 V, and E°redfor Zn2+(aq) is −0.76 V. Because
Fe2+(aq) has the more positive E°redof the two choices. Fe2+(aq) is the more easily reduced and
therefore plates out on the cathode. 189

Reduction and Oxidation