5 Steps to a 5 AP Chemistry

b. The cobalt is the anode. You get 1 point for this statement.

The reason Co is the anode is because the Co is oxidized. You get 1 point for this state-

ment, or if you say the Co loses electrons.

c. KCl cannot be substituted for KNO 3 in this case. This is worth 1 point.

The Cl−ion from the KCl would react with the silver ion to give insoluble AgCl. Give

yourself 1 point for this explanation.

d. The voltage would decrease. Give yourself 1 point for this answer.

The excess Co^2 +, from the Co(NO 3 ) 2 , would make the cell nonstandard. The Nernst

equation should be used to calculate the new voltage. The concentration of the cobalt

ions appears in the numerator of the logarithm term of the Nernst equation. This

makes the logarithm term more negative, yielding a lower voltage. Give yourself 1 point

for any answer relating to this. It might be helpful if you wrote out the Nernst equa-

tion for this cell.

e. At equilibrium the cell voltage would be 0 V. This is worth 1 point. At equilibrium no
work is done, so the potential must be zero. Give yourself 1 point for this answer.

There are a total of 10 points possible on these questions.

Second Free-Response Question

a. The following reduction half-reactions are provided on the AP exam:

The first reaction needs to be reversed, the silver half-reaction needs to be doubled, and

the reactions and voltages added:

You get 1 point for the correct equation. You also get 1 point for the correct cell voltage.

b. Use the equation log K=nE°/0.0592. (This equation is given on the AP exam.)

log K=(2 × 1.20)/0.0592

You get 1 point for entering the values into the correct equation. You get the point even

if you entered a wrong answer for E°from part a. The question does not ask you to do

any calculations, so there are no points for including any work beyond what is shown.

c. Q=[Cd^2 +]/[Ag+]^2 This answer is worth 1 point.

The remaining substances (Cd and Ag) are solids. Solids do not appear in Qexpressions.

You get 1 point for the explanation.

d. Use the equation G° = −nFE°. (This equation is given on the AP exam.)

G° = −nFE° = (−2)(96500)(1.20).

Cd Cd e V

2(Ag e Ag) V

2A

→+ + °=+

+→ °=+

+−

+−

(^22040)
1080

.

.

E

E

ggCd Cd Ag++→ +^221 E°=+. 20 V

Cd e Cd V

Ag e Ag V

(^22040)
1080
+−
+−

+→ °=−

+→ °=+

E

E

.

.

Electrochemistry  257