5 Steps to a 5 AP Chemistry

The above galvanic cell is constructed with a cadmium electrode in a 1.0 M Cd(NO 3 ) 2

solution in the left compartment, and a silver electrode in a 1.0 M AgNO 3 solution in the

right compartment. The salt bridge contains a KNO 3 solution. The cell voltage is positive.

a. What is the balanced net ionic equation for the reaction, and what is the cell

potential?

b. Show how to calculate the equilibrium constant for the cell.

c. Write the expression for Q that would be needed in the Nernst equation.

Explain why any substances from the net ionic equation do not appear in Q.

d. Show how to calculate the free energy for the reaction.

e. Identify the anode, the cathode, the oxidizing agent, and the reducing agent.

 Answers And Explanations

First Free-Response Question

a. The cell reaction is:

Give yourself 1 point if you got this correct. The physical states are not necessary.

The calculation of the cell potential may be done in different ways. Here is one method:

Give yourself 1 point for the correct answer regardless of the method used. The most

common mistake is to multiply the silver voltage by two. You do not get the point for an

answer of 1 V.

The half-reactions and their standard reductions potentials are supplied on the exam,

not in the problem as given here. You will be expected to find the appropriate half-reactions

in a table.

Co Co e V

Ag e Ag V

Co s

→+ °=+

+→ °=+

+

+−

+−

(^22028)
1080

E

E

.

().

() 2 221 Ag aq++()→Co^2 ()aq+°=+Ag s()E. 08 V

Co s()+→+ 22 Ag aq++()Co^2 ()aq Ag s()

V

V = voltmeter

256  Step 4. Review the Knowledge You Need to Score High