Question

You decide to construct a zinc/aluminum galvanic cell in which the electrodes are connected by a wire and the solutions are connected with a salt bridge. One electrode consists of an aluminum bar in a 1.0 M solution of aluminum nitrate. The other electrode consists of zinc bar in a 1.0 M solution of zinc nitrate.

a. Which electrode is the cathode and which is the anode?

b. What is the direction of electron flow?

c. What chemical reactions are occurring at each electrode?

d. What is the overall balanced chemical reaction?

e. After a period of time, will the bar of zinc become heavier, lighter or stay the same? What about the aluminum bar?

f. What is the standard cell potential?

Answer 1

Zn/Al

Al3+ + 3 e− <---> Al(s) −1.662

Zn2+ + 2 e− <---> Zn(s) −0.7618

The most negative will oxidize

Then:

Al3+ + 3 e− <---> Al(s) −1.662

Zn(s)<---> Zn2+ + 2 e− −0.7618

a)

cathode = reduction occurs here, this will be the aluminium

anode = oxidation takes place here, this will be the zinc

b)

the directino of flow is from anode (zinc) to cathode (aluminium)

c)

chemical reactions:

Al3+ + 3 e− <---> Al(s)

Zn(s)<---> Zn2+ + 2 e−

d)

Overal reaction:

2Al3+ + 6 e− <---> 2Al(s)

3Zn(s)<---> 3Zn2+ + 6 e−

Add:

2Al3+ + 6 e− 3Zn(s)<---> 2Al(s) + 3Zn2+ + 6 e−

Cancel e-

2Al3+ + 3Zn(s)<---> 2Al(s) + 3Zn2+

e)

the bar of zinc should become lighter, since it is being oxidized, all ions go in solution

the aluminium bar should increase in weight, since Al is beind reduced, deposits of solid aluminium increase the mass

f)

Ecell = −0.7618 - -1.662 = 0.9002 V