Portage Learning CHEM 104 Module 4 Exam

Portage Learning CHEM 104 Module 4 Exam

, Portage Learning CHEM 104 Module 4 Exam

Question 1
For the cell described by the following cell diagram.

Cr (s) | Cr+3 (aq, 1 M) || Al+3 (aq, 1 M) | Al (s)
Al+3 + 3e- → Al E0 = - 1.66 v
Cr+3 + 3e- → Cr E0 = - 0.73 v

(1) The anode half reaction is

(2) The cathode half reaction is

(3) The overall cell reaction is

(4) Show the calculation for the total cell potential

(5) State and explain whether the cell is voltaic or electrolytic
Your Answer:

1) Cr (s) → Cr+3 (aq) + 3e-

2) Al+3 (aq) + 3e- → Al (s)

3) Cr (s) + Al+3 (aq) → Cr+3 (aq) + Al (s)

4) E0 cell = E0 - E0
cathode anode


E0 cell = -1.66 - (-0.73)


E0 cell = -0.93 v

5) Since the total cell potential is negative, cell will not occur spontaneously and would have to be
electrolytic.