2 Al(s) + 3 Zn2+(aq) → 2 Al3+(aq) + 3 Zn(s)
a. Write the complete electron configuration (e.g., 1s2 2s2 . . .) for Zn2+.
1s2 2s2 2p6 3s2 3p6 3d10
b. Which species, Zn or Zn2+ , has the greater ionization energy? Justify your answer.
Zn2+ has the greater ionization energy.
The electron being removed from Zn2+ experiences a larger
effective nuclear charge than the electron being removed from Zn because Zn2+ has two fewer electrons shielding the nucleus.
It takes more energy to remove a negatively charged electron from a positive ion than from a neutral atom.
c. Identify the species that is oxidized in the reaction.
The diagram below shows a galvanic cell based on the reaction. Assume that the temperature is 25°C.
d. The diagram includes a salt bridge that is filled with a saturated solution of KNO3. Describe what happens in the salt bridge as the cell operates.
As the cell operates, NO3− ions flow toward the Al half-cell and K+ ions flow toward the Zn half-cell.
e. Determine the value of the standard voltage, E°, for the cell.
E° = (−0.76 V) − (−1.66 V) = 0.90 V
f. Indicate whether the value of the standard free-energy change, G°, for the cell reaction is positive, negative, or zero. Justify your answer.
G° is negative since E° is positive and ∆G° = –nFE°
G° must be negative because the reaction is spontaneous under standard conditions.
g. If the concentration of Al(NO3)3 in the Al(s)/Al3+(aq) half-cell is lowered from 1.0 M to 0.01 M at 25°C, does the cell voltage increase, decrease, or remain the same? Justify your answer.
Lowering [Al3+] causes an increase in the cell voltage. The value of Q will fall below 1.0 and the log term in the Nernst equation will become negative. This causes the value of Ecell to become more positive.
A decrease in a product concentration will increase the spontaneity of the reaction, increasing the value of Ecell.