Voltaic Cells Chemistry

Question

A voltaic cell using Cu²⁺/Cu and Al³⁺/Al half-cells is set up at standard conditions, and each compartment has a volume of 225 mL. What is the [Al³⁺] after the cell has delivered 0.120 A for 89.0 hours at 25.0 °C? (E° for Cu²⁺/Cu = 0.340 V and E° for Al³⁺/Al = -1.660 V.)

Answer 1

Writing the cell so that oxidation is at the anode and reduction at the cathode, you get

Cu| Cu2+(aq) || Al3+ | Al

The standard cell potential is the the reduction potential of the cathode minus the reduction potential of the anode).

So E^o_cell = -1.660 – 0.340 = -2.00 volts. This is negative so the reaction is not spontaneous. with Cu oxidizing to Cu2+ and Al3+ reducing to Al. The reaction is spontaneous in the opposite direction, so Al is oxidized to Al3+, thereby increasing the concentration of Al3+.

The reaction for Al3+ can be written Al (s) - 3e^- = Al 3+(aq).

The number of coulombs of charge delivered is 0.120 C/s x 6.24 e^- / C = 38448 C of electrons.

The number of moles of electrons is 38448 C x 6.24 x 10^18 e^- / C / 6.022 x 10^23 e/mole = 0.398 moles electrons. For every mole of electrons, only 1 mole of Al3+ was produced or 0.398 moles e^- / 3 = 0.1327 moles Al3+ produced which is a change of 0.1327 /225x1000 = 0.590 M in Al3+.

Since the initial concentration was 1M, the final concentration is 1.0 M + 0.590= 1.59 M which is the answer requested.

Check math.