A voltaic cell using Cu²⁺/Cu and Al³⁺/Al half-cells is set up at standard conditions

Question

and each compartment has a volume of 225 mL. What is the [Al³⁺] after the cell has delivered 0.120 A for 64.0 hours at 25.0 °C? (E° for Cu²⁺/Cu = 0.340 V and E° for Al³⁺/Al = -1.660 V.)

Answer 1

Cu²⁺(aq) + 2e ==> Cu(s) .. Eº = 0.340 V

Al³⁺(aq) + 3e- ==> Al(s) .. Eº = -1.660 V

Cu²⁺(aq) + 2e ==> Cu(s) x3

Al(s) ==> Al³⁺(aq) + 3e- x2

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3Cu²⁺(aq) + 6e- ==> 3Cu(s)

2Al(s) ==> 2Al³⁺(aq) + 6e-

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3Cu²⁺(aq) + 2Al(s) ==> 2Al³⁺(aq) + 3Cu(s) ... OVERALL REACTION

To find the [Al3+], we first need to find how many moles of Al3+ are formed.

0.120 A = 0.120 C/sec

64 hours x 60 min / hour = 3840 min x 60 sec / min = 230,400 sec

230,400 sec x 0.120 C / sec = 27648 C

27648 C x 1 mol e- / 96500 C = 0.2865 moles e-

0.2865 moles e- x 6 moles e- / 2 mols Al³⁺ = 0.8595 moles Al³⁺

Now we divide by the volume (in liters) to get [Al³⁺]...

0.8595 moles Al³⁺ / 0.225 L = 3.82 M