Question

A cell is constructed using a piece of Al in a 1.00 M Al(NO3)3 solution and a piece of Cu submerged in a Cu(NO3)2 solution.

Calculate the cell potential.

Write the overall reaction.

Which piece of metal gains mass and which loses mass?

Which reaction occurs at the anode and which reaction occurs at the cathode?

If the above cell was constructed using 0.250 M Al(NO₃)3 and 0.300 M Cu(NO₃)₂, calculate the cell potential.

Then calculate Delta G.

Answer 1

Al(s)/Al(NO3)3(1M)//Cu(NO3)2(1M)/Cu(s)

oxidation reaction at anode    [ Al(s) --------------------- Al+3(aq) + 3e-]x2

reduction reaction at cathode    [ Cu+2 + 2e- -------------- Cu(s)]x3

                              ---------------------------------------------------------------------------------------------

   Over all reaction                     2 Al(s) + 3 Cu+2(aq) ------------- 2 Al+3(aq) + 3 Cu(s)

                               -------------------------------------------------------------------------------------------

E0 of Al+3/Al = -1.66V                                       E0 of Cu+2/cu = +0.34V

E0cell = E0cathode - E0anode

E0cell = 0.34 - ( -1.66)

E0cell = 2.0V

n= 6e-                F= 96500 coloumbs

Ecell = E0cell - 0.0591/n log[Al+3]^2/[Cu+2[^3

Ecell = 2.0 - 0.0591/6 log(1/1)

Ecell = 2.0V

Al metal looses its mass at anode and Cu metal gains mass at cathode.

Al(s)/Al(NO3)3(0.250M)//Cu(NO3)2(0.300M)/Cu(s)

Ecell = 2.0 - 0.0591/6 log( 0.250)^2/(0.300)^3

Ecell = 1.996V

G= - nFEcell

G = - 6 x 96500 x 1.996

G = - 1155684J

G = - 115.57 KJ