At anode, Al--------->Al+3+3e-, Eo=1.66V (1) and Be+2+2e- ------->Be, Eo=-1.85V

Question

At anode, Al--------->Al+3+3e-, Eo=1.66V (1) and Be+2+2e- ------->Be, Eo=-1.85V (2)

Eq.1*2+Eq.2*3 gives 2Al+3Be+2-------->2Al+3+ 3Be, Eo= 1.66-1.85=-0.19V

deltaG=-nFE, n= no of electrons exchanged=6 and F= 96500coulumb/mole and E=-0.19V

deltaG=6*0.19*96500=110010 J/mole

at 300 deg.c= 300+273= 573K

E= Eo-(RT/nF)*logQ, Q= reaction coefficient = [Al+3]2/ [Be+2]3=1/(1*10-4)3=1012

logQ= log (1012)=12

E=-0.19- 8.314*573/(6*96500)*12=-0.29V, sinnce E is -ve make deltaG=-nFE postive nonspoontaneous at 300 deg.c as well.

Explanation / Answer

(VII) (1o points) For the following cell, Al (s) | A134 (1.00M) | | Be2+ (1.00 x 10-4M) | Be(s) (a) Write the overall net ionic equation (b) Calculate the cell potential, Eel, at 25°C, and predict the spontaneity of the overall reaction (c) Calculate the cell potential, Ecell. at 300°C, and predict the spontaneity of the overall reaction. The standard reduction potentials are: Al3+ (aq) + 3e- Al(s) E--1 .66 V Be2+ (aq) + 2e- Be(s) E=-1.85 V

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