In: Chemistry
For cell Al I Al3+ 1M II Be2+ 1x10^-4 M I Be (s) Write the overall net ionic equation Calculate the cell potential E cell at 25 C and predict the spontaneity of the overall reaction Calculate the cell potential at 300 degrees C and predict the spontaneity of the overall reaction The standard reduction potentials are Al- -1.66V Be -1.85 V
The cell reactions involved in this cell is
Anode is Be/Be2+ {since have high negative value }
Be ---------> Be^2+ + 2e.
Cathode is Al/Al^3+
Al^3+ + 3e ---------> Al.
Overall reaction
3 Be + 2 Al^3+ ---------> 3 Be^2+ + 2 Al
6 e transfer take place.
So Nernst equation is
E cell = Eocell - (2.303RT/6F)log [Be^2+]/[Al^3+]
Here
Eocell = Eocathode -Eoanode = - 1.66 V -(-1.88 V) = + 0.22 V
R = 8.314 V C K^-1 , T = temparature in K , n= no.of electrons,
F = faradays constant = 96485 C
So at 25 deg.c = 298 K,
2.303 RT/6F = 0.0591/6 = 0.00985 V
E cell = 0.22 V - 0.00985 log (1×10^-4)^3/(1)^2
= 0.22 V - (0.000985 )× log 10^-12 = 0.22 - (0.00985)×(-12) = 0.34 V.
At 300 deg.C = 573 K,
2.303 × 573 /6×96485 = 0.0190 V
E cell = 0.22 V - 0.0190 V log (1×10^-4)^3/(1)^2
= 0.22 V - (0.0023 V )× log 10^-12 = 0.22 - (0.0190 V)×(-12) = 0.45 V.
Since both E cell values are positive, both are spontaneous.