The cell reactions involved in this cell is Anode is Be/Be2+ {since have high ne

Question

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.

Explanation / Answer

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

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