18.5.5 CWK - K for Cell Reactions?

hey....for this question .....did you get the correct answer ....i sent you an e-mail regarding this but you didnt reply ....

http://answers.yahoo.com/question/index;_ylt=AsCAm...

i am extremely sorry for giving a wrong answer ......so do you have the correct answer ???

now for this question .........

M3+ + 3e- -------> M ....Eo = 0.29 V

N2+ + 2e- ----------> N ....Eo = 0.61

since reduction potential of N is greater than that of M ....so if a cell is constructed cathode will be N2+/N and anode will be M3+/M

cathode reaction will be :

N2+ + 2e- ---------> N ....(1)

and anode reaction will be :

M --------> M3+ + 3e- ...(2)

multiplying (1) by 3 and (2) by 2 and adding we get...

3N2+ + 6e- -------> 3N

2M -------> 2M3+ + 6e-

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

3N2+ + 2M -------> 3N + 2M3+

Eo(cell) = Eo(cathode) - Eo(anode) = 0.61 - 0.29 = 0.32 V

delta Go = -nFEo(cell)

where n = no.of electrons involved in electrode reaction = 6

F = 96500 C/mole

and Eo(cell) = 0.32 V

delta Go = - 6 X 96500 X 0.32 = -185280 CV/mole = -185280 j/mole ( as 1CV = 1 j)

and delta Go is related to equilibrium constant K by the following equation ...

delta Go = -2.303 X R X T X log K

where delta Go = - 185280 j/mole

R = 8.314 j/K/mole

T = 25 + 273 = 298 K

K = ?

-185280 = -2.303 X 8.314 X 298 X log K

log K = 185280/5705.848

log K = 32.472

taking antilog ...

K = 10^32.472 = 2.965 X 10^32

hope i am correct this time ....do mail me with that problems correct answer ....

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