Question

E^o_cell= -1.288V

Consider the following redox reaction, and answer the three questions below. 3 K+ (a q) + A1(s) rightarrow 3 K(s) + A13+(a q) Calculate E degree Number -1.288 v Calculate Delta G degree Number 372.876 kJ Calculate K. Number

n = 3

When I calculate it K = 5.37*10^-66 but for some reason that answer is wrong. What am I doing wrong?

Answer 1

Eo,cell = RT/(nF) lnK = (0.0257/z)lnK = (0.0592/n)logK

And also

dGo = - RTlnK

Then:

LogK = Eo,cell/(0.0592/n)= - 1.288V/0.01973V= - 65.28

K = 10^-65.28 =5.24x10^-66

Or

lnK = -dGo/(RT) = -372.9 kJ/ (0.008314 kJ/molK x 298) = - 150.5

K = e^{- 150.5} =4.35x10^-66

dGo = -nFEo,cell = -3 x 96485 C x (-1.288 V) = 372.8 kJ

Your calculation method is correct (see my verification), but :

-the limits of acceptance are exceeded only for the 3^rd answer

-small errors or rounding have high influence on log or ln values

-initial values of Eo are not from your computer data base (In my tables the two values are -2.925 and -1.66, so E0,cell will be -1.225 V !!)

If you can, report this problem. The answer limits must be a little bit larger.