Question

Be sure to answer all parts. What are E and AG of a redox reaction at 25°C for which n= 2 and K = 0.065? cell and AGO cell Agº -

Answer 1

We are given with

K (equilibrium constant) = 0.065

n = 2 (number of moles balanced from redox reaction)

Temperature = 250c = 25 +273.15 = 298.15 K

To find:

E0cell and $\Delta$ G0

Formulas to be used

E0cell (v) = (RT / nF) ln K ;   $\Delta$G0 (J) = -nFE0cell

R = 8.314 J mol-1 K-1

T = Temp (in Kelvin)

F = Faraday constant = 96,485 c/mol

Putting values:

- E all = RT lnk. hf - (8.314 I moty ) (298.15 x) (lu 0.065) (22 96,485 C mot (8.3147 (298,15) eu (0.065) (2) (96,4850) (9 478.6 1. J) (-2.733) (92970 c) = 0.0128) (-2.733) Ilc -=-0.035 JIC - -0.035 V

We get E0cell = -0.035 J/C = -0.035 V

-=-12 mote) (964 85 elmol. e-) (-00035J14) z (2) (9648.5.160.035) I 675 3.95 I

$\Delta$G0 = 6753.95 J