Answer to Question #249321 in General Chemistry for mytc

Question

. Calculate the concentration of the chromic ion in the cell which has an electrode potential of 0.688 at 298.15 K for the given reaction:

           Cr2O7 2 –(aq) + 14 H+ (aq) + 6 I – 1 (aq)   →    2 Cr 3+(aq) + 3I2(s) + 7 H2O(l)

         When       [Cr2O7 2 – ] = 2.25 M;   [H+] = 1.0 M ; [I – 1 ] = 1.0 M ;  [H2O] = 1.0 x 10 – 5 M

          Calculate the Gibbs free-energy  and the standard Gibbs Free-Energy occur in the system

Answer 1

1.0× 10^-5× 2.25/298

= 7.55× 10^-8V