Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32(aq), Ecell° 5 11.08 V(b) 6 Fe31(aq) 1 2 Cr31(aq) 1 7 H2O(l) S 6 Fe21(aq) 1 Cr2O722(aq) 1 14 H1(aq), Ecell° 5 21.29 V

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

          ............(1)

The given chemical equation follows:

Oxidation half reaction:         ( × 2)

Reduction half reaction:  

We are given:

Putting values in equation 1, we get:

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

The given chemical equation follows:

Oxidation half reaction:         ( × 6)

Reduction half reaction:  

We are given:

Putting values in equation 1, we get:

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

i'm pretty sure its b i might be wrong though!