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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 18.51

Determine the values of Eocell and Go for the following reactions.

(a) O2(g) + 4 I(aq) + 4 H+(aq)2 H2O(l) + 2 I2(s)

(b) Cr2O72–(aq) + 3 Cu(s) + 14 H+(aq)2 Cr3+(aq) + 3 Cu2+(aq) + 7 H2O(l)






Answer:

First, find the cell potential from standard reduction potentials. Then use the relationship Go = –nFEo to find the Gibb's Free Energy.

(a) O2(g) + 4 I(aq) + 4 H+(aq)2 H2O(l) + 2 I2(s)

Reduction: O2(g) + 4 H+(aq) + 4 e 2 H2O(l)     Eored = +1.229 V

Oxidation: 4 I(aq)2 I2(s) + 4 e    Eoox = –0.535 V

Eocell = +1.229 + –0.535 = +0.694 V

Go = –nFEo = –(4)(96485)(0.694) = –26800 J = –26.8 kJ



(b) Cr2O72–(aq) + 3 Cu(s) + 14 H+(aq)2 Cr3+(aq) + 3 Cu2+(aq) + 7 H2O(l)

Reduction: Cr2O72–(aq) + 14 H+(aq) + 6 e2 Cr3+(aq) + 7 H2O(l)    Eored = +1.33 V

Oxidation: Cu(s)Cu2+(aq) + 2 e    Eoox = –0.340 V

Eocell = +1.33 + –0.340 = +0.99 V

Go = –nFEo = –(6)(96485)(0.99) = –570000 J = –570 kJ