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

Problem 18.50

Rhodium is a rare metal used as a catalyst. The metal does not react with HCl(aq), but it does react with HNO3(aq), producing Rh3+(aq) and NO(g). Copper will displace Rh3+ from aqueous solution, but silver will not. Estimate a value of EoRh3+/Rh.

Consider the potentials for the reactions described:

(1)

Rh(s) + HCl(aq) NR  Eocell < 0

Eocell = Eooxidation + Eoreduction = – EoRh3+/Rh + EoH+/H2 < 0

Since 2 H+(aq) + 2 e H2(g)  Eoreduction = 0.00 V, then EoRh3+/Rh > 0 V.

(2)

Rh(s) + HNO3(aq) + 3 H+(aq) Rh3+(aq) + NO(g) + 2 H2O(l)  Eocell > 0

Eocell = Eooxidation + Eoreduction = – EoRh3+/Rh + EoHNO3/NO > 0

Since HNO3(aq) + 3 H+(aq) + 3 e NO(g) + 2 H2O(l)

Eoreduction = 0.956 V, then EoRh3+/Rh < 0.956 V.

(3)

Rh3+(aq) + Cu(s) Rh(s) + Cu2+(aq) Eocell > 0.

Eocell = Eooxidation + Eoreduction = EoRh3+/Rh + –EoCu2+/Cu > 0

Since Cu(s) Cu2+(aq) + 2 e  Eooxidation = –0.340 V, then EoRh3+/Rh > 0.340 V.

(4)

Rh3+(aq) + Ag(s) NR  Eocell < 0.

Eocell = Eooxidation + Eoreduction = EoRh3+/Rh + –EoAg+/Ag < 0

Since Ag(s) Ag+(aq) + e  Eooxidation = –0.800 V, then EoRh3+/Rh < 0.800 V.

Net conclusion:

0.340 V < EoRh3+/Rh < 0.800 V