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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:
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.
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.
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.
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.
0.340 V < EoRh3+/Rh < 0.800 V