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CuI(s)+e−→Cu(s)+I−(aq); E∘=−0.188 V Cu2+(aq)+I−+e−→CuI(s) E∘=0.868 V; Ksp(CuI)=10−12M2 and 298R ln 10F=0.059 If the equilibrium constant of the disproportionation reaction of Cu+ is Kc i.e. 2Cu+(aq)Kc⇌Cu(s)+Cu2+(aq); then find the value of (0.59×logKc)

Answer» CuI(s)+eCu(s)+I(aq); E=0.188 V
Cu2+(aq)+I+eCuI(s) E=0.868 V;
Ksp(CuI)=1012M2 and 298R ln 10F=0.059
If the equilibrium constant of the disproportionation reaction of Cu+ is Kc i.e.
2Cu+(aq)KcCu(s)+Cu2+(aq); then find the value of (0.59×logKc)


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