A Tl^+\|Tl couple was prepared by saturating 0.10 M – KBr with TlBr and allowing Tl^+ ions form the insoluble bromide to equilibrate. This couple was observed to have a potential –0.444 V with respect

Question

A Tl^+\|Tl couple was prepared by saturating 0.10 M – KBr with TlBr and allowing Tl^+ ions form the insoluble bromide to equilibrate. This couple was observed to have a potential –0.444 V with respect to Pb^2+ \| Pb couple in which Pb^2+ was 0.10 M. What is the Ksp of TlBr. [Given: E°_Pb2+\|Pb = –0.126 V, E°_Tl+\|Tl = –0.336 V, log 2.5 = 0.4, 2.303RT/F = 0.06]

Prashant Kotian · Accepted Answer

Correct answer: A. E_Pb = -0.126 - (0.06/2)*log(1/0.1) = -0.156 V. Since E_Tl - E_Pb = -0.444 V, E_Tl = -0.600 V. Also, E_Tl = E°_Tl - 0.06*log(1/[Tl^+]) = -0.336 + 0.06*log[Tl^+]. -0.600 = -0.336 + 0.06*log[Tl^+], so log[Tl^+] = -4.4. [Tl^+] = 4 * 10^-5 M. Ksp = [Tl^+][Br^-] = (4 * 10^-5)(0.10) = 4.0 * 10^-6. Therefore, correct answer is A.

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