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Class-12-science»Chemistry

The 2nd question-

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Anode : Ag (s) \leftrightarrow {Ag}^{+} (aq) + e^{-} Cathode : AgI (s) + e^{-} \leftrightarrow Ag (s) + I^{-} (aq) The cell producing these reactions would be Ag (s) | {Ag}^{+} (aq) | | I^{-} (aq) | AgI (s) | Ag (s) The standard cell potential would be E ° = E °_{I^{-} | AgI | Ag} - E °_{{Ag}^{+} | Ag} E °_{I^{-} | AgI | Ag} is calculated from above question, E °_{I^{-} | AgI | Ag} = - 0.15 V or, E ° = - 0.15 - 0.799 = - 0.949 V Now, nFE ° = {RTlnK}_{sp} or, K_{sp} = antilog (\frac{nFE °}{2.303 RT}) = antilog (\frac{nE °}{2.303 RT / F}) or, K_{sp} = antilog (\frac{- 0.949}{0.05915}) = antilog (- 16.0439) or, K_{sp} = 9.17 \times 10^{- 17} Hence, correct answer is (D)

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