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Heat and Work

The solubilit...

Question

The solubility product of silver iodide is $83 \times 10^{- 17}$ and the standard reduction potential of Ag, $A g^{+}$ electrode is +0.8 volts at $25^{o}$ C. The standard reduction potential of Ag, $A g I / I^{-}$ electrode from these data is?

- 0.30 V

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+ 0.15 V

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+ 0.10 V

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- 0.15 V

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Solution

The correct option is D $- 0.15 V$
$\begin{matrix} K_{S P} & of Ag I = 8, 3 \times 10^{- 17} E^{\circ} A g / A g^{+} & = - 0.8 \end{matrix}$
$\begin{matrix} E_{A g, A g I / I^{-}} & = E_{A g / A g^{+}}^{0} - \frac{R T}{n F} ln \frac{1}{K_{S P}} = - 0.8 - \frac{8.31 \times 295 \times 2.303}{1 \times 98500} log 8.3 \times 10^{- 17} ∴ & - 0.8 + 0.095 = - 0.15 V \end{matrix}$
$Hence, (D) is correct option$

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