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Oxidation States

A silver wire...

Question

A silver wire dipped in $0.1$ M $H C l$ solution saturated with $A g C l$ develops oxidation potential of $- 0.25 V .$ If $E_{A g / A g +}^{o} = - 0.799 V,$ the $K_{s p}$ of $A g C l$ in pure water will be :

2.95 \times 10^{- 11}

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5.1 \times 10^{- 11}

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3.95 \times 10^{- 11}

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1.95 \times 10^{- 11}

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Solution

The correct option is B $5.1 \times 10^{- 11}$
$Given$ $E_{e l} = - 0.25 V$ , $E_{e l}^{o} = - 0.799 V$

$Solution :$
$E_{e l} = E_{e l}^{0} - \frac{0.0592}{n l o g [A g +]}$
Substitute the values in the above expression
$- 0.25 V = - 0.799 V - \frac{0.0592}{1 l o g [A g +]}$
log[Ag+] = -9.2736
$[A g +] = 5.1 \times 10^{- 10}$
Hence the solubility product will be
$K_{s p} = [A g +] [C l -] = 5.1 \times 10^{- 10} \times 0.1$
$= 5.1 \times 10^{- 11}$

$Hence B is the correct option$

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A silver wire dipped in 0.1 M HCl solution saturated with AgCl develops oxidation potential of −0.25V. If EoAg/Ag+=−0.799V, the Ksp of AgCl in pure water will be :

A silver wire dipped in $0.1$ M $H C l$ solution saturated with $A g C l$ develops oxidation potential of $- 0.25 V .$ If $E_{A g / A g +}^{o} = - 0.799 V,$ the $K_{s p}$ of $A g C l$ in pure water will be :