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Q.

The standard reduction potential of a silver chloride electrode (metal-sparingly soluble salt electrode) is 0.209V and for silver electrodes $0.80 V$ . If the moles of AgCl than can dissolve in 10L of $0.01 M N a C l$ solution is represented as $10^{- x}$ . Then x is

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answer is 7.

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Detailed Solution

$E_{C l^{-} / A g C l / A g}^{0} = E_{A g^{+} / A g}^{0} + \frac{0.0591}{1} \log K_{S P} 0.209 = 0.80 + \frac{0.0591}{1} \log K_{S P} K_{S P} = 10^{- 10}$
Let solubility of AgCl in 0.01M solution is y then $10^{- 10} = y (y + 0.01) \Rightarrow y = 10^{- 8}$
$∴$ moles of AgCl dissolved in $10 L = 10^{- 8} \times 10 = 10^{- 7}$

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