The standard reduction potential of a silver chloride electrode (metal‐sparingly soluble salt electrode) is 0.209 V and for silver electrode is 0.80 V. If the moles of $A g C l$ that can dissolve in 10 L of a 0.01 M $N a C l$ solution is represented as $10^{- z}$ then find the value of Z.[use $\frac{2.303 RT}{F} = 0.0591$ ]

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Detailed Solution

$E_{C l^{-}}^{o} / AgCl / Ag = E_{Ag / Ag}^{o} + \frac{0.0591}{1} \log K_{s p}$

$0.209 = 0.80 + \frac{0.0591}{1} \log K_{s p}$

$K_{s p} = 10^{- 10};$ Let solubility of AgCl in 0.01 M solution is $x$

$10^{- 10} = x (x + 0.01)$

$x = 10^{- 8}$ $∴$ Moles of AgCl dissolved in

$10 L = 10^{- 8} \times 10 = 10^{- 7}$

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