Consider the cell Ag | AgBr| KBr||KCl | AgCl | Ag with EMF 0.059 V. Assume that [Br^–] = [Cl^–], Here, conclusion inferred may be

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The ratio of the simultaneous solubilities of AgCl and AgBr in pure water is 1000

Change in conc. of KCl will not effect EMF

It is concentration cell

$K_{sp} (AgCl) > K_{sp} (AgBr)$

answer is B, D.

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Detailed Solution

$\begin{array}{l} E_{RP} = E_{SRP}^{0} - (0 .0591 / 1) \log {({Ag}^{+})}_{a} / {({Ag}^{\oplus})}_{c} \\ E^{0} = 0 \\ [{Br}^{Θ}] = [{Cl}^{Θ}] Then K_{sp} AgCl > K_{sp} AgBr \end{array}$

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