For the galvanic cell,
$A g | A g C l (s), K C l (0.2 M) | | K B r (0.001 M), A g B r (s) | A g$
Calculate the emf generated and assign correct polarity to each electrode for the spontaneous process after taking into account the cell reaction at $25^{\circ} C$ .
Given, $K_{s p A g C l} = 2.8 \times 10^{- 10}; K_{s p A g B r} = 3.3 \times 10^{- 13}$ .

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Solution

$E_{c e l l} = E_{O x i d . p o t . L H S e l e c t r o d e} + E_{R e d . p o t . R H S e l e c t r o d e}$
$= E_{O x i d . p o t . A g / A g^{+}}^{\circ} - 0.0591 log [A g^{+}]_{L H S} + E_{R e d . p o t . A g^{+} / A g}^{\circ} + 0.0591 l o g [A g^{+}]_{R H S}$
$= 0.0591 log \frac{[A g^{+}]_{R H S}}{[A g^{+}]_{L H S}}$ [Since $E_{A g / A g^{+}}^{\circ} + E_{A g^{+} / A g}^{\circ} = 0$ ]
$= 0.0591 log \frac{\frac{K_{s p A g B r}}{[B r^{-}]}}{\frac{K_{s p A g C l}}{[C l^{-}]}}$
$= 0.0591 log \frac{3.3 \times 10^{- 13}}{0.001} \times \frac{0.2}{2.8 \times 10^{- 10}}$
$= 0.0371 v o l t$
The cell potential is negative; therefore, the cell reaction is non-spontaneous. For spontaneous reaction emf should be positive. Therefore, the correct cell reaction is
$A g | A g B r A n o d e; K B r | | K C l, A g C l C a t h o d e | A g$ .

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Q. For the galvanic cell,

A g | A g C l (s), K C l (0.2 M) | | K B r (0.001 M), A g B r (s) | A g

Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at

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Ag 1 AgCl(s), KCl(0.2 M)II KBr (0.001 M), AgBr(s) I Ag, the EMF (in V) generated for a spontaneous process after taking into account the cell reaction at $25^{\circ} C$ will be $37 \times 10^{- x}$ , then what is the value of x?

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K_{s p} (A g C l) = 2.8 \times 10^{- 10}

K_{s p} (A g B r) = 3.3 \times 10^{- 13}

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For the galvanic cell,Ag|AgCl(s),KCl(0.2 M)||KBr(0.001 M),AgBr(s)|AgCalculate the emf generated and assign correct polarity to each electrode for the spontaneous process after taking into account the cell reaction at 25∘C.Given, Ksp AgCl=2.8×10−10;Ksp AgBr=3.3×10−13.