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Nernst Equation

From the foll...

Question

From the following information, calculate the solubility product of AgBr.
$A g B r (s) + e^{-} \to A g (s) + B r^{⊖} (a q); E^{⊖} = 0.07 V$
$A g^{\oplus} (a q) + e^{-} \to A g (s); E^{⊖} = 0.80 V$

4 \times 10^{- 13}

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4 \times 10^{- 10}

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4 \times 10^{- 17}

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4 \times 10^{- 7}

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Solution

The correct option is A $4 \times 10^{- 13}$
Cathode : $A g B r (s) + e^{-} \to A g (s) + B r^{⊖} (a q); E_{r e d}^{⊖} = 0.07 V$
Anode : $A g (s) \to A g^{\oplus} (a q) + e^{-}; E_{r e d}^{⊖} = 0.08 V$
$A g B r (a q) \to A g^{\oplus} (a q) + B r^{⊖} (a q)$
At equilibrium : $E_{c e l l}^{⊖} = 0.07 - 0.8 = \frac{0.059}{1} l o g K_{s p}$
$\Rightarrow l o g K_{s p} = - 12.37 \Rightarrow K_{s p} = 4 \times 10^{- 13}$

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