Given the following half reactions Ag- aq -e- Ag s E0-0-799V AgCl s -e- Ag s -Cl- aq E0-0-222VWhat is the molar solubility of AgCl- -Given antilog -9-758 - 1-743- 10-10-

The correct option is B 1.32× 10^ 5 M AgCl_(s)+e→ Ag_(s)+Cl #160; #160; △ G_1^o= 1F(0.222) #160; Ag(s)→ Ag^+e^ AgCl→ Ag^+Cl^ △ G^2_o= ...

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Byju's Answer

Standard XII

Chemistry

Nernst Equation

Given the fol...

Question

Given the following half reactions
$A g^{+} (a q) + e^{-} \to A g (s)$ $E^{0} = + 0.799 V$
$A g C l (s) + e^{-} \to A g (s) + C l^{-} (a q)$ $E^{0} = + 0.222 V$
What is the molar solubility of AgCl? [Given antilog (-9.758) = $1.743 \times 10^{- 10}$ ]

4.46 \times 10^{- 9}

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1.32 \times 10^{- 5}

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1.79 \times 10^{- 10}

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0.577 M

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Solution

The correct option is B $1.32 \times 10^{- 5}$ M
$A g C l_{(s)} + e \to A g_{(s)} + C l △ G_{1}^{o} = - 1 F (0.222)$
$\frac{A g (s) \to A g^{+} e^{-}}{A g C l \to A g^{+} C l^{-}} △ G_{o}^{2} = - 1 F (- 0.7999)$
$Δ G_{3}^{0} = Δ G_{1}^{0} + Δ G_{0}^{2} = - 1 F (0.222) - 1 F (- 0.7999) = 0.5779 F = 0.5779 \times 96500 = 55767.35$
$Δ G_{3}^{0} = - R T l n K s p$
$55767.35 = - 8.314 \times 298 \times l n K_{s p}$
$l n K_{s p} = 22.51$
$K s p = 8.05 \times 10^{- 9}$
The molar solubility of $A g C l$ is $S = \sqrt{K s p} = \sqrt{8.05 \times 10^{- 9}} = 1.32 \times 10^{- 5} M$

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Given the following half reactionsAg+(aq)+e−→Ag(s) E0=+0.799VAgCl(s)+e−→Ag(s)+Cl−(aq) E0=+0.222VWhat is the molar solubility of AgCl? [Given antilog (-9.758) = 1.743×10−10]

Given the following half reactions
$A g^{+} (a q) + e^{-} \to A g (s)$ $E^{0} = + 0.799 V$
$A g C l (s) + e^{-} \to A g (s) + C l^{-} (a q)$ $E^{0} = + 0.222 V$
What is the molar solubility of AgCl? [Given antilog (-9.758) = $1.743 \times 10^{- 10}$ ]