CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The standard reduction potentials for two reactions are given below:
$${\text{AgCl}}\left( {\text{s}} \right){\text{ + }}{{\text{e}}^{\text{ - }}} \to {\text{Ag}}\left( {\text{s}} \right) + {\text{C}}{{\text{l}}^{\text{ - }}}\left( {{\text{aq}}} \right)$$      $${\text{E}}^\circ  = {\text{0}}{\text{.22V}}$$
$${\text{A}}{{\text{g}}^ + }\left( {aq} \right) + {e^ - } \to Ag\left( s \right)$$                    $${\text{E}}^\circ  = {\text{0}}{\text{.80V}}$$
The solubility product of $$AgCl$$ under standard conditions of temperature (298 K) is given by:


A
1.6×105
loader
B
1.5×108
loader
C
3.2×1010
loader
D
1.5×1010
loader

Solution

The correct option is C $$1.5 \times {10^{ - 10}}$$
Solution:- (D) $$1.5 \times {10}^{-10}$$
$${AgCl}_{\left( s \right)} + {e}^{-} \longrightarrow {Ag}_{\left( s \right)} + {{Cl}^{-}}_{\left( aq. \right)} \quad {E}^{°} = 0.22 \; V ..... \left( 1 \right)$$
$${{Ag}^{+}}_{\left( aq. \right)} + {e}^{-} \longrightarrow {Ag}_{\left( s \right)} \quad {E}^{°} = 0.80 \; V$$
$${Ag}_{\left( s \right)} \longrightarrow {{Ag}^{+}}_{\left( aq. \right)} + {e}^{-} ..... {E}^{°} = -0.80\; V ..... \left( 2 \right)$$
Adding equation $$\left( 1 \right) \& \left( 2 \right)$$, we get
$${AgCl}_{\left( s \right)} \longrightarrow {{Ag}^{+}}_{\left( aq. \right)} + {{Cl}^{-}}_{\left( aq. \right)} \quad {E}^{°} = - 0.58 \; V$$
From Nernst equation,
$${E}_{cell} = {{E}^{°}}_{cell} - \cfrac{RT}{nF} \ln{K}$$
$$0 = -0.58 - \cfrac{8.314 \times 298 \times 2.303}{1 \times 96500} \log{K}$$
$$\Rightarrow \log{K} = \cfrac{0.58}{0.0591} = -9.81$$
$$\Rightarrow K = 1.5 \times {10}^{-10}$$
Hence the solubility product of $$AgCl$$ is $$1.5 \times {10}^{-10}$$.

Chemistry

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image