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
B
1.5×108
C
3.2×1010
D
1.5×1010

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

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