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Question

Given the data at $$25^0\ C$$,
$$Ag + I^- \rightarrow AgI + e^-; E^0 = 0.152 V$$
$$Ag \rightarrow Ag^+ +e^-; E^0 = -0.800 V$$
What is the value of log $$K_{sp}$$ for $$AgI$$?
$$(Given:2.303 RT /F = 0.059 V)$$


A
16.13
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B
8.12
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C
+8.612
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D
37.83
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Solution

The correct option is C $$-16.13$$
At cathode: $$Ag^++e^- \longrightarrow Ag$$           $$E^o_{red}=0.800V$$
At anode: $$Ag+I^- \longrightarrow AgI+e^-$$    $$E^o_{oxd}=0.152V$$
               ______________________
Cell reaction $$Ag^++I^- \longrightarrow AgI$$     $$E^o_{cell}=0.8+0.152V= 0.952V$$
Note: The reaction will take place at cathode whose reduction potential is higher.

Using Nernst Equatiuon,
$$E_{cell}=E^o_{cell}-\cfrac {0.0591}{1}\log \cfrac {1}{[Ag^+][I^-]}$$

At $$Eqb^m, \Delta G=0$$, as $$\Delta G= -nF E_{cell}$$

so, $$E_{cell}=0$$
$$[Ag^+][I^-]=K_{sp}$$

$$\Rightarrow 0= E^o_{cell}-0.0591 \log \cfrac {1}{K_{sp}}$$

$$\Rightarrow 0= 0.952V+0.0591 \log K_{sp}$$

$$\Rightarrow \log K_{sp}=\cfrac {-0.952 V}{0.0591}=-16.13V$$

$$\Rightarrow \log K_{sp}= -16.13V$$

Chemistry

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