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
B
8.12
C
+8.612
D
37.83

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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