Redox Reactions And Electrochemistry Ques 340
Question: The solubility product of silver iodide is $ 8.3\times {10^{-17}} $ and the standard reduction potential of Ag/ $ A{g^{+}} $ electrode is + 0.8 volts at $ 25{}^\circ C. $ The standard reduction potential of Ag, $ AgI\text{/}{I^{-}} $ electrode from these data is
Options:
A) $ -,0.30,V $
B) + 0.16V
C) +0.10 V
D) $ -,0.16,V $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Anode reaction $ Ag\rightarrow A{g^{+}}+{e^{-}} $
Cathode reaction $ AgI+{e^{-}}\rightarrow Ag+{I^{-}} $
$ E_{cell}^{o}={{( E _{reduction}^{o} )} _{c}}-{{( E _{oxidation}^{o} )} _{a}} $ $ =E _{AgI/{I^{-}}}^{o}-E _{Ag/A{g^{+}}}^{o} $
$ =x-0.8V $ Cell is in equilibrium, thus $ E_{cell}=0 $
$ \therefore E_{cell}=E_{cell}^{o}-\frac{0.059}{1}\log [A{g^{+}}][{I^{-}}] $ $ 0=(x-0.8V)-\frac{0.06}{1}\log K_{sp}(0.059\cong 0.06) $ $ x-0.8V=0.06\log K_{sp} $ $ x-0.8V=0.06\log (8.3\times {10^{-17}}) $ $ =0.06,[log8.3-17\log 10] $ $ =0.06,[0.9191-17]=0.06\times (-16.1) $ $ x-0.8,V=-,0.966 $ $ X=( -0.966+0.8 )V=-0.16V $
