A) \[-\,0.30\,V\]
B) + 0.16V
C) +0.10 V
D) \[-\,0.16\,V\]
Correct Answer: D
Solution :
[d] Anode reaction \[Ag\rightleftharpoons A{{g}^{+}}+{{e}^{-}}\] Cathode reaction \[AgI+{{e}^{-}}\rightleftharpoons Ag+{{I}^{-}}\] \[E_{cell}^{o}={{\left( E_{reduction}^{o} \right)}_{c}}-{{\left( E_{oxidation}^{o} \right)}_{a}}\] \[=E_{AgI/{{I}^{-}}}^{o}-E_{Ag/A{{g}^{+}}}^{o}\] \[=x-0.8V\] Cell is in equilibrium, thus \[{{E}_{cell}}=0\] \[\therefore {{E}_{cell}}=E_{cell}^{o}-\frac{0.059}{1}\log [A{{g}^{+}}][{{I}^{-}}]\] \[0=(x-0.8V)-\frac{0.06}{1}\log {{K}_{sp}}(0.059\cong 0.06)\]\[x-0.8V=0.06\log {{K}_{sp}}\] \[x-0.8V=0.06\log (8.3\times {{10}^{-17}})\] \[=0.06\,[log8.3-17\log 10]\] \[=0.06\,[0.9191-17]=0.06\times (-16.1)\] \[x-0.8\,V=-\,0.966\] \[X=\left( -0.966+0.8 \right)V=-0.16V\]
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