A) \[+0.779\,V\]
B) \[+0.935\,V\]
C) \[+1.015\,V\]
D) \[+1.134\,V\]
Correct Answer: C
Solution :
At anode, \[P{{b}_{(s)}}\xrightarrow{\,}\,P{{b}^{2+}}+2{{e}^{-}}\] At cathode, \[\left( F{{e}^{3+}}+1{{e}^{-}}\xrightarrow{\,}F{{e}^{2+}} \right)\times 2\] Cell reaction: \[Pb+2F{{e}^{3+}}\xrightarrow{\ }\,P{{b}^{2+}}+2F{{e}^{2+}}\] \[{{E}_{cell}}=E_{cell}^{0}-\frac{0.0591}{2}\log \,\frac{[P{{b}^{2+}}]{{[F{{e}^{2+}}]}^{2}}}{{{[F{{e}^{3+}}]}^{2}}}\] \[=+0.897-\frac{0.0591}{2}\log \frac{(0.040){{(0.010)}^{2}}}{{{(0.20)}^{2}}}\] \[=0.897-0.0295\log \frac{0.04\times 0.01\times 0.01}{0.2\times 0.2}\] \[=0.897-0.0295\log {{10}^{-4}}\] \[=0.897-0.0295\times -4\log \,10\] \[=0.897-0.0295\times -4\] \[=1.015\,V\]
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