A) \[0.85V\]
B) \[0.79V\]
C) \[0.65V\]
D) \[0.35V\]
Correct Answer: B
Solution :
| the cell reaction is: \[\underset{\underset{Equib.conc.}{\mathop{initial}}\,}{\mathop{2Hg}}\,+\underset{\underset{1.0\times {{10}^{-3}}-2x}{\mathop{1.0\times {{10}^{-3}}M}}\,}{\mathop{2F{{e}^{3+}}}}\,\xrightarrow{{}}H\underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{g_{2}^{2+}}}\,+2\underset{\begin{smallmatrix} 0 \\ 2x \end{smallmatrix}}{\mathop{F}}\,{{e}^{2+}}\] |
| \[{{[F{{e}^{3+}}]}_{Equi}}=1.0\times {{10}^{-3}}-2x\] |
| \[=1.0\times {{10}^{-3}}\times \frac{5}{100}=5\times {{10}^{-5}}\operatorname{M}\Rightarrow x\]\[=4.75\times {{10}^{-4}}\operatorname{M}\] |
| \[[Hg_{2}^{2+}]=x=4.75\times {{10}^{-4}}M\] |
| \[[F{{e}^{2+}}]=2x=2\times 4.75\times {{10}^{-4}}=9.50\times {{10}^{-4}}M\] |
| Oxidation reaction: \[2Hg\xrightarrow{{}}Hg_{2}^{2+}+2{{e}^{-}}\] |
| Reduction reaction: \[2F{{e}^{3+}}+2{{e}^{-}}\xrightarrow{{}}2F{{e}^{2+}}\] |
| \[{{K}_{eq}}\]for the cell reaction \[=\frac{[Hg_{2}^{2}]{{[F{{e}^{2+}}]}^{2}}}{{{[F{{e}^{3+}}]}^{2}}}\] \[=\frac{4.75\times {{10}^{-4}}\times {{(9.5\times {{10}^{-4}})}^{2}}}{{{(5\times {{10}^{-5}})}^{2}}}=0.1715\operatorname{M}\] |
| \[{{E}_{cell}}=E_{cell}^{{}^\circ }-\frac{0.0591}{n}\log {{K}_{e}}\][At equib. \[{{E}_{\operatorname{cell}}}=0.0]\] |
| \[E_{cell}^{{}^\circ }=-\frac{0.0591}{2}\log 0.1715\] |
| \[{{E}^{{}^\circ }}_{\operatorname{Hg}_{2}^{2+}/Hg}=E{{{}^\circ }_{F{{e}^{3+}}/F{{e}^{2+}}}}-\frac{0.0591}{2}\log 0.1715\] |
| \[=0.77-0.0295\log 0.1715=0.79V.\] |
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