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. \[{{E}_{\operatorname{cell}}}=0.0]\] |
\[E_{cell}^{{}^\circ }=-\frac{0.0591}{2}\log 0.1715\] |
\[{{E}^{{}^\circ }}_{\operatorname{Hg}_{2}^{2+}/Hg}=E{{{}^\circ }_{F{{e}^{3+}}/F{{e}^{2+}}}}-\frac{0.0591}{2}\log 0.1715\] |
\[=0.77-0.0295\log 0.1715=0.79V.\] |
You need to login to perform this action.
You will be redirected in
3 sec