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BVP Medical BVP Medical Solved Paper-2004

  • question_answer
    \[C{{r}_{2}}O_{7}^{2-}+{{I}^{-}}\xrightarrow{{}}{{I}_{2}}+C{{r}^{3+}}\]\[{{E}^{o}}_{cell}=0.79V\]\[{{E}^{o}}_{C{{r}_{2}}O_{7}^{2-}}=1.33V,\,{{E}^{o}}_{{{I}_{2}}}\]is:

    A)  \[-0.10V\]          

    B)  \[+0.18V\]

    C)  \[-0.54V\]          

    D)  \[0.54V\]

    Correct Answer: D

    Solution :

                    \[\overset{+6}{\mathop{C{{r}_{2}}O_{7}^{2-}}}\,+{{I}^{-}}\xrightarrow{{}}{{I}_{2}}+C{{r}^{3+}}\] Here, \[{{I}^{-}}\] is undergoing oxidation while \[C{{r}_{2}}O_{7}^{2-}\] is undergoing reduction, hence \[{{I}_{2}}\] will form anode and \[C{{r}^{3+}}\] will form cathode in the cell. Hence,                 \[{{E}^{o}}_{cell}{{E}^{o}}_{cathode}-{{E}^{o}}_{anode}\]                 \[\therefore \]  \[0.79=1.33-{{E}^{o}}_{{{I}_{2}}}\]                 \[\therefore \]  \[{{E}^{o}}_{{{I}_{2}}}=1.33-0.79\]                                 \[=+0.54V\]


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