JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    What would be the reduction potential of an electrode at 298 K, which originally contained \[\text{1M}{{\text{K}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}\] solution in acidic buffer solution of pH = 1.0 and which was treated with 50% of the Sn necessary to reduce all \[\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{2-}\] to \[\text{C}{{\text{r}}^{3+}}\]. Assume pH of solution remains constant. Given: \[\text{E}_{\text{C}{{\text{r}}_{\text{2}}}\text{O}\frac{\text{2-}}{\text{7}}\text{/C}{{\text{r}}^{\text{3+}}}\text{,}{{\text{H}}^{\text{+}}}}^{\text{0}}\]\[=1.33V,\log 2=0.3,\] \[\frac{2.303RT}{F}=0.06\]

    A)  1.285 V               

    B)  1.193 V

    C)  1.187 V               

    D)  None of these

    Correct Answer: C

    Solution :

    \[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O\]
    Initial \[co{{n}^{n}}\] 1 0.1 0
    After reaction 0.5 0.1 M 1 M
    \[{{E}_{RP}}=E_{RP}^{0}-\frac{0.06}{6}\log \frac{{{[C{{r}^{3+}}]}^{2}}}{[C{{r}_{2}}O_{7}^{2-}]{{[{{H}^{+}}]}^{14}}}\] \[{{E}_{RP}}=1.33-\frac{0.06}{6}\log \frac{1}{(0.5){{(0.1)}^{14}}}\] \[=1.33-\frac{0.06}{6}\log (2\times {{10}^{14}})=1.187\,\text{V}\]


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