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NEET Sample Paper NEET Sample Test Paper-6

  • question_answer
    Consider the following cell reaction, \[2F{{e}_{(s)}}+{{O}_{2(g)}}+4{{H}^{+}}_{(aq)}\xrightarrow{{}}2F{{e}^{2+}}_{(aq)}+2{{H}_{2}}{{O}_{(l)}},{{E}^{o}}=1.67V,\] At \[\left[ F{{e}^{2+}} \right]={{10}^{-3}}M,\] \[P({{O}_{2}})=0.1\] atm and \[pH=3,\]the cell potential at \[{{25}^{o}}C\] is

    A)  \[1.47\,V\]                  

    B)  \[1.77V\]

    C)  \[1.87V\]                    

    D)  \[1.57V\]

    Correct Answer: D

    Solution :

    The half reaction are \[F{{e}_{(s)}}\xrightarrow{{}}F{{e}^{2+}}_{(aq)}+2{{e}^{-}}\times 2\] \[{{O}_{2(g)}}+4{{H}^{+}}+4{{e}^{-}}\xrightarrow{{}}2{{H}_{2}}O\] \[2F{{e}_{(s)}}+{{O}_{2(g)}}+4{{H}^{+}}\xrightarrow{{}}2F{{e}^{2+}}_{(aq)}+2{{H}_{2}}{{O}_{(l)}}\]\[E={{E}^{o}}-\frac{0.059}{4}\log \frac{{{\left( {{10}^{-3}} \right)}^{2}}}{{{\left( {{10}^{-3}} \right)}^{4}}\,(0.1)}=1.57\,V\]


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