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MGIMS WARDHA MGIMS WARDHA Solved Paper-2011

  • question_answer
    The hydrogen electrode is dipped in a solution of\[pH=3\]at\[25{}^\circ C\]. The potential of the cell would be (the value of 2.303 RT/F is 0.059 V)

    A)  0.177V                                

    B)  0.087V

    C)  \[-0.177V\]                       

    D)  0.059V

    Correct Answer: C

    Solution :

                     \[{{H}^{+}}+{{e}^{-}}\xrightarrow{{}}\frac{1}{2}{{H}_{2}}\] \[E={{E}^{o}}-\frac{0.059}{n}\log \frac{1}{[{{H}^{+}}]}=0-\frac{0.059}{1}pH\] \[E=-0.059\times 3=-0.177V\]


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