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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 43

  • question_answer
    The standard free energy change for the following reaction is - 210 KJ. What is me standard cell potential? \[2{{H}_{2}}{{O}_{2}}\left( aq \right)\xrightarrow{{}}2{{H}_{2}}O(\ell )+{{O}_{2}}\left( g \right)\]

    A) + 0.752          

    B)        + 1.09

    C) + 0.420           

    D)        + 0.640

    Correct Answer: B

    Solution :

    [b] n = 2 for given reaction, so \[\Delta G{}^\circ =-nFE_{cell}^{{}^\circ }\] \[-210\times 1000=-2\times 96500\times E_{cell}^{{}^\circ }\] \[E_{cell}^{{}^\circ }=1.09\,V\]


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