J & K CET Engineering J and K - CET Engineering Solved Paper-2011

  • question_answer The correct order of reducing character of alkali metals is

    A)  \[Rb<K<Na<Li\]

    B)  \[Li<Na<K<Rb\]

    C)  \[Na<K<Rb<Li\]

    D)  \[Rb<Na<K<Li\]

    Correct Answer: C

    Solution :

     The alkali metals are strong reducing agents, lithium the most and sodium the least powerful. The standard electrode potential \[({{E}^{o}})\] which measures the reducing power represents the overall change \[M(s)\xrightarrow{{}}M(g)\] Sublimation enthalpy \[M(g)\xrightarrow{{}}{{M}^{+}}(g)+{{e}^{-}}\] ionization enthalpy \[{{M}^{+}}(g)+{{H}_{2}}O\xrightarrow{{}}{{M}^{+}}(aq)\] hydration enthalpy With the small size of its ion, lithium has the highest hydration enthalpy which accounts for its high negative \[E{}^\circ \] value and its high reducing power.
    Metal Li Na K Rb
    \[{{E}^{\text{o}-}}/V\]for
    \[({{M}^{+}}/M)\] \[-3.04\] \[-2.714\] \[-2.925\] \[-2.930\]
    Hence, the order of reducing character of alkali metals is \[Na<K<Rb<Li.\]

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