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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    For the cell \[Z{{n}_{(s)}}|Z{{n}^{2+}}_{(aq)}||{{M}^{x+}}_{(aq)}|\,{{M}_{(s)}},\]different half cells and their standard electrode potentials are given below :
    \[{{M}^{x+}}_{(aq)}/\,{{M}_{(s)}}\] \[A{{u}^{+}}_{(aq)}\]\[/\,A{{u}_{(s)}}\] \[A{{u}^{+}}_{(aq)}/\]\[\,A{{g}_{(s)}}\] \[F{{e}^{3+}}_{(aq)}/\]\[F{{e}^{2+}}_{(aq)}\] \[F{{e}^{2+}}_{(aq)}/\]\[F{{e}_{(s)}}\]
    \[{{E}^{o}}_{{{M}^{x+}}/M}(V)\] 1.40 0.80 0.77 -0.44
    If \[{{E}^{o}}_{Z{{n}^{2+}}/Zn}=-0.76V,\]which cathode will give a maximum value of \[{{E}^{o}}_{cell}\] per electron transferred? [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[A{{u}^{3+}}/Au\]                            

    B) \[F{{e}^{3+}}/F{{e}^{2+}}\]

    C) \[A{{g}^{+}}/Ag\]                  

    D)   \[F{{e}^{2+}}/Fe\]

    Correct Answer: C

    Solution :

    \[{{E}^{o}}_{cell}={{E}^{o}}_{cathode}-{{E}^{o}}_{anode}\] (i) For \[A{{u}^{3+}}/Au\]: \[{{E}^{o}}_{cell}=1.40-(-0.76)=2.16V;\]\[\frac{2.16}{3}=0.72V\] (ii) For\[A{{g}^{+}}/Ag\]: \[{{E}^{o}}_{cell}=0.80-(-0.76)=1.56V\] (iii) For \[F{{e}^{3+}}/F{{e}^{2+}}:\] \[{{E}^{o}}_{cell}=0.77-(-0.76)=1.53V\] (iv) For \[F{{e}^{2+}}/Fe:\] \[{{E}^{o}}_{cell}=-0.44-(-0.76)=0.32V;\] \[\frac{0.32}{2}=0.16V\]\[{{E}^{o}}_{cell}\]is maximum for\[{{E}^{o}}_{A{{g}^{+}}_{(ag)}/A{{g}_{(s)}}.}\]


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