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NEET Sample Paper NEET Sample Test Paper-83

  • question_answer
    Calculate the value of \[E_{cell}^{{}^\circ }\] if standard reduction potential of \[F{{e}^{2\,+}}\] /Fe and \[S{{n}^{2\,+}}\]/Sn electrodes are - 0.44 V and - 0.14 V respectively. The cell reaction is\[{{\operatorname{Fe}}^{2\,+}}+\,\,Sn\,\,\xrightarrow{{}}\,\,Fe+S{{n}^{2\,+}}\]

    A)  0.42 V             

    B)  - 0.42 V

    C)  - 0.30 V                       

    D)  -1.10 V

    Correct Answer: C

    Solution :

    Half-cell reactions are (i) \[F{{e}^{2\,+}}+2e\xrightarrow{{}}\,Fe\]\[E_{F{{e}^{2+}}/Fe}^{{}^\circ }=-\,0.44\,V\] (ii) \[\operatorname{Sn} \xrightarrow{{}}\,\,S{{n}^{2+}}\,+\, 2e\]\[E_{Sn/S{{n}^{2+}}}^{{}^\circ }=+\,0.14\,V\] Adding Eqs. (i) and (ii), \[E{}^\circ =E_{F{{e}^{2+}}/Fe}^{{}^\circ }+E_{Sn/S{{n}^{2+}}}^{{}^\circ }\] \[=\,\,-0.44 +0.14=-\,0.30V\]


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