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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2006

  • question_answer
        The emf of the cell,\[({{E}_{Z{{n}^{2+}}/Zn}}=-0.76\,V)\]\[Zn/Z{{n}^{2+}}(1M)||C{{u}^{2+}}(1M)Cu\]\[({{E}_{C{{u}^{2+}}/Cu}}=+0.34V)\]will be

    A)  \[+1.10\text{ }V\]          

    B)  \[-1.10\text{ }V\]

    C)  \[+\text{ }0.42V\]          

    D)  \[-0.42V\]

    Correct Answer: A

    Solution :

                    Key Idea: (i) Find cathode and anode from given cell reaction. (ii) Calculate emf by formula \[{{E}_{cell}}={{E}_{c}}-{{E}_{a}}\] Given that                 \[Zn/Z{{n}^{2+}}||C{{u}^{2+}}/Cu\]        \[\therefore \]Zn is anode and Cu is cathode Given,   \[Z{{n}^{2+}}/Zn=-0.76\,V\]                 \[C{{u}^{2+}}/Cu=+0.34\,V\]                 \[{{E}_{cell}}={{E}_{c}}-{{E}_{a}}\] \[=0.34-(-0.76)\] \[=0.34+0.76\] \[=1.10V\]


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