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WB JEE Medical WB JEE Medical Solved Paper-2013

  • question_answer
    A button cell used in watches functions as following \[Zn(s)+A{{g}_{2}}O(s)+{{H}_{2}}O(l)2Ag(s)\] \[+Z{{n}^{2+}}(aq)+2O{{H}^{-}}(aq)\] If half cell potentials are \[Z{{n}^{2+}}(aq)+2{{e}^{-}}\to Zn(s);{{E}^{o}}=-0.76\,V\] \[A{{g}_{2}}O(s)+{{H}_{2}}O(l)+2{{e}^{-}}\] \[\to 2Ag(s)+2O{{H}^{-}}(aq),\] \[{{E}^{o}}=0.34\,V\] The cell potential will be

    A)  1.10V

    B)  0.42V

    C)  0.84V 

    D)  1.34V

    Correct Answer: A

    Solution :

    Anode is always the site of oxidation thus anode half cell is \[Z{{n}^{2+}}(aq)+2{{e}^{-}}\xrightarrow{{}}Zn(s);\]\[{{E}^{o}}=-0.76\,V\] Cathode hall cell is \[A{{g}_{2}}O(s)+{{H}_{2}}O(l)+2{{e}^{-}}\xrightarrow{{}}\]\[2\,Ag(s)+2O{{H}^{-}}(aq);{{E}^{o}}=0.34\,V\] \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] \[=0.34-(-0.76)=+\,1.10\,V\]


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