Solved papers for JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

done JEE Main Online Paper (Held on 9 April 2013) Total Questions - 1

• question_answer1)                 Electrode potentials \[\left( {{E}^{0}} \right)\] are given below:                 \[{{\operatorname{Cu}}^{+}}+\operatorname{Cu}=+0.52V,\]                 \[F{{e}^{3+}}/F{{e}^{2+}}=+0.77V,\]                 \[\frac{1}{2}{{\operatorname{I}}_{2}}\left( s \right)/{{I}^{-}}=+0.54V,\]                                                 \[A{{g}^{+}}/Ag=+0.88V.\]                 Based on the above potentials, strongest oxidizing agent will be:                   JEE Main Online Paper (Held On 09 April 2013)
  A)
                  \[{{\operatorname{Cu}}^{+}}\]                
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  B)
                  \[{{\operatorname{Fe}}^{3}}^{+}\]                
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  C)
                  \[{{\operatorname{Ag}}^{+}}\]                
  done clear
  D)
                  \[{{\operatorname{I}}_{2}}\]                
  done clear
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