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)