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Question
Use the data given in below find out which option the order of reducing power is correct.
`"E"_("Cr"_2"O"_7^(2-)//"Cr"^(3+))^⊖`= 1.33 V `"E"_("Cl"_2//"Cl"^-)^⊖` = 1.36 V
`"E"_("MnO"_4^-//"Mn"^(2+))^⊖` = 1.51 V `"E"_("Cr"^(3+)//"Cr")^⊖` = - 0.74 V
Options
\[\ce{Cr^{3+} < Cl– < Mn^{2+} < Cr}\]
\[\ce{Mn^{2+} < Cl– < Cr^{3+} < Cr}\]
\[\ce{Cr^{3+} < Cl– < Cr_2O_7^{2–} < MnO^{-}4}\]
\[\ce{Mn2+ < Cr3+ < Cl– < Cr}\]
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Solution
\[\ce{Mn^{2+} < Cl– < Cr^{3+} < Cr}\]
Explanation:
Lower the value of standard reduction potential greater will be the reducing power.
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