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Question
Use the data given in below find out the most stable oxidised species.
`E^0 (Cr_2O_1^(2-))/(Cr_(3+))` = 1.33 V `E^0 (Cl_2)/(Cl^-)` = 1.36 V
`E^0 (MnO_4^-)/(MN^(2+))` = 1.51 V `E^0 (Cr^(3+))/(Cr)` = – 0.74 V
Options
\[\ce{Cr^{3+}}\]
\[\ce{MnO^{-}4}\]
\[\ce{CrO^{2-}7}\]
\[\ce{Mn^{2+}}\]
Solution
\[\ce{Cr^{3+}}\]
Explanation:
\[\ce{Cr^{3+}/Cr}\] has most negative value of standard reduction potential. Hence, \[\ce{Cr^{3+}}\] is the most oxidized species.
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