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प्रश्न
Use the data given in below find out the most stable ion in its reduced form.
`"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
पर्याय
\[\ce{Cl–}\]
\[\ce{Cr^{3+}}\]
\[\ce{Cr}\]
\[\ce{Mn^{2+}}\]
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उत्तर
\[\ce{Mn^{2+}}\]
Explanation:
\[\ce{Mn^{2+}}\] is most stale in its reduced form due to highest E° value.
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संबंधित प्रश्न
Arrange the following metals in the order in which they displace each other from the solution of their salts.
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\[\ce{MnO^-_4 -> Mn^2+}\]
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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
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
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