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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Use the data given in below find out which option the order of reducing power is correct. E0cr2O12-Cr3+ = 1.33 V E0Cl2Cl- = 1.36 V E0MnO4-MN2+ = 1.51 V E0Cr3+Cr = – 0.74 V

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प्रश्न

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

पर्याय

  • \[\ce{Cr^{3+} < Cl– < Mn^{2+} < Cr}\]

  • \[\ce{Mn^{2+} < Cl– < Cr^{3+} < Cr}\]

  • \[\ce{Cr^{3+} < Cl– < Cr_2O_7^{2–} < MnO^{-}4}\]

  • \[\ce{Mn2+ < Cr3+ < Cl– < Cr}\]

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उत्तर

\[\ce{Mn^{2+} < Cl– < Cr^{3+} < Cr}\]

Explanation:

Lower the value of standard reduction potential greater will be the reducing power.

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पाठ 3: Electrochemistry - Exercises [पृष्ठ ३५]

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एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 12
पाठ 3 Electrochemistry
Exercises | Q I. 10. | पृष्ठ ३५

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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


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