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Two metals M1 and M2 have reduction potential values of −xV and +yV respectively. Which will liberate H2 and H2SO4.

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प्रश्न

Two metals M1 and M2 have reduction potential values of −xV and +yV respectively. Which will liberate H2 and H2SO4.

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उत्तर

Metals having negative reduction potential act as powerful reducing agents. Since M1 has – xV, therefore M1 easily liberates H2 in H2SO4. Metals having higher oxidation potential will liberate H2 from H2SO4. Hence, the metal M1 having +xV, oxidation potential will liberate H2 from H2SO4.

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अध्याय 9: Electro Chemistry - Evaluation [पृष्ठ ६६]

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सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 9 Electro Chemistry
Evaluation | Q 16. | पृष्ठ ६६

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`"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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`"E"_("MnO"_4^-//"Mn"^(2+))^⊖` = 1.51 V `"E"_("Cr"^(3+)//"Cr")^⊖` = - 0.74 V


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