Tamil Nadu Board of Secondary EducationHSC Science Class 12

Consider the following half cell reactions: MnX2++2eX−⟶Mn E° = –1.18 V MnX2+⟶MnX2++eX− E° = –1.51 V The E° for the reaction 3MnX2+⟶Mn+2MnX3+, and the possibility of the forward reaction are - Chemistry

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MCQ

Consider the following half cell reactions:

\[\ce{Mn^{2+} + 2e^- -> Mn}\] E0 = –1.18 V

\[\ce{Mn^{2+} -> Mn^{2+} + e^-}\] E0 = –1.51 V

The E0 for the reaction \[\ce{3Mn^{2+} -> Mn + 2Mn^{3+}}\], and the possibility of the forward reaction are respectively.

Options

  • 2.69 V and spontaneous

  • –2.69 V and non spontaneous

  • 0.33 V and Spontaneous

  • 4.18 V and non spontaneous

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Solution

–2.69 V and non spontaneous

Explanation:

\[\ce{Mn^{2+} + 2e^- -> Mn (E^0_{red}) = -1.18 V}\] 

\[\ce{2[MN^{2+} -> MN^{3+} + e^-] (E^0_{ox}) = -1.51 V}\]

 \[\ce{3Mn^{2+} -> Mn + 2Mn^{3+}}\] `"E"_"cell"^0` = ?

`"E"_"cell"^0` = `("E"_"ox"^0) + ("E"_"red"^0)`

= – 1.51 – 1.18 and non spontaneous

= – 2.69 V

Since E0 is – ve ∆G is +ve and the given forward cell reaction is non spontaneous.

Concept: Thermodynamics of Cell Reactions
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Chapter 9: Electro Chemistry - Evaluation [Page 62]

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