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

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

#### 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
Is there an error in this question or solution?
Chapter 9: Electro Chemistry - Evaluation [Page 62]

Share