Question

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

Answer 1

–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.