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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
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.
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