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EEJ MAIN Chemistry
QUESTION #7110
Question 1
Given the half-cell reactions:
1. $Mn^{2+} + 2e^{-} \rightarrow Mn$; $E^{0} = -1.18\text{ V}$
2. $Mn^{3+} + e^{-} \rightarrow Mn^{2+}$; $E^{0} = +1.51\text{ V}$
Determine the $E^{0}$ for the disproportionation reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ and its feasibility.
1. $Mn^{2+} + 2e^{-} \rightarrow Mn$; $E^{0} = -1.18\text{ V}$
2. $Mn^{3+} + e^{-} \rightarrow Mn^{2+}$; $E^{0} = +1.51\text{ V}$
Determine the $E^{0}$ for the disproportionation reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ and its feasibility.
Correct Answer Explanation
The target reaction is $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$. This is the sum of:
Reduction: $Mn^{2+} + 2e^{-} \rightarrow Mn$ ($E^{0}_{red} = -1.18\text{ V}$)
Oxidation: $2Mn^{2+} \rightarrow 2Mn^{3+} + 2e^{-}$ ($E^{0}_{ox} = -1.51\text{ V}$)
$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = -1.18 - 1.51 = -2.69\text{ V}$. Since $E^{0}_{cell}$ is negative, the reaction is non-spontaneous.
Reduction: $Mn^{2+} + 2e^{-} \rightarrow Mn$ ($E^{0}_{red} = -1.18\text{ V}$)
Oxidation: $2Mn^{2+} \rightarrow 2Mn^{3+} + 2e^{-}$ ($E^{0}_{ox} = -1.51\text{ V}$)
$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = -1.18 - 1.51 = -2.69\text{ V}$. Since $E^{0}_{cell}$ is negative, the reaction is non-spontaneous.
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