Redox Reactions And Electrochemistry Question 562
Question: Consider the following standard electrode potentials and calculate the equilibrium constant at 25° C for the indicated disproportion nation reaction:
$ 3M{n^{2+}}(aq)\xrightarrow{{}}Mn(s)+2M{n^{3+}}(aq) $ $ M{n^{3+}}( aq )+{e^{-}}\xrightarrow{{}}M{n^{2+}}( aq );{E^{{}^\circ }}=1.51V $ $ M{n^{2+}}(aq)+2{e^{-}}\xrightarrow{{}}Mn(s);E{}^\circ =-1.185V $
Options:
A) $ 1.2\times {10^{-43}} $
B) $ 2.4\times {10^{-73}} $
C) $ 6.3\times {10^{-92}} $
D) $ 1.5\times {10^{-62}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ 2M{n^{2+}}\xrightarrow{{}}2M{n^{3+}}+2{e^{-}},\Delta G_1^{{}^\circ } $ $ \frac{M{n^{2+}}+2{e^{-}}\to Mn,\Delta G_2^{{}^\circ }}{3M{n^{2+}}( aq )\to Mn( s )+2M{n^{3+}}( aq )} $ $ -2\times F\times E{{{}^\circ }_3}=-2\times F\times [ -1.51 ]-2\times F\times ( -1.185 ) $ $ E_3^{{}^\circ }=-2.695 $ $ E_3^{{}^\circ }=+\frac{0.0591}{2}log,K{ _{eq}};K{ _{eq}}\simeq 6.3\times {10^{-92}}. $
BETA
