Redox Reactions And Electrochemistry Question 186
Question: The standard reduction potentials for $ Z{n^{2+}} $ /Zn, $ N{i^{2+}} $ /Ni and $ F{e^{2+}} $ /Fe are $ -0.74, $ $ -0.22V $ and $ -0.44V $ respectively. The reaction $ X+{Y^{2+}}\to {X^{2+}}+Y $ Will be spontaneous when
Options:
A) X = Ni, Y = Fe
B) X = Ni, Y = Zn
C) X = Fe, Y = Zn
D) X = Zn, Y = Ni
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Answer:
Correct Answer: D
Solution:
[d] For spontaneous reaction, $ E_{cell}^{o}>0 $ $ X=Ni,Y=Fe $ $ Ni+F{e^{2+}}\to Fe+N{i^{2+}} $ $ E_{cell}^{o}=E_{Ni/N{i^{2+}}}^{o}+E_{F{e^{2+}}/Fe}^{o} $ $ =-E_{N{i^{2+}}/Ni}^{o}+E_{F{e^{2+}}/Fe}^{o} $ $ =-(-0.2)-0.4=-0.2V $ $ E_{cell}^{o}<0 $ thus, non-spontaneous
$ X=Ni,Y=Zn $ $ Ni+Z{n^{2+}}\to Zn+N{i^{2+}} $ $ E_{cell}^{o}=E_{Ni/N{i^{2+}}}^{o}+E_{Z{n^{2+}}/Zn}^{o} $ $ =-E_{N{i^{2+}}/Ni}^{0}+E_{Z{n^{2+}}/Zn}^{0} $ $ =-(-0.22)-0.74 $ $ =+0.22-0.74=-0.52V $ $ E_{cell}^{o}<0 $ thus, non-spontaneous
$ X=Fe,Y=Zn $ $ Fe+Z{n^{2+}}\to F{e^{2+}}+Zn $ $ E_{cell}^{o}=E_{Fe/F{e^{2+}}}^{o}+E_{Z{n^{2+}}/Zn}^{o} $ $ =-E_{F{e^{2+}}/Fe}^{o}+E_{Z{n^{2+}}/Zn}^{o} $ $ =-,(-0.44)-0.74 $ $ =+0.44-0.74=-0.3V $ $ E_{cell}^{o}<0 $ , Thus, non-spontaneous
$ X=Zn,Y=Ni $ $ Zn+N{i^{2+}}\to Z{n^{2+}}+Ni $ $ E_{cell}^{o}=E_{Zn/Z{n^{2+}}}^{o}+E_{N{i^{2+}}/Ni}^{o} $ $ =-E_{Z{n^{2+}}/Zn}^{o}+E_{N{i^{2+}}/Ni}^{o} $ $ =-(-0.74)-(0.22) $ $ =0.74-0.22=0.52V $ $ E_{cell}^{o}>0 $ Thus, spontaneous
BETA
