A) X = Ni, Y = Fe
B) X = Ni, Y = Zn
C) X = Fe, Y = Zn
D) X = Zn, Y = Ni
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
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