A) \[-\,0.19V\]
B) 0.38
C) 0.94V
D) \[-\,0.38V\]
Correct Answer: B
Solution :
\[E_{C{{u}^{2+}}/C{{u}^{+}}}^{o}=0.15\,V\] \[E_{C{{u}^{2+}}/C{{u}^{+}}}^{o}=0.34\,V\] \[C{{u}^{2+}}+{{e}^{-}}\xrightarrow{{}}C{{u}^{+}};\] \[{{E}^{o}}=+0.15V\] Or \[C{{u}^{+}}\xrightarrow{{}}C{{u}^{2+}}+{{e}^{-}};\] \[E_{1}^{o}=-0.15\,V\] \[\Delta G_{1}^{o}=-nFE_{1}^{o}=-1\times F\times (-0.15)=+0.15\,F\] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu;\] \[E_{2}^{o}=0.34\,V\] \[C{{u}^{+}}+e\xrightarrow{{}}Cu;\] \[E_{3}^{o}=?\] \[\Delta G_{2}^{o}=-2\times 0.34\times F\] \[\Delta G_{3}^{o}=-1\times E_{3}^{o}\times F\] \[\Delta G_{3}^{o}=\Delta G_{1}^{o}+\Delta G_{2}^{o}\] \[-E_{3}^{o}F=+0.15F-0.68\,F\] \[E_{3}^{o}=+0.53\,V\] For disproportionation reaction of\[C{{u}^{+}}\]for half-cell reaction: \[C{{u}^{+}}+{{e}^{-}}\xrightarrow{{}}Cu;\] \[{{E}^{o}}=+0.53\,V\] \[C{{u}^{+}}\xrightarrow{{}}C{{u}^{2+}}+\overline{e};\] \[{{E}^{o}}=-0.15\,V\] \[\therefore \]\[2C{{u}^{2+}}\xrightarrow{{}}C{{u}^{2+}}+Cu;\] \[E_{cell}^{o}=0.38\,V\]
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