A) 0.38 V
B) 0.53 V
C) 0.19 V
D) 0.49 V
Correct Answer: B
Solution :
[b] \[C{{u}^{2+}}+{{e}^{-}}\xrightarrow{{}}C{{u}^{+}};\] \[E_{1}^{{}^\circ }=0.15V;\Delta G_{1}^{{}^\circ }=-{{n}_{1}}E_{1}^{{}^\circ }F\] \[C{{u}^{2+}}2e\xrightarrow{{}}Cu;\] \[E_{2}^{{}^\circ }=0.34V;\Delta G_{2}^{{}^\circ }=-{{n}_{2}}E_{2}^{{}^\circ }F\] On subracting eq. (i) from eq. (ii) we get \[C{{u}^{+}}+{{e}^{-}}\xrightarrow{{}}Cu;\Delta G_{2}^{{}^\circ }=\Delta G{}^\circ =\Delta G_{2}^{{}^\circ }-\Delta G_{1}^{{}^\circ }\] \[-nE{}^\circ F=-({{n}_{2}}E{}^\circ F-{{n}_{1}}E_{1}^{{}^\circ }F)\] \[E{}^\circ =\frac{{{n}_{2}}E_{2}^{{}^\circ }F-{{n}_{1}}E_{1}^{{}^\circ }F}{nF}\] \[=\frac{2\times 0.34-0.15}{1}=0.53V\]
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