Redox Reactions And Electrochemistry Question 547
Question: Given $ E{{{}^\circ } _{C{u^{2+}}/Cu}}=0.34V,E{{{}^\circ } _{C{u^{2+}}/Cu^+}}=0.15V $ Standard electrode potential for the half cell $ C{u^{+}}/Cu $ is
Options:
A) 0.38 V
B) 0.53 V
C) 0.19 V
D) 0.49 V
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ C{u^{2+}}+{e^{-}}\xrightarrow{{}}C{u^{+}}; $
$ E_1^{{}^\circ }=0.15V;\Delta G_1^{{}^\circ }=-n_1E_1^{{}^\circ }F $
$ C{u^{2+}}2e\xrightarrow{{}}Cu; $
$ E_2^{{}^\circ }=0.34V;\Delta G_2^{{}^\circ }=-n_2E_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_2E{}^\circ F-n_1E_1^{{}^\circ }F) $ $ E{}^\circ =\frac{n_2E_2^{{}^\circ }F-n_1E_1^{{}^\circ }F}{nF} $ $ =\frac{2\times 0.34-0.15}{1}=0.53V $
BETA
