A) - 0.38 V
B) + 0.49 V
C) + 0.38 V
D) - 0.19 V
Correct Answer: C
Solution :
From given data \[(From\,\Delta {{G}^{o}}=-n{{E}^{o}}F)\] \[(i)\,Cu(s)\xrightarrow[{}]{{}}C{{u}^{2+}}(aq)+2{{e}^{-}}\] \[\Delta G_{1}^{o}=-2\times (-0.34)\times F\] \[(ii)\,C{{u}^{2+}}(aq)+{{e}^{-}}\xrightarrow[{}]{{}}C{{u}^{+}}(aq)\] \[\Delta G_{2}^{o}=-1\times (0.15)\,\,F\] On addition \[Cu(s)\xrightarrow[{}]{{}}C{{u}^{2+}}(aq)+{{e}^{-}},\,\,\,\Delta G_{3}^{o}=-1\times {{E}^{o}}\times F\] \[\Delta G_{3}^{o}=\Delta G_{1}^{o}+\Delta G_{2}^{o}\] \[=(-2\times -0.34\times F)+(-1\times 0.15\times F)\] \[=\,+\,0.68\,F-0.15\,F=0.53\,F\] \[or\,{{E}^{o}}=-0.53\,V\] Reaction \[2C{{u}^{+}}(aq)\rightleftharpoons C{{u}^{2+}}(aq)+Cu(s)\,{{E}^{o}}=?\] So, \[C{{u}^{2+}}(aq)+{{e}^{-}}\rightleftharpoons Cu(s),\,{{E}^{o}}=+0.53\,V\] \[\frac{C{{u}^{+}}(aq)C{{u}^{2+}}(aq)+{{e}^{-}};\,\,\,\,{{E}^{o}}=-0.15V}{2C{{u}^{+}}(aq)C{{u}^{2+}}(aq)+Cu(s),\,{{E}^{o}}=+0.38V}\]
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