A) ? 0.49 V
B) 0.49 V
C) ? 0.38 V
D) 0.38 V
Correct Answer: D
Solution :
The required reaction \[(C{{u}^{++}}+Cu\to 2C{{u}^{+}})\] can be obtained by using the following reactions. \[C{{u}^{++}}+{{e}^{-}}\to C{{u}^{+}};\,\,\,E_{C{{u}^{++}}/C{{u}^{+}}}^{o}=0.15\,\,V\,\,\] ?..(i) \[C{{u}^{++}}+2{{e}^{-}}\to Cu;\,\,\,\,E_{C{{u}^{++}}/Cu}^{o}=0.\,34\,\,V\,\] ?..(ii) Multiplying eq. (i) by 2 we get \[2C{{u}^{++}}+2{{e}^{-}}\to 2C{{u}^{+}}\] ?..(iii) \[\Delta {{G}_{1}}=-nFE=-2\times F\times 0.15\] \[C{{u}^{++}}+2{{e}^{-}}\to Cu\,\,\,\] ?..(iv) \[\Delta {{G}_{2}}=-nFE=-2\times F\times 0.34\] Subtract the eq. (iv) from (iii) \[C{{u}^{++}}+Cu\to 2C{{u}^{+}}\] \[\Delta {{G}_{3}}=-\,nFE=-1\times F\times {{E}^{o}}\] Also \[\Delta {{G}_{3}}=\Delta {{G}_{1}}-\Delta {{G}_{2}}\] \[-1F{{E}^{o}}=(-\,2F\times 0.15)-(-\,2F\times 0.34)\] \[{{E}^{o}}=-\,0.38\] This is the value for the reaction \[C{{u}^{++}}+Cu\to 2C{{u}^{+}}\] But the given reaction is just reverse of it \\[{{E}_{\text{cell}}}\] for given reaction = + 0.38V.
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