A) 0.500 V
B) 0.325 V
C) 0.650 V
D) 0.150 V
Correct Answer: B
Solution :
[B]\[\frac{\begin{matrix} C{{u}^{2+}}+1{{e}^{-}}\to C{{u}^{+}} & E_{1}^{0}=0.15\,v\,\,\Delta \,\,G_{1}^{0}=-\,{{n}_{1}}\,\,E_{1}^{0}\,\,F \\ C{{u}^{2+}}+1{{e}^{-}}\to Cu & E_{2}^{0}=0.50\,v\,\,\Delta \,\,G_{2}^{0}=-\,{{n}_{2}}\,\,E_{2}^{0}\,\,F \\ \end{matrix}}{\begin{matrix} C{{u}^{2+}}+2{{e}^{-}}\to Cu & \Delta \,G{}^\circ \\ \end{matrix}=\Delta G{{{}^\circ }_{1}}+\Delta G{{{}^\circ }_{2}}}\] |
\[(-\,1)n\,\,E{}^\circ F=(-\,1)\,{{n}_{1}}\,\,E_{1}^{0}F+(-\,1)\,{{n}_{2}}E_{2}^{0}F\] |
\[E{}^\circ =\frac{{{n}_{1}}E_{1}^{0}+{{n}_{2}}E_{2}^{0}}{n}=\frac{0.15\times 1+0.50\times 1}{2}\] |
\[=0.325V\] |
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