A) -1.532V
B) -1.503V
C) 1.532V
D) -3.06V
Correct Answer: A
Solution :
[a] : \[2A{{g}_{(s)}}\xrightarrow[{}]{{}}2A{{g}^{+}}(0.1M)+2{{e}^{-}}\] \[2{{e}^{-}}+Z{{n}^{2+}}(0.1M)\xrightarrow[{}]{{}}Z{{n}_{(s)}}\] \[\underline{\overline{2A{{g}_{(s)}}+Z{{n}^{2+}}(0.1M)\xrightarrow[{}]{{}}2A{{g}^{+}}(0.1M)+Z{{n}_{(s)}}}}\] \[\therefore \]\[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{2}{{\log }_{10}}\frac{{{[A{{g}^{+}}]}^{2}}}{[Z{{n}^{2+}}]}\] \[\therefore \]\[{{E}_{cell}}=-1.562-\frac{0.0591}{2}{{\log }_{10}}\frac{{{(0.1)}^{2}}}{0.1}\] \[=-1.562-0.03{{\log }_{10}}{{10}^{-1}}\] \[\Rightarrow \] \[{{E}_{cell}}=-1.562+0.03=-1.532V\]
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