A) The forward reaction is spontaneous
B) The backward reaction is spontaneous
C) \[{{E}_{cell}}=0.163\,\nu \]
D) \[{{E}_{cell}}=1.585\,\nu \]
Correct Answer: B
Solution :
[b] \[{{E}_{cell}}=E_{cell}^{{}^\circ }-\frac{0.0592}{2}\log \frac{[Hg_{2}^{2+}(aq)]}{{{[A{{g}^{+}}(aq)]}^{2}}}\] \[=(0.80-0.785)-\frac{0.0592}{2}log\frac{{{10}^{-1}}}{{{({{10}^{-3}})}^{2}}}=-0.133V\] hence backward reaction is spontaneous.You need to login to perform this action.
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