Redox Reactions And Electrochemistry Question 471
Question: Given that: $ E{{{}^\circ } _{A{g^{+}}/Ag}}~=0.80V $ and $ [A{g^{+}}]={10^{-3}}M; $ $ E{{{}^\circ } _{Hg_2^{2+}/Hg}}=0.785V $ and $ [Hg_2^{2+}]={10^{-1}}M $ which is true for the cell reaction $ 2Hg(l)+2A{g^{+}}(aq)\to 2Ag(s)+Hg_2^{2+}(aq)- $
Options:
A) The forward reaction is spontaneous
B) The backward reaction is spontaneous
C) $ E_{cell}=0.163,\nu $
D) $ E_{cell}=1.585,\nu $
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Answer:
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.
BETA
