Redox Reactions And Electrochemistry Ques 736
Question: For the following cell reaction
$ Pb(s)+Hg_2SO_4(s)PbSO_4(s)+2Hg(l) $ $ E_{cell}^{o}=0.92V $ $ K_{sp}(PbSO_4)=2\times {10^{-8}}, $ $ K_{sp}(Hg_2SO_4)=1\times {10^{-6}} $ Hence, $ E_{cell} $ is
Options:
A) 0.92 V
B) 0.89 V
C) 1.04 V
D) 0.95 V
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ Q=\frac{[P{b^{2+}}]}{[Hg_2^{2+}]}=\frac{\sqrt{K_{sp}(PbSO_4)}}{\sqrt{K_{sp}(Hg_2SO_4)}} $
$ =\frac{\sqrt{2\times {10^{-8}}}}{\sqrt{1\times {10^{-6}}}} $
$ =\sqrt{2\times {10^{-2}}}=0.14 $
$ E_{cell}=E_{cell}^{o}-\frac{0.059}{2}\log 0.14 $
$ =0.92-0.0295log0.14 $
$ =0.95V $
