| For the following cell reaction |
| \[Pb(s)+H{{g}_{2}}S{{O}_{4}}(s)PbS{{O}_{4}}(s)+2Hg(l)\] \[E_{cell}^{o}=0.92V\] |
| \[{{K}_{sp}}(PbS{{O}_{4}})=2\times {{10}^{-8}},\]\[{{K}_{sp}}(H{{g}_{2}}S{{O}_{4}})=1\times {{10}^{-6}}\] |
| Hence, \[{{E}_{cell}}\] is |
A) 0.92 V
B) 0.89 V
C) 1.04 V
D) 0.95 V
Correct Answer: D
Solution :
| [d] \[Q=\frac{[P{{b}^{2+}}]}{[Hg_{2}^{2+}]}=\frac{\sqrt{{{K}_{sp}}(PbS{{O}_{4}})}}{\sqrt{{{K}_{sp}}(H{{g}_{2}}S{{O}_{4}})}}\] |
| \[=\frac{\sqrt{2\times {{10}^{-8}}}}{\sqrt{1\times {{10}^{-6}}}}\] |
| \[=\sqrt{2\times {{10}^{-2}}}=0.14\] |
| \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.059}{2}\log 0.14\] |
| \[=0.92-0.0295\text{ }log\text{ }0.14\] |
| \[=0.95V\] |
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