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EMF of following cell is 0.67V at 298K $Pt/H_2(1atm)/H^+(pH=X)\parallel KCl(1N)/Hg_2Cl_2(s)/Hg$.Calculate pH of anode compartment.Given : $E^0_{Cl^-/Hg^2Cl_2/Hg}=0.28V$.

$\begin{array}{1 1}(a)\;5\\(b)\;6.6\\(c)\;6\\(d)\;7\end{array}$

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At anode :$H_2\rightarrow 2H^++2e^-$
At cathode :$Hg_2Cl_2+2e^-\rightarrow 2Hg+2Cl^-$
Net reaction :$H_2+Hg_2Cl_2\rightarrow 2H^++2Hg+2Cl^-$
$\therefore E_{cell}=E^0_{Cl^-/Hg_2Cl_2/Hg}-E^0_{H^+/H_2}-\large\frac{0.059}{2}$$\log \large\frac{[Cl^-]^2[H^+]^2}{pH}$
$\therefore pH=6.61$
Hence (b) is the correct answer.
answered Feb 17, 2014 by sreemathi.v

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