$\frac{540}{ t _2}=\frac{\ln \frac{10}{4}}{\ln 10}$
$\frac{540}{ t _2}=\frac{\log 10-\log 4}{\log 10}$
$\frac{540}{ t _2}=\frac{1-0.6}{1}$
$\Rightarrow \frac{540}{ t _2}=0.4$
$\Rightarrow t _2=\frac{540}{0.4}=1350\,sec$
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$\mathop {{H_2}(Pt)}\limits_{1\,atm} \left| {\begin{array}{*{20}{c}}
{H_{aq}^ + }\\
{pH = 4}
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
{A{g^ + }}\\
{xM}
\end{array}} \right|\begin{array}{*{20}{c}}
{Ag}\\
{}
\end{array}$
What is the value of $x$ ? ........... $\mathrm{M}$
( give : $E_{A{g^ + }|Ag}^o = + 0.8$ $\frac{{2.303\,RT}}{F} = 0.06 $)
