$Pt(s)|\mathop {{H_2}(g)}\limits_{1\ atm} |HA\,\,\,\,\,\mathop {1\,M}\limits_{({K_a} = {{10}^{ - 7}})} \,\,||\,HB\,\,\,\,\,\mathop {1\,M}\limits_{({K_1} = {{10}^{ - 5}})}$
$\,|\mathop{{H_2}(g)}\limits_{1\ atm} |Pi(s)$
$\underline{2\text{H}_{\text{cathode }}^{+}+2{{\text{e}}^{-}}\to {{\text{H}}_{2}}(\text{g})}$
$\underline{{{\text{H}}_{2}}(\text{g})+2\text{H}_{\text{cathode }}^{+}\to 2\text{H}_{\text{anode }}^{+}+{{\text{H}}_{2}}(\text{g})}$
$\text{Q}=\frac{\left[ {{\text{H}}^{+}} \right]_{\text{anode }}^{2}}{\left[ {{\text{H}}^{+}} \right]_{\text{cathode }}^{2}}$ $=\frac{{{10}^{-7}}}{{{10}^{-5}}}={{10}^{-2}}$
$\mathrm{n}=2$
$\mathrm{E}_{\text {cell }}^{\circ}=0$
$\mathrm{E}_{\text {cell }}=0-\frac{0.06}{2} \log \,10^{-2}=0.06\, \mathrm{V}$
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The time taken for $A$ to become $1 / 4^{\text {th }}$ of its inital concentration is twice the time taken to become $1 / 2$ of the same. Also, when the change of concentration of $B$ is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is . . . . .
$CoCl _{3} . xNH _{3}+ AgNO _{3}( aq ) \rightarrow$
If two equivalents of $AgCl$ precipitate out, then the value of $x$ will be $....$
Choose the correct answer from the options given below: