$Pt(s)|\mathop {{H_2}(g)}\limits_{1\ atm} |HA\,\,\,\,\,\mathop {1\,M}\limits_{({K_a} = {{10}^{ - 7}})} \,\,||\,HB\,\,\,\,\,\mathop {1\,M}\limits_{({K_1} = {{10}^{ - 5}})}$
$\,|\mathop{{H_2}(g)}\limits_{1\ atm} |Pi(s)$
$\underline{2\text{H}_{\text{cathode }}^{+}+2{{\text{e}}^{-}}\to {{\text{H}}_{2}}(\text{g})}$
$\underline{{{\text{H}}_{2}}(\text{g})+2\text{H}_{\text{cathode }}^{+}\to 2\text{H}_{\text{anode }}^{+}+{{\text{H}}_{2}}(\text{g})}$
$\text{Q}=\frac{\left[ {{\text{H}}^{+}} \right]_{\text{anode }}^{2}}{\left[ {{\text{H}}^{+}} \right]_{\text{cathode }}^{2}}$ $=\frac{{{10}^{-7}}}{{{10}^{-5}}}={{10}^{-2}}$
$\mathrm{n}=2$
$\mathrm{E}_{\text {cell }}^{\circ}=0$
$\mathrm{E}_{\text {cell }}=0-\frac{0.06}{2} \log \,10^{-2}=0.06\, \mathrm{V}$
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$A.$ $CH _3 COOH$
$B.$ $F _3 C - COOH$
$C.$ $ClCH _2- COOH$
$D.$ $FCH _2- COOH$
$E.$ $BrCH _2- COOH$
Choose the correct answer from the options given below:
$2 H _2( g )+2 NO ( g ) \rightarrow N _2( g )+2 H _2 O ( g )$
which following the mechanism given below:
$2 NO ( g ) \underset{ k _{-1}}{\stackrel{ k _1}{\rightleftharpoons}} N _2 O _2( g )$
$N _2 O _2( g )+ H _2( g ) \stackrel{ k _2}{\rightleftharpoons} N _2 O ( g )+ H _2 O ( g )$
$N _2 O ( g )+ H _2( g ) \stackrel{ k _3}{\rightleftharpoons} N _2( g )+ H _2 O ( g )$
(fast equilibrium)
(slow reaction)
(fast reaction)
The order of the reaction is