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# Calculate EMF of cell Pt,$H_2(1\;atm)/CH_3COOH(0.1M)\parallel NH_3(aq)(0.01N)/H_2\;atm$.Given :$K_a(CH_3COOH)=1.8\times 10^{-5},K_b(NH_3)=1.8\times 10^{-5}$

$\begin{array}{1 1}(a)\;0.45V\\(b)\;-0.45V\\(c)\;0.90V\\(d)\;-0.90V\end{array}$

Can you answer this question?

$E^o_{H^+/H_2}=0$
$E_{cell}=(E_{redu})_{cathode}-(E_{redu})_{anode}$
$CH_3COOH\rightleftharpoons H^++CH_3COO^-$
$K_a=\large\frac{C\alpha^2}{1-\alpha}$
$1-\alpha\approx 1$
So $\alpha=\sqrt{\large\frac{K_a}{C}}$
$\alpha=\sqrt{\large\frac{1.8\times 10^{-5}}{0.1}}$
$\quad=1.34\times 10^{-2}$
$[H^+]=C\alpha=1.34\times 10^{-3}$
$P^H=-\log [H^+]=2.87$
$E_{cell}=-0.059\times P^H$
$\quad\;\;\;=-0.0059\times 2.81=-0.169V$
$NH_3+H_2O\rightleftharpoons NH_4OH$
So $\alpha=\sqrt{\large\frac{K_b}{C}}$
$\alpha=\sqrt{\large\frac{1.8\times 10^{-5}}{0.01}}$
$\quad=4.2\times 10^{-2}$
$[OH^-]=C\alpha=4.2\times 10^{-4}$
$P^{OH}=3.37$
$P^H=10.62$
$E_{cell}=-0.059\times P^H$
$\qquad=-0.62V$
Now $E_{cell}=(E_{redu})_{cathode}-(E_{redu})_{anode}$
$\qquad\qquad=-0.62+0.169$
$\qquad\qquad=-0.45V$
Hence (b) is the correct answer.
answered Dec 11, 2013