Question

Calculate the e.m.f. of cell in Volts.

$\begin{array}{c}P{t}_{H_2}\\ 1atm\end{array}\left|\begin{array}{c}C{H}_{3}COOH\\ 0.1M\end{array}\right|\left|\begin{array}{c}N{H}_{4}OH\\ 0.01M\end{array}\right|\begin{array}{c}P{t}_{H_2}\\ 1atm\end{array}$

$K_a$ for $C{H}_{3}COOH=1.8\times {10}^{-5}$;

$K_b$ for $N{H}_{4}OH=1.8\times {10}^{-5}$

If answer is $x$, then write as nearest integer to $10|x|$.

Answer 1

L.H.S. : $C{H}_{3}COO\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$
$\therefore \left[H^{+}\right]=c\alpha =\sqrt[c]{c\left(K_a/c\right)}=\sqrt{\left(K_{a}c\right)}$
$\left[H^{+}\right]=\sqrt{(1.8\times {10}^{-5}\times 0.1)}$
$=1.342\times {10}^{-3}$ mol litre${}^{-1}$
R.H.S. : $N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
$\left[O{H}^{-}\right]=c\alpha =\sqrt[c]{c\left(K_b/c\right)}=\sqrt{(K_b.c)}$
$\left[O{H}^{-}\right]=\sqrt{(1.8\times {10}^{-5}\times 0.01)}$
$=0.424\times {10}^{-3}$ mol litre${}^{-1}$
$\left[H^{+}\right]={10}^{-14}/\left[O{H}^{-}\right]={10}^{-14}/(0.424\times {10}^{-3})$
$=2.359\times {10}^{-11}$ mol litre${}^{-1}$
For dissociation of weak acid or weak base, refer chapter of ionic equilibrium.
The cell reactions are :
$\frac{1}{2}{H}_2\to H^{+}+e$ Anode (or L.H.S.)
$H^{+}+e\to \frac{1}{2}{H}_2$ Cathode (or R.H.S.)
$=E_{O{P}_H}^o-\frac{0.059}{1}log\frac{\left[H^{+}\right]L.H.S.}{P_{H_2}}+E_{R{P}_H}^o+\frac{0.059}{1}log\frac{\left[H^{+}\right]R.H.S.}{P_{H_2}}$
or $E_{cell}=\frac{0.059}{1}log\frac{\left[H^{+}\right]R.H.S.}{\left[H^{+}\right]L.H.S.}(\because P_{H_2}=1atm)$
$\therefore E_{cell}=\frac{0.059}{1}log\frac{2.359\times {10}^{-11}}{1.342\times {10}^{-3}}$
$=-0.4575V$