Chemistry - Electrochemistry Question with Solution | TestHub

$Pt(s)/\underset{(1\mathrm{bar})}{{\mathrm{H}}_2(\mathrm{g})\mid }\left(\underset{(100\mathrm{mL},0.2\mathrm{M})}{{\mathrm{H}}_3{\mathrm{PO}}_4}+\underset{(100\mathrm{mL},0.1\mathrm{M})}{{\mathrm{Na}}_3{\mathrm{PO}}_4}\right)\Vert \left(\underset{(100\mathrm{mL},0.1\mathrm{M})}{{\mathrm{CH}}_3\mathrm{COOH}}+\underset{(1\mathrm{bar})}{{\mathrm{H}}_2(\mathrm{g})/\mathrm{Pt}(\mathrm{s})}\right)$ From Nernst Equation

$\begin{array}{rlr}& {\mathrm{E}}_{\text{cell }}=-0.06\times \log \frac{{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{a}}}{{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{c}}}& \end{array}$

$or{\mathrm{E}}_{cell}=\left[(\mathrm{pH}{)}_{\mathrm{a}}-(\mathrm{pH}{)}_{\mathrm{c}}\right]\times 0.06$ ...(1)

${\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{PO}}_4^{3-}$$⟶{}^{{}_{}}{\mathrm{H}}_2{\mathrm{PO}}_4^{-}+{\mathrm{HPO}}_4^{2-}$

20 10 0 0

10 0 10 10

Again

${\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{HPO}}_4^{2-}⟶2{\mathrm{H}}_2{\mathrm{PO}}_4^{-}$

10 10 10

0 0 30 $\text{displaylines}\begin{array}{ll}& \\ & \\ & \\ \therefore (\mathrm{pH}{)}_{\mathrm{a}}=& \frac{{\mathrm{pK}}_{{\mathrm{a}}_1}+{\mathrm{pK}}_{{\mathrm{a}}_2}}{2}=\frac{4+8}{2}=6\\ & \end{array}$ Also, at cathode

${\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}^{+}$ $\alpha =\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{C}}}=\sqrt{\frac{10^{-5}}{0.1}}=10^{-2},(\mathrm{pH}{)}_{\mathrm{a}}=10^{-3}$

$\therefore (\mathrm{pH}{)}_{\mathrm{c}}=3$

${\mathrm{E}}_{\text{cell }}=0.06[6-3]$

$=0.18\mathrm{V}$

(II) If $100\mathrm{mL},0.1\mathrm{M}\mathrm{NaOH}$ is added in anode compartments

${\mathrm{H}}_3{\mathrm{PO}}_4$, ${\mathrm{PO}}_4^{3-}$, NaOH

$20\mathrm{m}.\mathrm{mol}$ $10\mathrm{m}.\mathrm{mol}$ $10\mathrm{m}.\mathrm{mol}$

${\mathbf{H}}_3{\mathbf{PO}}_4+{\mathbf{OH}}^{-}⟶$ ${\mathbf{H}}_2{\mathbf{PO}}_4^{-}+{\mathbf{H}}_2\mathbf{O}$

$20$ $10$ $0$

$10$ $0$ $10$

Also,

So, it is a buffer.

$(\mathrm{pH}{)}_{\mathrm{a}}={\mathrm{pK}}_{{\mathrm{a}}_2}+\log \frac{\left[{\mathrm{HPO}}_4^{2-}\right]}{\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{2-}\right]}=8+\log \frac{10}{20}$

$\therefore (\mathrm{pH}{)}_{\mathrm{a}}=7.7$

${\mathrm{E}}_{\text{cell }}=0.06[7.7-3]$

$=0.282\mathrm{V}$

(III) $50\mathrm{mL},0.1\mathrm{M}\mathrm{NaOH}$ is added in cathode compartment

$\therefore \mathrm{P}$ it is also a buffer.

$(\mathrm{pH}{)}_{\mathrm{c}}=5+\log \frac{5}{5}=5$

$\therefore {\mathrm{E}}_{\text{cell }}=0.06[6-5]$

$=0.06\mathrm{V}$

(IV) $100\mathrm{mL},0.1\mathrm{M}\mathrm{HCl}$ is added in anode compartment

${\mathrm{PO}}_4^{3-}$$+{\mathrm{H}}^{+}$ ⟶ ${\mathrm{HPO}}_4^{2-}$

10 10 0

0 0 10

$H_{3}P{O}_4+{\mathrm{HPO}}_4^{2-}\to 2{\mathrm{H}}_2{\mathrm{PO}}_4^{-}$

20 10 0

10 0 20

$\therefore (\mathrm{pH}{)}_{\mathrm{a}}=4+\log 2$

= 4.3

${\mathrm{E}}_{\text{cell }}=0.06[4.3-3]$

$=0.06\times 1.3$

$=0.078\mathrm{V}$