Chemistry - Electrochemistry Question with Solution | TestHub

A) $\mathrm{Pt}\left|{\mathrm{H}}_2(0.1\mathrm{Bar})\right|{\mathrm{H}}^{+}(0.1\mathrm{M})\Vert {\mathrm{H}}^{+}(1\mathrm{M})\left|{\mathrm{H}}_2(0.01\mathrm{Bar})\right|\mathrm{Pt}$

It is a concentration cell, reaction quotient $(\mathrm{Q})=\frac{(0.1{)}^2(0.01)}{(1{)}^2(0.1)}=10^{-3}$.

$\therefore {\mathrm{E}}_{\text{cell }}>0$

B) $\mathrm{Ag}\left|\mathrm{AgCl}(\mathrm{s}),\mathrm{KCl}(0.1\mathrm{M})\Vert {\mathrm{Ag}}^{+}(0.01\mathrm{M})\right|\mathrm{Ag}$, it is not concentration cell.

Reaction quotient $(\mathrm{Q})=\frac{10^{-10}}{(0.1)(0.01)}=10^{-7}.\therefore {\mathrm{E}}_{\text{cell }}>0$

C) $\mathrm{Cu}\left|{\mathrm{Cu}}^{2+}(0.1\mathrm{M})\Vert {\mathrm{Cu}}^{2+}(0.01\mathrm{M})\right|\mathrm{Cu}$, it is concentration cell.

Reaction quotient $(\mathrm{Q})=\frac{0.1}{(0.01)}=10.\therefore {\mathrm{E}}_{\text{cell }}<0$

D) $\mathrm{Pt}\left|{\mathrm{Cl}}_2(1\mathrm{Bar})\right|\mathrm{HCl}(0.1\mathrm{M})\Vert \mathrm{NaCl}(0.1\mathrm{M})\left|{\mathrm{Cl}}_2(1\mathrm{Bar})\right|\mathrm{Pt}$, it is concentration cell.

Reaction quotient $(\mathrm{Q})=\frac{(1)(0.1{)}^2}{(1)(0.1{)}^2}=1.\therefore {\mathrm{E}}_{\text{cell }}=0$