Chemistry - Electrochemistry Question with Solution | TestHub

The Nernst equation, $E=E^{\circ }-\frac{RT}{nF}\ln Q$, governs cell voltage. To increase $E$, $Q$ must decrease.

A. For $\mathrm{Ag}(\mathrm{s})|AgN{\mathrm{O}}_3(\mathrm{aq},{\mathrm{m}}_{\mathrm{L}})||AgN{\mathrm{O}}_3(\mathrm{aq},{\mathrm{m}}_{\mathrm{R}})|Ag(\mathrm{s})$, the cell reaction is $\mathrm{AgN}{\mathrm{O}}_3(\mathrm{aq},{\mathrm{m}}_{\mathrm{R}})\rightleftharpoons \mathrm{AgN}{\mathrm{O}}_3(\mathrm{aq},{\mathrm{m}}_{\mathrm{L}})$. Increasing ${\mathrm{m}}_{\mathrm{L}}$ increases the product concentration, shifting equilibrium left, increasing $Q$, and decreasing $E$.

B. For $\mathrm{Pt}(\mathrm{s})|{\mathrm{H}}_2(\mathrm{g},{\mathrm{P}}_1)|HCl(\mathrm{aq})|{\mathrm{H}}_2(\mathrm{g},{\mathrm{P}}_2)|Pt(\mathrm{s})$, the cell reaction is ${\mathrm{H}}_2(\mathrm{g},{\mathrm{P}}_1)\rightleftharpoons {\mathrm{H}}_2(\mathrm{g},{\mathrm{P}}_2)$. Increasing ${\mathrm{P}}_1$ (reactant) shifts equilibrium right, decreasing $Q$, and increasing $E$.

C. For $\mathrm{Pt}(\mathrm{s})|{\mathrm{K}}_4\mathrm{Fe}(\mathrm{CN}{)}_6(\mathrm{aq}),{\mathrm{K}}_3\mathrm{Fe}(\mathrm{CN}{)}_6(\mathrm{aq})||M{\mathrm{n}}^{2+}(\mathrm{aq}),{\mathrm{H}}^{+}(\mathrm{aq})|Mn{\mathrm{O}}_2(\mathrm{s})|Pt(\mathrm{s})$, the overall reaction is $2\mathrm{Fe}(\mathrm{CN}{)}_6^{4-}(\mathrm{aq})+\mathrm{Mn}{\mathrm{O}}_2(\mathrm{s})+4{\mathrm{H}}^{+}(\mathrm{aq})\rightleftharpoons 2Fe(\mathrm{CN}{)}_6^{3-}(\mathrm{aq})+\mathrm{M}{\mathrm{n}}^{2+}(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$. Decreasing pH means increasing $[{\mathrm{H}}^{+}]$, which is a reactant. This shifts equilibrium right, decreasing $Q$, and increasing $E$.

D. For $\mathrm{Pt}(\mathrm{s})|F{\mathrm{e}}^{2+}(\mathrm{aq}),\mathrm{F}{\mathrm{e}}^{3+}(\mathrm{aq})||S{\mathrm{n}}^{4+}(\mathrm{aq}),\mathrm{S}{\mathrm{n}}^{2+}(\mathrm{aq})|Pt(\mathrm{s})$, the cell reaction is $2\mathrm{F}{\mathrm{e}}^{2+}(\mathrm{aq})+\mathrm{S}{\mathrm{n}}^{4+}(\mathrm{aq})\rightleftharpoons 2F{\mathrm{e}}^{3+}(\mathrm{aq})+\mathrm{S}{\mathrm{n}}^{2+}(\mathrm{aq})$. Adding $\mathrm{FeC}{\mathrm{l}}_3$ increases $[\mathrm{F}{\mathrm{e}}^{3+}]$ (product), shifting equilibrium left, increasing $Q$, and decreasing $E$.

Therefore, options B and C increase the cell voltage.