An electrochemical cell is set up as follows: Pt \ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Electrodes and Nernst Equation The given electrochemical cell consists of two hydrogen gas electrodes connected in a concentration cell arrangement: $$\ce{Pt | H2(s) (1 atm) | 0.001 M HCl || 0.1 M HA | H2(s) (1 atm) | Pt}$$ *(Note: The symbol $\ce{H2(s)}$ printed in the cell notation is a typographical error in the source and represents gaseous hydrogen, $\ce{H2(g)}$).* The reduction half-reaction at each hydrogen electrode is: $$\ce{2H+ (aq) + 2e- <=> H2(g)}$$ The reduction potential ($E$) of a hydrogen electrode at $298\text{ K}$ and $P_{\ce{H2}} = 1\text{ atm}$ is given by the Nernst equation: $$E = E^\circ_{\ce{H+/H2}} - \frac{2.303 RT}{2F} \log \frac{P_{\ce{H2}}}{[\ce{H+}]^2}$$ Since $E^\circ_{\ce{H+/H2}} = 0\text{ V}$ and $P_{\ce{H2}} = 1\text{ atm}$, the expression simplifies to: $$E = 0 - \frac{0.0591}{2} \log \frac{1}{[\ce{H+}]^2}$$ $$E = 0.0591 \log [\ce{H+}] = -0.0591 \text{ pH}$$ Step 2 - Determine pH of the Anode Compartment (\ce{HCl}) Hydrochloric acid (\ce{HCl}) is a strong acid that dissociates completely in aqueous solution: $$\ce{HCl(aq) -> H+(aq) + Cl-(aq)}$$ Therefore, the concentration of hydrogen ions in the anode compartment ($[\ce{H+}]_1$) is equal to the molar concentration of the \ce{HCl} solution: $$\text{Formula: } [\ce{H+}]_1 = C_{\ce{HCl}}$$ $$\text{Substitution: } [\ce{H+}]_1 = 0.001\text{ M}$$ $$\text{Calculation: } [\ce{H+}]_1 = 10^{-3}\text{ M} \implies \text{pH}_1 = -\log(10^{-3}) = 3$$ Step 3 - Determine pH of the Cathode Compartment (\ce{HA}) The weak monobasic acid \ce{HA} has a concentration $C = 0.1\text{ M}$ and a $pK_a = 5$. The acid dissociation constant ($K_a$) is: $$K_a = 10^{-pK_a} = 10^{-5}$$ Using the standard approximation for a weak acid (since the degree of dissociation $\alpha \ll 1$): $$\text{Formula: } [\ce{H+}]_2 = \sqrt{K_a \cdot C}$$ $$\text{Substitution: } [\ce{H+}]_2 = \sqrt{10^{-5} \times 0.1}$$ $$\text{Calculation: } [\ce{H+}]_2 = \sqrt{10^{-6}} = 10^{-3}\text{ M} \implies \text{pH}_2 = -\log(10^{-3}) = 3$$ Step 4 - Calculate the Cell Potential ($E_{\text{cell}}$) The overall EMF of the concentration cell ($E_{\text{cell}}$) is the difference between the reduction potentials of the cathode and anode half-cells: $$\text{Formula: } E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = \frac{2.303 RT}{F} \log \frac{[\ce{H+}]_{\text{cathode}}}{[\ce{H+}]_{\text{anode}}}$$ $$\text{Substitution: } E_{\text{cell}} = 0.0591 \log \left(\frac{10^{-3}}{10^{-3}}\right)$$ $$\text{Calculation: } E_{\text{cell}} = 0.0591 \log(1) = 0\text{ V}$$ $$\boxed{E_{\text{cell}} = 0\text{ V}}$$ Since both compartments have an identical hydrogen ion concentration ($10^{-3}\text{ M}$) and therefore identical pH values ($3$), their electrode potentials are exactly equal, leading to a net cell EMF of zero. This corresponds to option (c).