A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10^ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Cell and Electrode Reactions An electrochemical cell containing two hydrogen electrodes with different ion concentrations is a **hydrogen concentration cell**. By convention, the negative electrode is the **anode** (where oxidation occurs), and the positive electrode is the **cathode** (where reduction occurs). The half-reactions at each electrode are: * **At the Anode (Negative Electrode - Oxidation):** $$\ce{\frac{1}{2} H2(g, 1 atm) -> H+(aq, anode) + e^-}$$ * **At the Cathode (Positive Electrode - Reduction):** $$\ce{H+(aq, cathode) + e^- -> \frac{1}{2} H2(g, 1 atm)}$$ Summing these half-reactions gives the net cell reaction: $$\ce{H+(aq, cathode) <=> H+(aq, anode)}$$ For this process, the number of moles of electrons transferred per mole of reaction is: $$n = 1$$ Step 2 - Derive the Nernst Equation for the Cell Since both half-cells are hydrogen electrodes, their standard reduction potentials are identical ($E^\circ_{\ce{H+/H2}} = 0\text{ V}$). Thus, the standard cell potential ($E^\circ_{\text{cell}}$) is: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0\text{ V} - 0\text{ V} = 0\text{ V}$$ Using the Nernst equation at $25^\circ\text{C}$ ($298\text{ K}$): $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$$ Where $Q$ is the reaction quotient defined by: $$Q = \frac{[\ce{H+}]_{\text{anode}}}{[\ce{H+}]_{\text{cathode}}}$$ Substituting $E^\circ_{\text{cell}} = 0\text{ V}$ and $n = 1$ into the Nernst equation: $$E_{\text{cell}} = 0 - 0.0591 \log \left( \frac{[\ce{H+}]_{\text{anode}}}{[\ce{H+}]_{\text{cathode}}} \right)$$ $$E_{\text{cell}} = 0.0591 \log \left( \frac{[\ce{H+}]_{\text{cathode}}}{[\ce{H+}]_{\text{anode}}} \right)$$ Step 3 - Substitute the Given Values We are given: * EMF of the cell ($E_{\text{cell}}$) = $0.118\text{ V}$ * Hydrogen ion concentration at the negative electrode ($[\ce{H+}]_{\text{anode}}$) = $10^{-6}\text{ M}$ Let the concentration of hydrogen ions at the positive electrode ($[\ce{H+}]_{\text{cathode}}$) be $x$. Substitute these parameters into our Nernst equation: $$0.118\text{ V} = 0.0591\text{ V} \log \left( \frac{x}{10^{-6}\text{ M}} \right)$$ Step 4 - Calculate the Hydrogen Ion Concentration ($x$) Rearranging the equation to solve for the logarithmic term: $$\log \left( \frac{x}{10^{-6}} \right) = \frac{0.118}{0.0591}$$ $$\log \left( \frac{x}{10^{-6}} \right) \approx 2$$ Taking the base-10 antilogarithm of both sides: $$\frac{x}{10^{-6}} = 10^2$$ $$x = 10^2 \times 10^{-6}\text{ M}$$ $$x = \boxed{10^{-4}\text{ M}}$$ Step 5 - Evaluate the Options * **Option (A) is incorrect:** If $[\ce{H+}]_{\text{cathode}} = 10^{-6}\text{ M}$, both half-cells would have identical concentrations, resulting in $E_{\text{cell}} = 0\text{ V}$. * **Option (B) is incorrect:** If $[\ce{H+}]_{\text{cathode}} = 10^{-3}\text{ M}$, the potential would be $0.0591 \log(10^3) \approx 0.177\text{ V}$. * **Option (C) is correct:** As calculated, a cathode concentration of $10^{-4}\text{ M}$ yields exactly the given EMF of $0.118\text{ V}$. * **Option (D) is incorrect:** If $[\ce{H+}]_{\text{cathode}} = 10^{-5}\text{ M}$, the potential would be $0.0591 \log(10^1) \approx 0.059\text{ V}$. $$\text{Correct Option: } \boxed{\text{C}}$$