The following electrochemical cell has been set up: Pt(s)\ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Reconstruct the Correct Cell representation and Identify the Half-Cell Reactions First, let us address a formatting issue: in the original question, the vertical bars ($\mid$ and $\parallel$) representing phase boundaries and salt bridges in the cell diagram were misparsed as table delimiters. The actual electrochemical cell is represented as: $$\ce{Pt(s) \mid Fe^3+, Fe^2+ (a = 1) \parallel Ce^4+, Ce^3+ (a = 1) \mid Pt(s)}$$ The given standard reduction potentials at $298\text{ K}$ are: * Standard reduction potential of the iron redox couple: $$E^\circ_{\ce{Fe^3+/Fe^2+}} = 0.77\text{ V}$$ * Standard reduction potential of the cerium redox couple: $$E^\circ_{\ce{Ce^4+/Ce^3+}} = 1.61\text{ V}$$ Step 2 - Determine the Cathode, Anode, and the Direction of Current Flow In a galvanic cell, the electrode with the higher standard reduction potential undergoes reduction and acts as the cathode, while the electrode with the lower standard reduction potential undergoes oxidation and acts as the anode. Comparing the potentials: $$E^\circ_{\ce{Ce^4+/Ce^3+}} (1.61\text{ V}) > E^\circ_{\ce{Fe^3+/Fe^2+}} (0.77\text{ V})$$ Thus: * **Cathode (Reduction half-reaction on the right):** $$\ce{Ce^4+(aq) + e^- -> Ce^3+(aq)}$$ * **Anode (Oxidation half-reaction on the left):** $$\ce{Fe^2+(aq) -> Fe^3+(aq) + e^-}$$ The spontaneous overall cell reaction is: $$\ce{Fe^2+(aq) + Ce^4+(aq) -> Fe^3+(aq) + Ce^3+(aq)}$$ The standard cell potential is: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = 1.61\text{ V} - 0.77\text{ V} = +0.84\text{ V}$$ * **Flow of Electrons:** Electrons are released during oxidation at the anode ($\ce{Fe}$ electrode side) and flow through the external circuit to the cathode ($\ce{Ce}$ electrode side). * **Flow of Conventional Current:** By convention, the direction of electric current is defined as opposite to the flow of electrons. Therefore, conventional electric current flows from the **Ce electrode to the Fe electrode**. Step 3 - Analyze the Current Change Over Time According to the Nernst equation, the cell potential ($E_{\text{cell}}$) at any given moment is: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{F} \ln Q$$ $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{F} \ln \left(\frac{[\ce{Fe^3+}][\ce{Ce^3+}]}{[\ce{Fe^2+}][\ce{Ce^4+}]}\right)$$ As the spontaneous cell reaction proceeds over time: 1. The reactants ($\ce{Fe^2+}$ and $\ce{Ce^4+}$) are consumed, so their concentrations decrease. 2. The products ($\ce{Fe^3+}$ and $\ce{Ce^3+}$) are formed, so their concentrations increase. 3. As a result, the reaction quotient $Q = \frac{[\ce{Fe^3+}][\ce{Ce^3+}]}{[\ce{Fe^2+}][\ce{Ce^4+}]}$ increases, which causes $E_{\text{cell}}$ to decrease continuously. According to Ohm's law, the electric current ($I$) flowing through the ammeter is directly proportional to the cell potential ($E_{\text{cell}}$): $$I = \frac{E_{\text{cell}}}{R}$$ Since $E_{\text{cell}}$ decreases over time as the system approaches equilibrium, the electric current **decreases with time**. Step 4 - Evaluate the Options and Correct the Parsing Discrepancy Based on our thermodynamic analysis, the correct behavior of the cell is: **Ce electrode to Fe electrode, decrease**. The actual options printed in the standard curriculum are: * **(a) Ce electrode to Fe electrode, decrease** * **(b) Ce electrode to Fe electrode, increase** * **(c) Fe electrode to Ce electrode, decrease** * **(d) Fe electrode to Ce electrode, increase** This indicates that option (a) is the correct answer. *(Note: The "Correct Answer: B" and the existing hint provided in the prompt were mismatched from question 51 due to a database parsing error. For the correct question 61, the answer is option a.)* $$\text{Correct Option: } \boxed{\text{A}}$$