[Four-digit Integer] Determine potential (in mV) of the cell: Pt — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify Anode and Cathode Half-Cells and Write Balanced Equations The given notation of the electrochemical cell is: $$\text{Pt} \mid \ce{Fe^2+}(0.75\text{ M}), \ce{Fe^3+}(0.75\text{ M}) \parallel \ce{Cr2O7^{2-}}(2\text{ M}), \ce{Cr^{3+}}(4\text{ M}), \ce{H^+}(1\text{ M}) \mid \text{Pt}$$ According to standard conventions: * **Anode Half-Cell (Left - Oxidation):** The iron couple undergoes oxidation. The balanced anode half-reaction is: $$\ce{Fe^2+(aq) -> Fe^3+(aq) + e^-}$$ * **Cathode Half-Cell (Right - Reduction):** The dichromate couple in acidic medium undergoes reduction. The balanced cathode half-reaction is: $$\ce{Cr2O7^{2-}(aq) + 14H^+(aq) + 6e^- -> 2Cr^3+(aq) + 7H2O(l)}$$ To equalize the number of electrons transferred, we multiply the oxidation half-reaction by $6$: $$\ce{6Fe^2+(aq) -> 6Fe^3+(aq) + 6e^-}$$ Adding the two balanced half-reactions yields the overall cell reaction: $$\ce{Cr2O7^{2-}(aq) + 14H^+(aq) + 6Fe^2+(aq) -> 2Cr^3+(aq) + 6Fe^3+(aq) + 7H2O(l)}$$ The total number of moles of electrons transferred in this balanced reaction is: $$n = 6$$ Step 2 - Calculate the Standard Cell Potential ($E^\circ_{\text{cell}}$) The standard potential of the cell is the difference between the standard reduction potentials of the cathode and the anode: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ Substitute the given standard reduction potentials ($E^\circ_{\ce{Cr2O7^{2-}/Cr^3+}} = 1.35\text{ V}$ and $E^\circ_{\ce{Fe^3+/Fe^2+}} = 0.77\text{ V}$): $$E^\circ_{\text{cell}} = 1.35\text{ V} - 0.77\text{ V}$$ $$E^\circ_{\text{cell}} = 0.58\text{ V}$$ Step 3 - Set Up and Calculate the Reaction Quotient ($Q$) The reaction quotient ($Q$) for the overall cell reaction is expressed as: $$Q = \frac{[\ce{Cr^3+}]^2 [\ce{Fe^3+}]^6}{[\ce{Cr2O7^{2-}}] [\ce{H^+}]^{14} [\ce{Fe^2+}]^6}$$ Since the concentrations of iron(II) and iron(III) ions are equal ($[\ce{Fe^2+}] = [\ce{Fe^3+}] = 0.75\text{ M}$), their terms cancel out completely: $$\frac{[\ce{Fe^3+}]^6}{[\ce{Fe^2+}]^6} = \frac{(0.75)^6}{(0.75)^6} = 1$$ Therefore, the expression for $Q$ simplifies to: $$Q = \frac{[\ce{Cr^3+}]^2}{[\ce{Cr2O7^{2-}}] [\ce{H^+}]^{14}}$$ Substitute the given active concentrations ($[\ce{Cr^3+}] = 4\text{ M}$, $[\ce{Cr2O7^{2-}}] = 2\text{ M}$, and $[\ce{H^+}] = 1\text{ M}$): $$Q = \frac{4^2}{2 \times (1)^{14}}$$ $$Q = \frac{16}{2}$$ $$Q = 8$$ Step 4 - Apply the Nernst Equation to Calculate the Non-Standard Cell Potential ($E_{\text{cell}}$) At $298\text{ K}$, the Nernst equation for the overall cell potential is: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} Q$$ Substitute the standard cell potential $E^\circ_{\text{cell}} = 0.58\text{ V}$, number of electrons $n = 6$, slope parameter $\frac{2.303 RT}{F} = 0.06\text{ V}$, and $Q = 8$: $$E_{\text{cell}} = 0.58\text{ V} - \frac{0.06\text{ V}}{6} \log_{10}(8)$$ $$E_{\text{cell}} = 0.58\text{ V} - 0.01\text{ V} \times \log_{10}(2^3)$$ $$E_{\text{cell}} = 0.58\text{ V} - 0.01\text{ V} \times (3 \log_{10} 2)$$ Using the given value $\log_{10} 2 = 0.3$: $$E_{\text{cell}} = 0.58\text{ V} - 0.01\text{ V} \times (3 \times 0.3)$$ $$E_{\text{cell}} = 0.58\text{ V} - 0.01\text{ V} \times 0.9$$ $$E_{\text{cell}} = 0.58\text{ V} - 0.009\text{ V}$$ $$E_{\text{cell}} = 0.571\text{ V}$$ Step 5 - Convert the Potential to Millivolts To express the potential of the cell in millivolts ($\text{mV}$): $$E_{\text{cell}} = 0.571\text{ V} \times 1000\text{ mV/V}$$ $$E_{\text{cell}} = 571\text{ mV}$$ Since the question requires a four-digit integer format: $$\boxed{0571}$$