If the solution of is replaced by a solution, then the magnitude of the cell potential would be: — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Nature of a Concentration Cell A metal concentration cell consists of two half-cells of the exact same metal-metal ion electrode but operating at different electrolyte concentrations [1]. The IUPAC cell representation is: $$\ce{M(s) \mid M^+(aq; \text{anode}) \parallel M^+(aq; \text{cathode}) \mid M(s)}$$ At the anode, oxidation occurs: $$\ce{M(s) -> M^+(aq; } C_{\text{anode}}\ce{) + e^-}$$ At the cathode, reduction occurs: $$\ce{M^+(aq; } C_{\text{cathode}}\ce{) + e^- -> M(s)}$$ The net cell reaction is the transfer of metal ions from the higher concentration (cathode) to the lower concentration (anode): $$\ce{M^+(aq; } C_{\text{cathode}}\ce{) -> M^+(aq; } C_{\text{anode}}\ce{)}$$ In this process, the number of moles of electrons transferred is $n = 1$. The standard electromotive force ($E^\circ_{\text{cell}}$) of any concentration cell is always zero because both electrodes are identical: $$E^\circ_{\text{cell}} = 0\text{ V}$$ Step 2 - Apply the Nernst Equation The cell potential ($E_{\text{cell}}$) at temperature $T$ is calculated using the Nernst equation: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q$$ Substituting $E^\circ_{\text{cell}} = 0$, $n = 1$, and the reaction quotient $Q = \frac{[\ce{M^+}]_{\text{anode}}}{[\ce{M^+}]_{\text{cathode}}}$: $$E_{\text{cell}} = -\frac{RT}{F} \ln \left( \frac{[\ce{M^+}]_{\text{anode}}}{[\ce{M^+}]_{\text{cathode}}} \right)$$ We can express the magnitude of the cell potential $|E_{\text{cell}}|$ as: $$|E_{\text{cell}}| = \frac{RT}{F} \ln \left( \frac{[\ce{M^+}]_{\text{cathode}}}{[\ce{M^+}]_{\text{anode}}} \right)$$ Let us define the temperature-dependent term as a constant $C = \frac{RT}{F}$: $$|E_{\text{cell}}| = C \ln \left( \frac{[\ce{M^+}]_{\text{cathode}}}{[\ce{M^+}]_{\text{anode}}} \right)$$ Step 3 - Substitute the Initial Conditions to Determine the Constant We are given the initial concentrations and the initial cell potential magnitude: * $[\ce{M^+}]_{\text{anode}} = 0.05\text{ M}$ * $[\ce{M^+}]_{\text{cathode}} = 1\text{ M}$ * $|E_{\text{cell}}| = 70\text{ mV}$ Substitute these values into the magnitude equation: $$70\text{ mV} = C \ln \left( \frac{1}{0.05} \right)$$ $$70\text{ mV} = C \ln (20)$$ This gives us the relationship: $$C \ln (20) = 70\text{ mV}$$ Step 4 - Calculate the Magnitude of the New Cell Potential In the new state, the $0.05\text{ M}$ solution of $\ce{M^+}$ at the anode is replaced by a $0.0025\text{ M}$ solution, while the cathode concentration remains $1\text{ M}$: * $[\ce{M^+}]_{\text{anode, new}} = 0.0025\text{ M}$ * $[\ce{M^+}]_{\text{cathode}} = 1\text{ M}$ Using the magnitude equation for the new cell potential ($|E_{\text{cell, new}}|$): $$|E_{\text{cell, new}}| = C \ln \left( \frac{1}{0.0025} \right)$$ $$|E_{\text{cell, new}}| = C \ln (400)$$ Since $400 = 20^2$, we can apply the logarithmic power rule $\ln(x^y) = y \ln(x)$: $$|E_{\text{cell, new}}| = C \ln (20^2)$$ $$|E_{\text{cell, new}}| = 2 C \ln (20)$$ Using our initial relationship where $C \ln (20) = 70\text{ mV}$: $$|E_{\text{cell, new}}| = 2 \times 70\text{ mV}$$ $$|E_{\text{cell, new}}| = 140\text{ mV}$$ Step 5 - Evaluate and Explain Each Option * **Option (a) is incorrect:** $35\text{ mV}$ is half of the original cell potential. This would happen if the concentration gradient decreased to $\sqrt{20}$. * **Option (b) is incorrect:** $70\text{ mV}$ is the original potential. Diluting the anode solution increases the concentration gradient, which must change the potential. * **Option (c) is correct:** As mathematically demonstrated, replacing the anode concentration with $0.0025\text{ M}$ doubles the logarithmic factor, resulting in a cell potential of exactly $140\text{ mV}$. * **Option (d) is incorrect:** $700\text{ mV}$ is ten times the original potential, which assumes a linear relation rather than the correct logarithmic dependence. $$\text{Correct Option: } \boxed{C}$$