The standard EMF of a Daniel cell at 298 K is E1. When the concentration of is 1.0 M and that of is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Cell Reactions of a Daniell Cell A Daniell cell consists of a zinc electrode immersed in a zinc sulphate ($\ce{ZnSO4}$) solution (anode) and a copper electrode immersed in a copper sulphate ($\ce{CuSO4}$) solution (cathode). The individual half-cell reactions and the net chemical cell reaction are: * **Oxidation at the Anode (Negative Electrode):** $$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}$$ * **Reduction at the Cathode (Positive Electrode):** $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$$ * **Net Cell Reaction:** $$\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$$ The number of moles of electrons transferred in this balanced redox process is: $$n = 2$$ Step 2 - Formulate the Nernst Equation The electromotive force (EMF) of the cell at a temperature of $298\text{ K}$ is given by the Nernst equation: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} Q$$ Where: * $E_{\text{cell}}$ is the actual (non-standard) cell potential. * $E^\circ_{\text{cell}}$ is the standard cell potential. * $Q$ is the reaction quotient, defined as the ratio of product ion concentration to reactant ion concentration: $$Q = \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]}$$ * At standard temperature $T = 298\text{ K}$, the term $\frac{2.303 RT}{F}$ is constant and is approximated as: $$\frac{2.303 RT}{F} \approx 0.0591\text{ V}$$ Substituting these parameters into the Nernst equation: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591\text{ V}}{2} \log_{10} \left( \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]} \right)$$ Step 3 - Substitute the Concentration Values and Calculate the EMF Difference We are given the following parameters and concentrations at $298\text{ K}$: * Standard EMF ($E^\circ_{\text{cell}}$) = $E_1$ * Non-standard EMF ($E_{\text{cell}}$) = $E_2$ * Zinc sulphate concentration, which equals the zinc ion concentration: $$[\ce{Zn^{2+}}] = [\ce{ZnSO4}] = 1.0\text{ M}$$ * Copper sulphate concentration, which equals the copper ion concentration: $$[\ce{Cu^{2+}}] = [\ce{CuSO4}] = 0.01\text{ M} = 10^{-2}\text{ M}$$ Now, substitute these concentration values with units into our formulated Nernst equation: $$E_2 = E_1 - \frac{0.0591\text{ V}}{2} \log_{10} \left( \frac{1.0\text{ M}}{0.01\text{ M}} \right)$$ $$E_2 = E_1 - 0.02955\text{ V} \log_{10}(100)$$ $$E_2 = E_1 - 0.02955\text{ V} \log_{10}(10^2)$$ $$E_2 = E_1 - 0.02955\text{ V} \times 2$$ $$E_2 = E_1 - 0.0591\text{ V}$$ Step 4 - Determine the Relationship between $E_1$ and $E_2$ From our calculation in Step 3, the relation is: $$E_2 = E_1 - 0.0591\text{ V}$$ Rearranging to find the difference between the two cell potentials: $$E_1 - E_2 = 0.0591\text{ V}$$ Since $0.0591\text{ V} > 0$: $$E_1 - E_2 > 0 \implies \boxed{E_1 > E_2}$$ Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** $E_1 = E_2$ only when the reaction quotient is $1$, which occurs when $[\ce{Zn^{2+}}] = [\ce{Cu^{2+}}]$. Because the concentrations are unequal, their potentials cannot be equal. * **Option (B) is incorrect:** The potential $E_2 = E_1 - 0.0591\text{ V}$. Given that the standard potential $E_1$ of a Daniell cell is approximately $+1.10\text{ V}$, the actual potential is $E_2 \approx 1.04\text{ V}$, which is far from zero. * **Option (C) is correct:** As calculated, the standard EMF ($E_1$) is greater than the actual EMF ($E_2$) because the reaction quotient $Q = 100 > 1$, which thermodynamically opposes the forward cell reaction. * **Option (D) is incorrect:** $E_1 < E_2$ would only be true if the reactant concentration ($[\ce{Cu^{2+}}]$) were greater than the product concentration ($[\ce{Zn^{2+}}]$), which would make the logarithmic correction term positive. $$\text{Correct Option: } \boxed{\text{C}}$$