$^{\circ}$: The standard cell potential (or standard electrode potential), measured in volts (V). This is the potential when all reactants and products are in their standard states (1 M concentration for solutions, 1 atm pressure for gases, 298 K temperature). $R$: The ideal gas constant*, $8.314 \text{ J mol}^{-1} \text{ K}^{-1}$. $T$: The absolute temperature* in Kelvin (K). $n$: The number of moles of electrons transferred* in the balanced redox reaction. $F$: Faraday's constant*, $96500 \text{ C mol}^{-1}$. $Q$: The reaction quotient*. For a generic reaction $aA + bB \rightleftharpoons cC + dD$, $Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}$. It represents the ratio of product concentrations to reactant concentrations at any given time, raised to the power of their stoichiometric coefficients. Pure solids and liquids are not included in $Q$. 1. (c) A zinc rod is placed in a 0.1 M solution of zinc sulphate at 27 °C. Calculate the potential of the electrode at this temperature, assuming 76 % dissociation of zinc sulphate and $E^{\circ}(\text{Zn}^{2+}/\text{Zn}) = 0.52 \text{ V}$. Step 1: Write the half-reaction and determine the number of electrons transferred. The reduction half-reaction for the zinc electrode is: $$\text{Zn}^{2+}(aq) + 2e^- \rightleftharpoons \text{Zn}(s)$$ From this, the number of moles of electrons transferred is $n = 2$. Step 2: Calculate the actual concentration of $\text{Zn}^{2+}$ ions. Given that the initial concentration of $\text{ZnSO}_4$ is $0.1 \text{ M}$ and its dissociation is $76\%$: $$[\text{Zn}^{2+}] = 0.1 \text{ M} \times 0.76 = 0.076 \text{ M}$$ Step 3: Convert the temperature to Kelvin. Given temperature $T = 27 \text{ °C}$. $$T = 27 + 273 = 300 \text{ K}$$ Step 4: Apply the Nernst equation for the half-cell. The Nernst equation is $E = E^{\circ} - \frac{RT}{nF} \ln Q$. For the reduction of $\text{Zn}^{2+}$ to $\text{Zn}(s)$, the reaction quotient $Q = \frac{1}{[\text{Zn}^{2+}]}$ (since $\text{Zn}(s)$ is a pure solid). So, the Nernst equation becomes: $$E = E^{\circ} - \frac{RT}{nF} \ln \left(\frac{1}{[\text{Zn}^{2+}]}\right)$$ $$E = E^{\circ} + \frac{RT}{nF} \ln [\text{Zn}^{2+}]$$ Step 5: Substitute the given values into the Nernst equation. Given: $E^{\circ}(\text{Zn}^{2+}/\text{Zn}) = 0.52 \text{ V}$ $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ $T = 300 \text{ K}$ $n = 2$ $F = 96500 \text{ C mol}^{-1}$ $[\text{Zn}^{2+}] = 0.076 \text{ M}$ $$E = 0.52 \text{ V} + \frac{(8.314 \text{ J mol}^{-1} \text{ K}^{-1})(300 \text{ K})}{(2)(96500 \text{ C mol}^{-1})} \ln (0.076)$$ Step 6: Calculate the value. First, calculate the term $\frac{RT}{nF}$: $$\frac{RT}{nF} = \frac{(8.314)(300)}{(2)(96500)} = \frac{2494.2}{193000} \approx 0.012923 \text{ V}$$ Next, calculate $\ln(0.076)$: $$\ln(0.076) \approx -2.5770$$ Now, substitute these values back into the equation for $E$: $$E = 0.52 \text{ V} + (0.012923 \text{ V})(-2.5770)$$ $$E = 0.52 \text{ V} - 0.03331 \text{ V}$$ $$E = 0.48669 \text{ V}$$ Rounding to three significant figures: $$E \approx 0.487 \text{ V}$$ The potential of the electrode is $\boxed{\text{0.487 V}}$. That's 2 down. 3 left today — send the next one.