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Solve and answer the following.
Calculate the emf and Δ G° for the cell reaction at 25°C: Zn(s) | Zn²⁺(aq, 0·1 M) ∥ Cd²⁺(aq, 0·01…
(i)[3]
Calculate the emf and $\Delta G^\circ$ for the cell reaction at $25^\circ\text{C}$:
$\text{Zn}(s) \mid \text{Zn}^{2+}(aq, 0\cdot1\text{ M}) \parallel \text{Cd}^{2+}(aq, 0\cdot01\text{ M}) \mid \text{Cd}(s)$
Given $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0\cdot763\text{ V}$ and $E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0\cdot403\text{ V}$.
(ii)[2]
Define the following terms:
(1) Equivalent conductivity
(2) Corrosion of metals
Answer
Answer (i)
Formulas used: Standard emf of a cell: E0cell = E0cathode - E0anode with E0anode = -0.763 V, E0cathode = -0.403 V, so E0cell = 0.36 V Nernst equation at 298 K: E = E0 - 0.0591/n*log10(Q) with Q = 10, n = 2, E0 = 0.36 V, so E = 0.3305 V Gibbs energy from emf: dG = -n*F*E with E = 0.36 V, F = 96500 C mol^-1, n = 2, so dG = -69480 J mol^-1
Cell reaction: $\mathrm{Zn + Cd^{2+} \rightarrow Zn^{2+} + Cd}$, $n = 2$
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0\cdot403 - (-0\cdot763) = 0\cdot36\text{ V}$
$E_{cell} = E^\circ - \dfrac{0\cdot0591}{2}\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cd^{2+}}]} = 0\cdot36 - \dfrac{0\cdot0591}{2}\log\dfrac{0\cdot1}{0\cdot01} = 0\cdot3305\text{ V}$
$\Delta G^\circ = -nFE^\circ = -2\times96500\text{ C mol}^{-1}\times0\cdot36\text{ V} = -69480\text{ J mol}^{-1}$
Final answer: E = 0.3305 V; dG0 = -69.48 V; kJ mol^-1
Answer (ii)
(1) Equivalent conductivity is the conductance of all the ions produced by one gram equivalent of the electrolyte dissolved in a given volume of solution, kept between two parallel electrodes 1 cm apart: $\Lambda_{eq} = \dfrac{\kappa\times1000}{N}$.
(2) Corrosion is the slow eating away of a metal by the action of air, moisture or chemicals on its surface, forming its oxide, carbonate or sulphide (e.g. rusting of iron).