Electrochemistry and Ionic Equilibria Questions and Answers

Answer: A

The most negative standard electrode potential belongs to Li⁺(aq) + e^- → Li(s).

Lithium has the most negative standard reduction potential of all, about -3.05 V, because it loses its electron very readily in water (its small ion is strongly hydrated). It is therefore the strongest reducing agent in aqueous solution.

Na⁺/Na is also strongly negative (about -2.71 V) but less so than lithium. F2/F^- is the most positive of all, about +2.87 V, and I2/I^- (about +0.54 V) and Cu⁺/Cu (about +0.52 V) are positive.

Answer: B

The limiting ionic conductance of Cl^- is 76.3 S cm² mol^-1.

By Kohlrausch's law, the limiting molar conductance of a salt is the sum of the limiting conductances of its ions, each multiplied by the number of those ions in the formula. CaCl2 gives one Ca²⁺ and two Cl^-:

\[ \begin{aligned} \Lambda^{\circ}_{m} &= \lambda^{\circ}_{\text{Ca}^{2+}} + 2\lambda^{\circ}_{\text{Cl}^{-}} \\ 271.6 &= 119 + 2\lambda^{\circ}_{\text{Cl}^{-}} \\ \lambda^{\circ}_{\text{Cl}^{-}} &= 152.6 / 2 = 76.3 \end{aligned} \]

152.6 is the answer you get by forgetting that there are two chloride ions per formula unit.

Answer: D

The volume of oxygen is 11.2 dm³.

At the anode, water is oxidised: \(2\text{H}_{2}\text{O} \rightarrow \text{O}_{2} + 4\text{H}^{+} + 4e^{-}\). One mole of \(\text{O}_{2}\) needs 4 F.

\[ \begin{aligned} n(\text{O}_{2}) &= \frac{2}{4} = 0.5\ \text{mol} \\ V &= 0.5 \times 22.4 = 11.2\ \text{dm}^{3} \end{aligned} \]

22.4 dm³ is the volume of hydrogen released at the cathode by the same 2 F, since each \(\text{H}_{2}\) needs only 2 electrons.

Answer: E

The EMF of the cell is 3.0 V (2.99 V).

The cell reaction is \(\mathrm{Mg + 2Ag^+ \rightarrow Mg^{2+} + 2Ag}\), so 2 electrons are transferred. Apply the Nernst equation:

\[ \begin{aligned} E &= E^\circ - \frac{0.06}{2}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^+}]^{2}} \\ &= 3.17 - 0.03\log\frac{10^{-2}}{(10^{-4})^{2}} \\ &= 3.17 - 0.03 \times 6 \\ &= 3.17 - 0.18 = 2.99 \approx 3.0 \text{ V} \end{aligned} \]

3.17 V is the standard value and ignores the concentrations. Forgetting to square [Ag⁺] gives \(\log 100 = 2\) and 3.11 V, and adding instead of subtracting the correction gives 3.35 V.

Answer: A

The mass of aluminium deposited is 0.67 g.

The charge passed is \(Q = It\), and each Al³⁺ ion needs 3 electrons, so 3 F deposit one mole (27 g) of Al.

\[ \begin{aligned} Q &= 2 \times 3600 = 7200 \text{ C} \\ m &= \frac{27 \times 7200}{3 \times 96500} \\ &= \frac{194400}{289500} \approx 0.67 \text{ g} \end{aligned} \]

1.34 g is exactly double the correct mass, and 0.067 g and 0.134 g are factor-of-ten slips; remember that each Al³⁺ ion takes three electrons.

Answer: B

The pH is about 2.52.

\[ \begin{aligned} \mathrm{pH} &= -\log(3 \times 10^{-3}) \\ &= 3 - \log 3 = 3 - 0.4771 \\ &= 2.5229 \approx 2.52 \end{aligned} \]

3.47 comes from adding \(\log 3\) instead of subtracting it, and 2.32 or 2.47 from rounding slips. A solution with \([\mathrm{H^+}]\) between \(10^{-3}\) and \(10^{-2}\) M must have pH between 2 and 3, which rules out 3.47 and 5.52 at once.

Answer: B

The emf is 0.93 V.

Two electrons are transferred (n = 2). By the Nernst equation at 298 K:

\[ \begin{aligned} E &= E^{\circ} - \frac{0.059}{2}\log\frac{[\mathrm{Ni^{2+}}]}{[\mathrm{Ag^{+}}]^{2}} \\ &= 1.05 - 0.0295\log\frac{0.04}{(0.002)^{2}} \\ &= 1.05 - 0.0295 \times 4 = 0.932\ \text{V} \end{aligned} \]

The ratio inside the logarithm is \(\frac{0.04}{4 \times 10^{-6}} = 10^{4}\). Forgetting to square \([\mathrm{Ag^+}]\), or adding the correction, leads to the other values.

Answer: A

The quantity of electricity is 386000 Coulombs.

The equation shows 4 electrons for each O₂, so reducing one mole of oxygen needs 4 moles of electrons, that is 4 faradays:

\[ Q = 4F = 4 \times 96500 = 386000\ \text{C} \]

The cell potential of 1.23 V is not needed; it would be used to find the energy, not the charge. 96500 C is one faraday, enough for only a quarter of a mole of O₂.

Answer: C

The metal is palladium.

In a hydrogen-oxygen fuel cell the gases are bubbled through porous carbon electrodes into the electrolyte. Finely divided platinum or palladium is incorporated into the electrodes as a catalyst, to increase the rate of the electrode reactions.

Nickel is used as a catalyst in some alkaline cells and in hydrogenation, but the standard answer for these electrodes is platinum or palladium. Zinc, tin and magnesium are reactive metals that would be consumed, not catalysts.

Answer: C

The solubility is \(2 \times 10^{-5}\) mol lit^-1.

For AX₂ dissolving to give A²⁺ and 2X^-, a solubility \(s\) gives \([A^{2+}] = s\) and \([X^{-}] = 2s\):

\[ \begin{aligned} K_{sp} &= s(2s)^{2} = 4s^{3} \\ s^{3} &= \frac{3.2 \times 10^{-14}}{4} = 8 \times 10^{-15} \\ s &= 2 \times 10^{-5} \text{ mol lit}^{-1} \end{aligned} \]

Forgetting the 4 (using \(K_{sp} = s^{3}\)) gives about \(3.2 \times 10^{-5}\), and the \(10^{-3}\) and \(10^{-2}\) options misplace the cube root of the power of ten.