Electrochemistry and Ionic Equilibria Questions and Answers

Answer: C

The EMF is 3.02 V.

The cell reaction is Mg + 2Ag⁺ → Mg²⁺ + 2Ag, with n = 2. Apply the Nernst equation.

\[ \begin{aligned} E &= E^{0} - \frac{0.06}{2}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^{+}}]^{2}} \\ &= 3.17 - 0.03\log\frac{0.10}{(10^{-3})^{2}} \\ &= 3.17 - 0.03 \times 5 = 3.02\ \text{V} \end{aligned} \]

3.32 V adds the correction (3.17 + 0.15) instead of subtracting it, and 3.17 V is the standard EMF, which would apply only at unit concentrations. Forgetting to square [Ag⁺] gives log 100 = 2 and 3.11 V, which is not offered.

Answer: C

The electrolyte is 38% H₂SO₄ aqueous solution.

A lead storage battery has a lead anode and a lead dioxide cathode dipped in sulphuric acid of about 38% by mass (density about 1.3 g mL^-1). During discharge both electrodes turn to PbSO₄ and the acid is used up, which is why its density is a measure of the state of charge.

Hydrochloric acid is not used, because PbCl₂ would form and the cell reactions depend on the sulphate. 10% acid is too dilute and 60% too concentrated for the standard cell.

Answer: B

\(K_{sp}\) is \(108\,S^{5}\).

Each formula unit dissolving gives 3 X ions and 2 Y ions, so a solubility S gives [X] = 3S and [Y] = 2S:

\[ K_{sp} = (3S)^{3}(2S)^{2} = 27S^{3} \times 4S^{2} = 108S^{5} \]

The options with \(S^{2}\), \(S^{4}\) or \(S^{6}\) use the wrong total number of ions (the power of S is 3 + 2 = 5), and 27 drops the factor 4 from \((2S)^{2}\).

Answer: C

\(E_{cell}\) is 1.159 V.

Two electrons are transferred. By the Nernst equation,

\[ \begin{aligned} E &= E^{0} - \frac{0.059}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \\ &= 1.1 - 0.0295\log\frac{0.001}{0.1} \\ &= 1.1 + 0.059 = 1.159\ \text{V} \end{aligned} \]

The low Zn²⁺ concentration pushes the reaction forward, so E is above \(E^{0}\). 1.041 V gets the sign of the log term wrong, and 1.218 V uses n = 1.

Note the half in \(\frac{0.059}{n}\): with n = 2 the log term changes E by 0.0295 V per factor of ten in the concentration ratio, and the ratio here is a hundredfold.

Answer: D

The highest standard potential is for \(\mathrm{F_{2}(g) + 2e^{-} \rightarrow 2F^{-}}\).

Fluorine is the strongest oxidising agent, with \(E^{0} = +2.87\) V. The others, in volts, are Co³⁺/Co²⁺ +1.81, Cl₂/Cl⁻ +1.36, MnO₂/Mn²⁺ +1.23 and AgCl/Ag +0.22.

All are positive, but F₂ is the largest, which is why fluorine oxidises everything else on the list.

Answer: B

The hydroxide ion concentration is \(10^{-10}\) mol dm^-3.

At 25 °C, pH + pOH = 14 (from \(K_{w} = [\mathrm{H^{+}}][\mathrm{OH^{-}}] = 10^{-14}\)):

\[ \begin{aligned} \mathrm{pOH} &= 14 - 4 = 10 \\ [\mathrm{OH^{-}}] &= 10^{-10}\ \text{mol dm}^{-3} \end{aligned} \]

\(10^{-4}\) is the hydrogen ion concentration, not the hydroxide. The solution is acidic, so \([\mathrm{OH^{-}}]\) must be much smaller than \(10^{-7}\), which also rules out \(10^{-2}\) and \(10^{-6}\).

Answer: E

The hydrogen ion concentration is \(0.9 \times 10^{-5}\) M.

4 g of NaOH is 0.1 mol. It neutralises 0.1 mol of the acetic acid, forming 0.1 mol more acetate, so the buffer now holds 0.9 mol acid and 1.1 mol salt in 1 L. For an acidic buffer:

\[ \begin{aligned} [\mathrm{H^{+}}] &= K_{a} \times \frac{[\text{acid}]}{[\text{salt}]} \\ &= 1.1 \times 10^{-5} \times \frac{0.9}{1.1} = 0.9 \times 10^{-5}\ \text{M} \end{aligned} \]

The added base lowers \([\mathrm{H^{+}}]\) only slightly from \(1.1 \times 10^{-5}\) M, which is the buffer action. Every other option is above \(1.1 \times 10^{-5}\) M, which would mean adding a base made the solution more acidic.

Answer: E

The cell constant is 0.6 cm^-1.

Conductivity, conductance and cell constant \(G^{*} = \frac{l}{A}\) are related by \(\kappa = \frac{G^{*}}{R}\), so:

\[ \begin{aligned} G^{*} &= \kappa R \\ &= 0.012\ \text{ohm}^{-1}\text{cm}^{-1} \times 50\ \text{ohm} \\ &= 0.6\ \text{cm}^{-1} \end{aligned} \]

Dividing instead of multiplying gives \(0.012/50 = 2.4 \times 10^{-4}\), not offered. 0.02, 0.5, 0.8 and 0.1 cm^-1 do not follow from these data.

Answer: D

The compound is Bisphenol.

One way of preventing corrosion is to cover the metal surface with paint or with certain chemicals, and the standard textbook example of such a chemical is bisphenol (used in epoxy coatings). The coating keeps air and moisture away from the metal.

Phenol, benzene, acetone and nitrophenol do not form protective films; acetone and benzene are volatile solvents.

The other ways of preventing corrosion in the same account are galvanising (coating with a more active metal such as zinc) and sacrificial protection with magnesium or zinc.

Answer: A

The hydrogen ion concentration is \(1 \times 10^{-11}\) M.

pH is defined from the hydrogen ion concentration whatever the solute, so

\[ [\text{H}^{+}] = 10^{-\text{pH}} = 10^{-11}\ \text{M} \]

That the solute is a weak base only explains why the pH is above 7. \(10^{-3}\) M is the hydroxide ion concentration, from pOH = 14 - 11 = 3, which is the usual trap. \(10^{-14}\) is \(K_{w}\) itself, and the other values do not follow from pH 11.