Electrochemistry and Ionic Equilibria Questions and Answers

Answer: C

The key gives Electrochemical reaction at equilibrium. The question is loosely worded, and options B and D are also true statements about EMF, so this rests on the thermodynamic reading of the term.

The EMF of a cell is the potential difference measured reversibly, with no net current, so the cell reaction is balanced by an opposing potential and runs only infinitesimally. That reversible condition is what links EMF to the free energy of the reaction and, through the standard EMF, to its equilibrium constant:

\[ \Delta G = -nFE \]

\[ \ln K = \frac{nFE^{\circ}}{RT} \]

Strictly, when the cell reaction itself has reached chemical equilibrium the EMF falls to zero, so the keyed wording describes the reversible limit and the link to K, not a nonzero reading at equilibrium. A half cell (A) has an electrode potential, not an EMF, which needs two electrodes. A complete cell (B) and the potential difference of the cell (D) describe what is measured, which is why a candidate can defend them.

Answer: A

The voltage rises by raising [Cu²⁺] or/and decreasing [Cd²⁺].

The cell reaction is Cd + Cu²⁺ → Cd²⁺ + Cu, with n = 2. The Nernst equation is

\[ \begin{aligned} E &= E^{\circ} \\ &- \frac{RT}{2F}\ln\frac{[\mathrm{Cd}^{2+}]}{[\mathrm{Cu}^{2+}]} \end{aligned} \]

E increases when the ratio [Cd²⁺]/[Cu²⁺] gets smaller: more of the reactant ion Cu²⁺, or less of the product ion Cd²⁺.

Raising [Cd²⁺] or decreasing [Cu²⁺] makes the ratio larger and lowers the voltage. Raising both by the same factor leaves the ratio, and the voltage, unchanged.

Answer: C

The plot is (C): E falls steeply at first, then levels off.

The calomel electrode reaction is Hg₂Cl₂ + 2e⁻ → 2Hg + 2Cl⁻, so its potential is

\[ E = E^{\circ} - \frac{RT}{F}\ln[\mathrm{Cl}^{-}] \]

E decreases as [Cl⁻] increases, and logarithmically: a given rise in concentration lowers E a lot when [Cl⁻] is small and only a little when it is large. Plotted against [Cl⁻] itself (not its logarithm), that is a curve falling steeply and flattening out, as in (C).

(A) and (B) show E rising with chloride, the wrong direction. (D) is a straight line, which is what E would look like against log[Cl⁻], not against [Cl⁻].

Answer: B

The buffer capacity is 2.

Buffer capacity β is the amount of strong base, in moles per litre of buffer, needed to change the pH by one unit. The paper prints the pH change as '0-2', which can only mean 0.2: that is the value with which the keyed figure works out.

\[ \begin{aligned} [\mathrm{OH^{-}}] \text{ added} &= \frac{0.1\ \text{mol}}{0.250\ \text{L}} = 0.4\ \text{M} \\ \beta &= \frac{0.4}{0.2} = 2 \end{aligned} \]

Leaving out the conversion to a litre basis gives 0.1/0.2 = 0.5, and reading the change as 0.1 pH unit gives 4; neither 1 nor 3 follows from the data.

Answer: C

The isoelectric point is about 4.8.

At the isoelectric point the protein carries no net charge, so it does not move in an electric field: its drift speed is zero. The table shows the speed changing sign between pH 4.6 (+0.25 μm/s) and pH 5.0 (−0.25 μm/s). The speeds are equal and opposite, so by linear interpolation the zero falls halfway between:

\[ \text{pI} \approx \frac{4.6 + 5.0}{2} = 4.8 \]

This matches the known pI of bovine serum albumin, about 4.7 to 4.9. At pH 4.1 the protein still moves with positive speed, at 5.2 it is already moving the other way, and at 7.0 it is strongly negative, so none of those is the point of zero mobility.

Answer: A

The pH is 5.

Ammonium sulfate is the salt of a weak base and a strong acid, so its solution is acidic through hydrolysis of NH₄⁺. 0.05 M (NH₄)₂SO₄ gives 0.1 M NH₄⁺. Taking K_b = 10⁻⁵ for ammonia (printed '(NH)'):

\[ \begin{aligned} K_{a} &= \frac{K_{w}}{K_{b}} = \frac{10^{-14}}{10^{-5}} = 10^{-9} \\ [\mathrm{H^{+}}] &= \sqrt{K_{a}C} = \sqrt{10^{-9}\times 0.1} = 10^{-5} \\ \text{pH} &= 5 \end{aligned} \]

Forgetting that each formula unit gives two NH₄⁺ and using 0.05 M gives pH 5.15, still closest to 5. pH 6 would need a far weaker acid, and pH 3 or 4 a much stronger one than NH₄⁺.

Answer: B

The predominant species are H₂PO₄⁻, H₂PO₄⁻, HPO₄²⁻, PO₄³⁻ (option B, which prints the first formula with its subscript misplaced).

Each species dominates in the pH range between the two pKa values that bound it: H₃PO₄ below 2.12, H₂PO₄⁻ from 2.12 to 7.20, HPO₄²⁻ from 7.20 to 12.36, and PO₄³⁻ above 12.36.

• pH 3.0 and pH 5.0 both lie between 2.12 and 7.20, so H₂PO₄⁻ dominates at both.
• pH 10.0 lies between 7.20 and 12.36: HPO₄²⁻.
• pH 13.0 is above 12.36: PO₄³⁻.

Option A puts H₃PO₄ at pH 3.0, but pH 3.0 is already above pKa₁. Options C and D also start with H₃PO₄ and contain impossible or wrong species for pH 5.

Answer: C

The key's answer is the Debye-Huckel extended law.

The ionic strength of 1.0 M H₂SO₄ is far above the range of the limiting law, log γ_± = −A|z₊z₋|√I, which treats ions as point charges and holds only for very dilute solutions (I below about 0.01). The extended law divides by (1 + Ba√I), allowing for the finite size of the ions, and so gives a better estimate at higher ionic strength.

Even the extended law is only approximate at 1 M, but of the options it is the most accurate. 'Debye-Huckel theory of acid and base' and 'Debye-Huckel theory for strong acids' are not names of separate formulations.

Answer: C

The NHE is Pt | H⁺ (1M HCl) | H₂(g) (1 atm).

The normal hydrogen electrode is the older practical reference: platinised platinum in a one normal (1 N) solution of a strong monoprotic acid, with hydrogen gas at 1 atm bubbling over it. 1 M HCl is 1 N in H⁺.

The cell with H⁺ at unit activity is the standard hydrogen electrode (SHE), the thermodynamic definition, which is not the same as a fixed concentration. 1 M H₂SO₄ is 2 N in H⁺, and H₃PO₄ is a weak acid that does not supply 1 M H⁺.

Answer: D

On rapid discharge the cell will give out maximum entropy.

The maximum useful (electrical) work a cell can deliver equals -ΔG, and it is obtained only when the cell is discharged reversibly, infinitely slowly, with no current-driven losses. A rapid discharge is strongly irreversible: large currents cause resistive heating and overpotentials, so much of the free energy is dissipated as heat and the entropy generated is at its largest.

The other options describe the opposite, slow limit: small entropy generation and maximum energy delivered. Enthalpy is not what is converted to useful work in any case; the limit is set by the Gibbs energy.