Electrochemistry and Ionic Equilibria Questions and Answers

Answer: C

The cell potential is the reduction potential of the cathode minus that of the anode. Copper is reduced and nickel is oxidised, so

\[ E^\circ_{cell} = 0.34 - (-0.25) = 0.59\ \text{V} \]

Two electrons are transferred (\(n = 2\)), and the Faraday constant is 96500 C/mol.

\[ \begin{aligned} \Delta G^\circ = -nFE^\circ \\ &= -2 \times 96500 \times 0.59 \\ &= -113870\ \text{J mol}^{-1} \end{aligned} \]

The sign is negative because the reaction is spontaneous. The value \(-56935\) comes from using \(n = 1\) and the positive values from the wrong sign. The answer is option C.

Answer: C

The pH of the solution is 8.0.

The cell has the hydrogen electrode on the left (anode) and the calomel electrode on the right (cathode). With H₂ at 1 bar the hydrogen electrode potential is -0.0591 pH, so:

\[ \begin{aligned} E_{cell} &= E_{cal} - E_{H} = 0.2802 + 0.0591\,\text{pH} \\ \text{pH} &= \frac{0.7530 - 0.2802}{0.0591} \\ &= \frac{0.4728}{0.0591} = 8.0 \end{aligned} \]

12.7 is 0.7530 / 0.0591, which forgets to subtract the calomel electrode's potential. 7.4 and 4.7 do not reproduce the measured EMF when substituted back: 0.2802 + 0.0591 × 7.4 = 0.717 V, not 0.7530 V.

Answer: B

The equilibrium constant is 10³³⁸.

Ten electrons are transferred in the balanced equation (2 Mn from +7 to +2). The cell potential is the cathode (permanganate) minus the anode (oxalate) standard potential:

\[ \begin{aligned} E^{\circ}_{cell} &= 1.51 - (-0.49) = 2.00\ \text{V} \\ \log K &= \frac{nE^{\circ}}{0.0591} = \frac{10 \times 2.00}{0.0591} \\ &= 338 \end{aligned} \]

The other exponents do not follow from n = 10 and a cell potential of 2.00 V. A common slip is to subtract the two potentials (1.51 - 0.49 = 1.02 V), which gives log K of about 173, also not an option; 10³⁸ is far too small for a ten-electron reaction with a 2 V driving force.

Answer: A

The masses are 0.469 g of silver and 0.138 g of copper.

Charge passed and moles of electrons:

\[ \begin{aligned} Q &= 0.7 \times 600 = 420\ \mathrm{C} \\ n_{e} &= \frac{420}{96485} = 4.353 \times 10^{-3}\ \mathrm{mol} \end{aligned} \]

Ag⁺ takes one electron and Cu gives two:

\[ \begin{aligned} m_{\mathrm{Ag}} &= 4.353 \times 10^{-3} \times 107.9 = 0.469\ \mathrm{g} \\ m_{\mathrm{Cu}} &= \frac{4.353 \times 10^{-3}}{2} \times 63.55 = 0.138\ \mathrm{g} \end{aligned} \]

0.069 g of copper divides by 2 twice, and 0.235 g of silver treats Ag⁺ as if it needed two electrons per atom.

Answer: B

The order is \(\mathrm{II} \gt \mathrm{I} \gt \mathrm{III}\).

The question attributes the differences between the cells to their liquid junctions, so compare the junction potential each adds, taken as the potential of the right-hand solution relative to the left. A junction charges the side towards which the faster ion runs.

• II (0.1 M HCl | 0.01 M HCl): H⁺ (transport number about 0.82) diffuses into the dilute side much faster than Cl^-, making the right side positive by roughly \(0.64 \times 59\) mV, about +38 mV.
• I (0.1 M KCl | 0.1 M NaCl): common anion, and K⁺ is only slightly more mobile than Na⁺, so the right side is a few millivolts positive.
• III (0.01 M KCl | 0.01 M HCl): H⁺ runs leftwards into the KCl far faster than K⁺ runs right, making the left side positive, a negative contribution of roughly -27 mV.

So E_cell falls in the order II, I, III. The options treating I and III as equal ignore the large biionic junction in III, and \(\mathrm{I} \gt \mathrm{II} \gt \mathrm{III}\) misses that the tenfold HCl gradient in II gives the largest junction of all.

Answer: C

The pH of the buffer is about 5.1.

After mixing, the total volume is 200 ml, so [acid] = 0.01/0.2 = 0.05 M and [acetate] = 0.02/0.2 = 0.10 M. The Henderson-Hasselbalch equation gives

\[ \begin{aligned} pK_{a} &= -\log(1.7 \times 10^{-5}) = 4.77 \\ pH &= pK_{a} + \log\frac{[\text{A}^{-}]}{[\text{HA}]} \\ &= 4.77 + \log 2 = 5.07 \end{aligned} \]

Because both solutions are diluted into the same volume, only the ratio of moles (0.02 to 0.01) matters. 4.5 is the result of inverting the ratio, and 4.1 or 5.5 do not follow from a pK_a of 4.77 with a 2:1 salt-to-acid ratio.

Answer: A

The copper nitrate concentration is about \(4.1 \times 10^{-4}\) M.

For the half reaction \(\text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu}\), the Nernst equation at 298 K is

\[ \begin{aligned} E &= E^{0} + \frac{0.0591}{2}\log[\text{Cu}^{2+}] \\ 0.24 &= 0.34 + 0.02955\log[\text{Cu}^{2+}] \\ \log[\text{Cu}^{2+}] &= -0.10/0.02955 = -3.38 \\ [\text{Cu}^{2+}] &= 4.1 \times 10^{-4}\ \text{M} \end{aligned} \]

The measured potential is 0.10 V below the standard value, so the concentration must be well below 1 M. Using n = 1 instead of 2 gives about \(2.0 \times 10^{-2}\) M, a common slip; the other values do not satisfy the equation.

Answer: C

The value of x is 389.

Kohlrausch's law of independent migration lets the limiting molar conductance of a weak acid be built from strong electrolytes. Adding HBr and CH₂BrCOOK and subtracting KBr leaves exactly H⁺ and CH₂BrCOO^-:

\[ \begin{aligned} \Lambda^{0}_{\text{acid}} &= \Lambda^{0}_{HBr} + \Lambda^{0}_{CH_2BrCOOK} - \Lambda^{0}_{KBr} \\ &= 427.95 + 112.72 - 151.64 \\ &= 389.03 \end{aligned} \]

in units of 10^-4 S m² mol^-1, which rounds to 389.

467 is 427.95 + 151.64 - 112.72, which swaps the roles of KBr and the potassium salt, and 164 is 427.95 - 151.64 - 112.72, which subtracts both salts. 195 is about half the correct value. Only the combination that cancels K⁺ and Br^- leaves the acid's own ions.

Answer: A

The buffer is \(CH_{3}COOH + CH_{3}COONa\).

A buffer needs a weak acid together with its conjugate base (or a weak base with its conjugate acid), so that added \(H^{+}\) is taken up by the base and added \(OH^{-}\) by the acid. Acetic acid is a weak acid and sodium acetate supplies its conjugate base, acetate.

Acetic acid with NaCl has no conjugate base, since chloride is the conjugate base of a strong acid and does not take up protons. NaOH with NaCl is a strong base and a neutral salt. NaOH with sodium acetate is a strong base plus a weak base, with no acid to absorb added base.

Answer: B

The conductivity first falls and then rises, the V-shaped plot in option (B).

At the start the solution holds Na⁺ and OH^-, and OH^- has a very high ionic mobility. Each drop of HCl neutralises some OH^- (H⁺ + OH^- → H₂O) and puts in its place a Cl^- ion, which conducts much less. So conductivity drops steadily until the equivalence point (1 mL of the 10 mM acid for 10 mL of 1 mM base).

After that, the extra drops add H⁺, the most mobile ion of all, together with Cl^-, so the conductivity climbs steeply.

Plot (A) has the trend upside down. Plots (C) and (D) level off, which would need the acid to stop adding ions after the equivalence point; excess HCl keeps raising the conductivity.